Chapter 8 Algebraic Expressions and Identities

Solution:

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We are asked to add the following algebraic expressions:

Expression 1: $7xy + 5yz – 3zx$

Expression 2: $4yz + 9zx – 4y$

Expression 3: $–3xz + 5x – 2xy$ (Note that $zx$ and $xz$ are the same term, we can write $–3xz$ as $–3zx$). So, Expression 3 is $–2xy + 5x – 3zx$.

To add algebraic expressions, we combine the like terms. Like terms are terms with the same variables raised to the same powers.

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Horizontal Method:

Write all the expressions in a single line with addition signs between them:

$(7xy + 5yz – 3zx) + (4yz + 9zx – 4y) + (–2xy + 5x – 3zx)$

Remove the parentheses:

$7xy + 5yz – 3zx + 4yz + 9zx – 4y – 2xy + 5x – 3zx$

Group the like terms together:

$(7xy – 2xy) + (5yz + 4yz) + (–3zx + 9zx – 3zx) + (–4y) + (5x)$

Combine the coefficients of each group of like terms:

$(7 - 2)xy + (5 + 4)yz + (-3 + 9 - 3)zx - 4y + 5x$

$5xy + 9yz + 3zx - 4y + 5x$

The sum is $5xy + 9yz + 3zx + 5x - 4y$.

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Column Method:

Write the expressions in rows, aligning the like terms in columns. Leave space for terms that are not present in an expression (or write them with a coefficient of 0).

Add the expressions column by column:

Adding the coefficients in each column:

• $xy$ terms: $7 + (-2) = 5$
• $yz$ terms: $5 + 4 = 9$
• $zx$ terms: $-3 + 9 + (-3) = -6 + 9 = 3$
• $x$ terms: $0 + 0 + 5 = 5$
• $y$ terms: $0 + (-4) + 0 = -4$

The sum is $5xy + 9yz + 3zx + 5x - 4y$.

Solution:

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We are asked to subtract the expression $5x^2 – 4y^2 + 6y – 3$ from the expression $7x^2 – 4xy + 8y^2 + 5x – 3y$.

This means we need to calculate:

$(7x^2 – 4xy + 8y^2 + 5x – 3y) - (5x^2 – 4y^2 + 6y – 3)$

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Horizontal Method:

Write the first expression, followed by a minus sign, and then the second expression in parentheses:

$(7x^2 – 4xy + 8y^2 + 5x – 3y) - (5x^2 – 4y^2 + 6y – 3)$

To subtract the second expression, change the sign of each term inside the parentheses and then remove the parentheses:

$7x^2 – 4xy + 8y^2 + 5x – 3y - 5x^2 + 4y^2 - 6y + 3$

Group the like terms together:

$(7x^2 - 5x^2) + (-4xy) + (8y^2 + 4y^2) + (5x) + (-3y - 6y) + (3)$

Combine the coefficients of each group of like terms:

$(7 - 5)x^2 - 4xy + (8 + 4)y^2 + 5x + (-3 - 6)y + 3$

$2x^2 - 4xy + 12y^2 + 5x - 9y + 3$

The result of the subtraction is $2x^2 - 4xy + 12y^2 + 5x - 9y + 3$.

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Column Method:

Write the expression to be subtracted from first, then write the expression to be subtracted below it, aligning like terms. Indicate the subtraction operation and change the signs of the terms in the second expression before adding column by column.

Subtracting the coefficients in each column (or adding after changing signs):

• $x^2$ terms: $7 - 5 = 2$
• $xy$ terms: $-4 - 0 = -4$
• $y^2$ terms: $8 - (-4) = 8 + 4 = 12$
• $x$ terms: $5 - 0 = 5$
• $y$ terms: $-3 - 6 = -9$
• Constant terms: $0 - (-3) = 0 + 3 = 3$

The result is $2x^2 - 4xy + 12y^2 + 5x - 9y + 3$.

Solution:

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To add algebraic expressions, we combine the coefficients of the like terms.

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(i) Add $ab – bc$, $bc – ca$, $ca – ab$

The given expressions are $ab – bc$, $bc – ca$, and $ca – ab$.

Horizontal Method:

Sum = $(ab – bc) + (bc – ca) + (ca – ab)$

Remove the parentheses and group the like terms:

Sum = $(ab - ab) + (-bc + bc) + (-ca + ca)$

Combine the coefficients of the like terms:

Sum = $(1 - 1)ab + (-1 + 1)bc + (-1 + 1)ca$

Sum = $0ab + 0bc + 0ca$

Sum = $0$.

Column Method:

Write the expressions in rows, aligning the like terms in columns:

Adding the coefficients in each column gives 0 for each term.

The sum is $0ab + 0bc + 0ca = 0$.

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(ii) Add $a – b + ab$, $b – c + bc$, $c – a + ac$

The given expressions are $a – b + ab$, $b – c + bc$, and $c – a + ac$.

Horizontal Method:

Sum = $(a – b + ab) + (b – c + bc) + (c – a + ac)$

Remove the parentheses and group the like terms:

Sum = $(a - a) + (-b + b) + (-c + c) + (ab) + (bc) + (ac)$

Combine the coefficients of the like terms:

Sum = $(1 - 1)a + (-1 + 1)b + (-1 + 1)c + ab + bc + ac$

Sum = $0a + 0b + 0c + ab + bc + ac$

Sum = $ab + bc + ac$.

Column Method:

Write the expressions in rows, aligning the like terms in columns:

Adding the coefficients in each column:

• $a$ terms: $1 + 0 + (-1) = 0$
• $b$ terms: $-1 + 1 + 0 = 0$
• $c$ terms: $0 + (-1) + 1 = 0$
• $ab$ terms: $1 + 0 + 0 = 1$
• $bc$ terms: $0 + 1 + 0 = 1$
• $ac$ terms: $0 + 0 + 1 = 1$

The sum is $0a + 0b + 0c + ab + bc + ac = ab + bc + ac$.

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(iii) Add $2p^2q^2 – 3pq + 4$, $5 + 7pq – 3p^2q^2$

The given expressions are $2p^2q^2 – 3pq + 4$ and $5 + 7pq – 3p^2q^2$.

Horizontal Method:

Sum = $(2p^2q^2 – 3pq + 4) + (5 + 7pq – 3p^2q^2)$

Remove the parentheses and group the like terms:

Sum = $(2p^2q^2 – 3p^2q^2) + (-3pq + 7pq) + (4 + 5)$

Combine the coefficients of the like terms:

Sum = $(2 - 3)p^2q^2 + (-3 + 7)pq + (4 + 5)$

Sum = $-1p^2q^2 + 4pq + 9$

Sum = $-p^2q^2 + 4pq + 9$.

Column Method:

Write the expressions in rows, aligning the like terms in columns:

Adding the coefficients in each column:

• $p^2q^2$ terms: $2 + (-3) = -1$
• $pq$ terms: $-3 + 7 = 4$
• Constant terms: $4 + 5 = 9$

The sum is $-p^2q^2 + 4pq + 9$.

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(iv) Add $l^2 + m^2$, $m^2 + n^2$, $n^2 + l^2$, $2lm + 2mn + 2nl$

The given expressions are $l^2 + m^2$, $m^2 + n^2$, $n^2 + l^2$, and $2lm + 2mn + 2nl$.

Horizontal Method:

Sum = $(l^2 + m^2) + (m^2 + n^2) + (n^2 + l^2) + (2lm + 2mn + 2nl)$

Remove the parentheses and group the like terms:

Sum = $(l^2 + l^2) + (m^2 + m^2) + (n^2 + n^2) + 2lm + 2mn + 2nl$

Combine the coefficients of the like terms:

Sum = $(1 + 1)l^2 + (1 + 1)m^2 + (1 + 1)n^2 + 2lm + 2mn + 2nl$

Sum = $2l^2 + 2m^2 + 2n^2 + 2lm + 2mn + 2nl$

We can take out the common factor 2:

Sum = $2(l^2 + m^2 + n^2 + lm + mn + nl)$.

Column Method:

Write the expressions in rows, aligning the like terms in columns:

Adding the coefficients in each column:

• $l^2$ terms: $1 + 0 + 1 + 0 = 2$
• $m^2$ terms: $1 + 1 + 0 + 0 = 2$
• $n^2$ terms: $0 + 1 + 1 + 0 = 2$
• $lm$ terms: $0 + 0 + 0 + 2 = 2$
• $mn$ terms: $0 + 0 + 0 + 2 = 2$
• $nl$ terms: $0 + 0 + 0 + 2 = 2$

The sum is $2l^2 + 2m^2 + 2n^2 + 2lm + 2mn + 2nl$, which can be written as $2(l^2 + m^2 + n^2 + lm + mn + nl)$.

Solution:

To subtract one algebraic expression from another, we change the sign of each term of the expression to be subtracted and then add the resulting expression to the other expression.

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(a) Subtract $4a – 7ab + 3b + 12$ from $12a – 9ab + 5b – 3$

This means we calculate $(12a – 9ab + 5b – 3) - (4a – 7ab + 3b + 12)$.

Horizontal Method:

$(12a – 9ab + 5b – 3) - (4a – 7ab + 3b + 12)$

Remove the parentheses and change the sign of each term in the second expression:

$= 12a – 9ab + 5b – 3 - 4a + 7ab - 3b - 12$

Group the like terms:

$= (12a - 4a) + (-9ab + 7ab) + (5b - 3b) + (-3 - 12)$

Combine the coefficients:

$= (12 - 4)a + (-9 + 7)ab + (5 - 3)b + (-3 - 12)$

$= 8a + (-2)ab + 2b + (-15)$

The result is $8a - 2ab + 2b - 15$.

Column Method:

Write the expression from which we subtract first. Below it, write the expression to be subtracted, aligning like terms. Change the sign of each term in the expression being subtracted and add.

Subtracting the coefficients in each column (equivalent to adding after changing signs):

• $a$ terms: $12 - 4 = 8$
• $ab$ terms: $-9 - (-7) = -9 + 7 = -2$
• $b$ terms: $5 - 3 = 2$
• Constant terms: $-3 - 12 = -15$

The result is $8a - 2ab + 2b - 15$.

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(b) Subtract $3xy + 5yz – 7zx$ from $5xy – 2yz – 2zx + 10xyz$

This means we calculate $(5xy – 2yz – 2zx + 10xyz) - (3xy + 5yz – 7zx)$.

Horizontal Method:

$(5xy – 2yz – 2zx + 10xyz) - (3xy + 5yz – 7zx)$

Remove the parentheses and change the sign of each term in the second expression:

$= 5xy – 2yz – 2zx + 10xyz - 3xy - 5yz + 7zx$

Group the like terms:

$= (5xy - 3xy) + (-2yz - 5yz) + (-2zx + 7zx) + (10xyz)$

Combine the coefficients:

$= (5 - 3)xy + (-2 - 5)yz + (-2 + 7)zx + 10xyz$

$= 2xy + (-7)yz + 5zx + 10xyz$

The result is $2xy - 7yz + 5zx + 10xyz$.

Column Method:

Write the expression from which we subtract first. Below it, write the expression to be subtracted, aligning like terms. Change the sign of each term in the expression being subtracted and add.

Subtracting the coefficients in each column (equivalent to adding after changing signs):

• $xy$ terms: $5 - 3 = 2$
• $yz$ terms: $-2 - 5 = -7$
• $zx$ terms: $-2 - (-7) = -2 + 7 = 5$
• $xyz$ terms: $10 - 0 = 10$

The result is $2xy - 7yz + 5zx + 10xyz$.

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(c) Subtract $4p^2q – 3pq + 5pq^2 – 8p + 7q – 10$ from $18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q$

This means we calculate $(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q) - (4p^2q – 3pq + 5pq^2 – 8p + 7q – 10)$.

Horizontal Method:

$(18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q) - (4p^2q – 3pq + 5pq^2 – 8p + 7q – 10)$

Remove the parentheses and change the sign of each term in the second expression:

$= 18 – 3p – 11q + 5pq – 2pq^2 + 5p^2q - 4p^2q + 3pq - 5pq^2 + 8p - 7q + 10$

Group the like terms:

$= (5p^2q - 4p^2q) + (-2pq^2 - 5pq^2) + (5pq + 3pq) + (-3p + 8p) + (-11q - 7q) + (18 + 10)$

Combine the coefficients:

$= (5 - 4)p^2q + (-2 - 5)pq^2 + (5 + 3)pq + (-3 + 8)p + (-11 - 7)q + (18 + 10)$

$= 1p^2q + (-7)pq^2 + 8pq + 5p + (-18)q + 28$

The result is $p^2q - 7pq^2 + 8pq + 5p - 18q + 28$.

Column Method:

Write the expression from which we subtract first. Below it, write the expression to be subtracted, aligning like terms. Change the sign of each term in the expression being subtracted and add.

Subtracting the coefficients in each column (equivalent to adding after changing signs):

• $p^2q$ terms: $5 - 4 = 1$
• $pq^2$ terms: $-2 - 5 = -7$
• $pq$ terms: $5 - (-3) = 5 + 3 = 8$
• $p$ terms: $-3 - (-8) = -3 + 8 = 5$
• $q$ terms: $-11 - 7 = -18$
• Constant terms: $18 - (-10) = 18 + 10 = 28$

The result is $p^2q - 7pq^2 + 8pq + 5p - 18q + 28$.

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Solution:

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The area of a rectangle is calculated by multiplying its length and breadth.

Area = Length $\times$ Breadth

We will fill in the 'area' column of the table by calculating the product of the given length and breadth for each row.

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Calculations:

For the second row:

Length = $9y$

Breadth = $4y^2$

Area = $9y \times 4y^2 = (9 \times 4) \times (y \times y^2) = 36 \times y^{1+2} = 36y^3$

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For the third row:

Length = $4ab$

Breadth = $5bc$

Area = $4ab \times 5bc = (4 \times 5) \times (a \times b \times b \times c) = 20 \times a \times b^{1+1} \times c = 20ab^2c$

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For the fourth row:

Length = $2l^2m$

Breadth = $3lm^2$

Area = $2l^2m \times 3lm^2 = (2 \times 3) \times (l^2 \times l \times m \times m^2) = 6 \times l^{2+1} \times m^{1+2} = 6l^3m^3$

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Completed Table:

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Solution:

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The volume of a rectangular box is given by the formula:

Volume = Length $\times$ Breadth $\times$ Height

We will calculate the volume for each set of dimensions provided in the table.

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(i) Length = 2ax, Breadth = 3by, Height = 5cz

Volume = $(2ax) \times (3by) \times (5cz)$

Multiply the coefficients and the variables separately:

Volume = $(2 \times 3 \times 5) \times (a \times x \times b \times y \times c \times z)$

Volume = $30 \times (abcxyz)$

Volume = $30abcxyz$

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(ii) Length = m²n, Breadth = n²p, Height = p²m

Volume = $(m^2n) \times (n^2p) \times (p^2m)$

Group the terms with the same base and add their exponents:

Volume = $(m^2 \times m) \times (n \times n^2) \times (p \times p^2)$

Volume = $m^{2+1} \times n^{1+2} \times p^{1+2}$

Volume = $m^3 \times n^3 \times p^3$

Volume = $m^3n^3p^3$

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(iii) Length = 2q, Breadth = 4q², Height = 8q³

Volume = $(2q) \times (4q^2) \times (8q^3)$

Multiply the coefficients and combine the variable terms:

Volume = $(2 \times 4 \times 8) \times (q \times q^2 \times q^3)$

Volume = $64 \times q^{1+2+3}$

Volume = $64q^6$

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Summary of Volumes:

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Solution:

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To find the product of monomials, we multiply the coefficients and multiply the variables separately, combining the powers of the same variables.

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(i) 4, 7p

Product $= 4 \times 7p$

Product $= (4 \times 7) \times p$

Product $= 28p$

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(ii) – 4p, 7p

Product $= (-4p) \times (7p)$

Product $= (-4 \times 7) \times (p \times p)$

Product $= -28 \times p^{1+1}$

Product $= -28p^2$

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(iii) – 4p, 7pq

Product $= (-4p) \times (7pq)$

Product $= (-4 \times 7) \times (p \times pq)$

Product $= -28 \times (p^{1+1}q)$

Product $= -28p^2q$

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(iv) 4p³, – 3p

Product $= (4p^3) \times (-3p)$

Product $= (4 \times -3) \times (p^3 \times p)$

Product $= -12 \times p^{3+1}$

Product $= -12p^4$

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(v) 4p, 0

Product $= (4p) \times (0)$

Any expression multiplied by zero is zero.

Product $= 0$

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Solution:

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The area of a rectangle is given by the product of its length and breadth.

Area = Length $\times$ Breadth

We need to find the product of each pair of monomials given as length and breadth.

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(i) Length = p, Breadth = q

Area = $p \times q = pq$

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(ii) Length = 10m, Breadth = 5n

Area = $(10m) \times (5n)$

Area = $(10 \times 5) \times (m \times n)$

Area = $50mn$

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(iii) Length = 20x², Breadth = 5y²

Area = $(20x^2) \times (5y^2)$

Area = $(20 \times 5) \times (x^2 \times y^2)$

Area = $100x^2y^2$

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(iv) Length = 4x, Breadth = 3x²

Area = $(4x) \times (3x^2)$

Area = $(4 \times 3) \times (x \times x^2)$

Area = $12 \times x^{1+2}$

Area = $12x^3$

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(v) Length = 3mn, Breadth = 4np

Area = $(3mn) \times (4np)$

Area = $(3 \times 4) \times (m \times n \times n \times p)$

Area = $12 \times m \times n^{1+1} \times p$

Area = $12mn^2p$

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The areas of the rectangles are:

(i) $pq$

(ii) $50mn$

(iii) $100x^2y^2$

(iv) $12x^3$

(v) $12mn^2p$

Solution:

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To complete the table, we need to multiply each monomial from the first column (Second Monomial $\downarrow$) by each monomial from the first row (First Monomial $\rightarrow$). When multiplying monomials, we multiply the numerical coefficients and add the exponents of the same variables.

For example, the entry in the cell for row '2x' and column '-5y' is the product $(2x) \times (-5y)$.

$(2x) \times (-5y) = (2 \times -5) \times (x \times y) = -10xy$.

The entry in the cell for row '-5y' and column '3x²' is the product $(-5y) \times (3x^2)$.

$(-5y) \times (3x^2) = (-5 \times 3) \times (y \times x^2) = -15x^2y$. (This matches the example given in the question).

We perform similar multiplications for all the cells.

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The completed table shows the product of each pair of monomials as calculated by multiplying the coefficients and adding the exponents of the corresponding variables.

The volume (V) of a rectangular box is obtained by multiplying its length (l), breadth (b), and height (h). The formula is:

$V = l \times b \times h$

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(i) For length = $5a$, breadth = $3a^2$, and height = $7a^4$

The given dimensions are:

Length (l) = $5a$

Breadth (b) = $3a^2$

Height (h) = $7a^4$

The volume of the rectangular box is:

$V_1 = (5a) \times (3a^2) \times (7a^4)$

Multiplying the numerical coefficients:

$5 \times 3 \times 7 = 15 \times 7 = 105$

Multiplying the variable parts (using the rule $x^m \times x^n = x^{m+n}$):

$a \times a^2 \times a^4 = a^1 \times a^2 \times a^4 = a^{(1+2+4)} = a^7$

Combining these results:

$V_1 = 105 \times a^7 = 105a^7$

Thus, the volume of the rectangular box is $105a^7$ cubic units.

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(ii) For length = $2p$, breadth = $4q$, and height = $8r$

The given dimensions are:

Length (l) = $2p$

Breadth (b) = $4q$

Height (h) = $8r$

The volume of the rectangular box is:

$V_2 = (2p) \times (4q) \times (8r)$

Multiplying the numerical coefficients:

$2 \times 4 \times 8 = 8 \times 8 = 64$

Multiplying the variable parts:

$p \times q \times r = pqr$

Combining these results:

$V_2 = 64 \times pqr = 64pqr$

Thus, the volume of the rectangular box is $64pqr$ cubic units.

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(iii) For length = $xy$, breadth = $2x^2y$, and height = $2xy^2$

The given dimensions are:

Length (l) = $xy$

Breadth (b) = $2x^2y$

Height (h) = $2xy^2$

The volume of the rectangular box is:

$V_3 = (xy) \times (2x^2y) \times (2xy^2)$

Multiplying the numerical coefficients:

$1 \times 2 \times 2 = 4$

Multiplying the variable parts (using the rule $x^m \times x^n = x^{m+n}$ for each variable):

For variable $x$: $x \times x^2 \times x = x^1 \times x^2 \times x^1 = x^{(1+2+1)} = x^4$

For variable $y$: $y \times y \times y^2 = y^1 \times y^1 \times y^2 = y^{(1+1+2)} = y^4$

Combining these results:

$V_3 = 4 \times x^4 \times y^4 = 4x^4y^4$

Thus, the volume of the rectangular box is $4x^4y^4$ cubic units.

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(iv) For length = $a$, breadth = $2b$, and height = $3c$

The given dimensions are:

Length (l) = $a$

Breadth (b) = $2b$

Height (h) = $3c$

The volume of the rectangular box is:

$V_4 = (a) \times (2b) \times (3c)$

Multiplying the numerical coefficients:

$1 \times 2 \times 3 = 6$

Multiplying the variable parts:

$a \times b \times c = abc$

Combining these results:

$V_4 = 6 \times abc = 6abc$

Thus, the volume of the rectangular box is $6abc$ cubic units.

To obtain the product of the given monomials, we multiply their numerical coefficients and then multiply their variable parts. For variables with exponents, we use the rule $x^m \times x^n = x^{m+n}$.

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(i) Product of $xy, yz, zx$

The given monomials are $xy$, $yz$, and $zx$.

Product $= (xy) \times (yz) \times (zx)$

Group like variables together:

Product $= (x \times x) \times (y \times y) \times (z \times z)$

Product $= (x^{1+1}) \times (y^{1+1}) \times (z^{1+1})$

Product $= x^2 \times y^2 \times z^2$

Product $= x^2y^2z^2$

Thus, the product is $x^2y^2z^2$.

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(ii) Product of $a, – a^2, a^3$

The given monomials are $a$, $-a^2$, and $a^3$.

Product $= (a) \times (-a^2) \times (a^3)$

The numerical coefficients are $1$, $-1$, and $1$. Their product is $1 \times (-1) \times 1 = -1$.

The variable parts are $a$, $a^2$, and $a^3$. Their product is $a^1 \times a^2 \times a^3 = a^{(1+2+3)} = a^6$.

Combining these results:

Product $= -1 \times a^6 = -a^6$

Thus, the product is $-a^6$.

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(iii) Product of $2, 4y, 8y^2, 16y^3$

The given monomials are $2$, $4y$, $8y^2$, and $16y^3$.

Product $= (2) \times (4y) \times (8y^2) \times (16y^3)$

Multiply the numerical coefficients:

$2 \times 4 \times 8 \times 16 = 1024$.

Multiply the variable parts:

$y \times y^2 \times y^3 = y^1 \times y^2 \times y^3 = y^{(1+2+3)} = y^6$. (Note: the first monomial '2' has no y term, or $y^0=1$)

Combining these results:

Product $= 1024 \times y^6 = 1024y^6$

Thus, the product is $1024y^6$.

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(iv) Product of $a, 2b, 3c, 6abc$

The given monomials are $a$, $2b$, $3c$, and $6abc$.

Product $= (a) \times (2b) \times (3c) \times (6abc)$

Multiply the numerical coefficients:

$1 \times 2 \times 3 \times 6 = 2 \times 3 \times 6 = 6 \times 6 = 36$.

Multiply the variable parts (group like variables):

For variable $a$: $a \times a = a^1 \times a^1 = a^{(1+1)} = a^2$.

For variable $b$: $b \times b = b^1 \times b^1 = b^{(1+1)} = b^2$.

For variable $c$: $c \times c = c^1 \times c^1 = c^{(1+1)} = c^2$.

Combining these results:

Product $= 36 \times a^2 \times b^2 \times c^2 = 36a^2b^2c^2$

Thus, the product is $36a^2b^2c^2$.

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(v) Product of $m, – mn, mnp$

The given monomials are $m$, $-mn$, and $mnp$.

Product $= (m) \times (-mn) \times (mnp)$

The numerical coefficients are $1$, $-1$, and $1$. Their product is $1 \times (-1) \times 1 = -1$.

Multiply the variable parts (group like variables):

For variable $m$: $m \times m \times m = m^1 \times m^1 \times m^1 = m^{(1+1+1)} = m^3$.

For variable $n$: $n \times n = n^1 \times n^1 = n^{(1+1)} = n^2$.

For variable $p$: $p = p^1$.

Combining these results:

Product $= -1 \times m^3 \times n^2 \times p^1 = -m^3n^2p$

Thus, the product is $-m^3n^2p$.

(i) $x (x – 3) + 2$ for $x = 1$

Simplification:

First, simplify the expression $x(x-3) + 2$.

Use the distributive property: $a(b-c) = ab - ac$.

$x(x-3) + 2 = x \cdot x - x \cdot 3 + 2$

$= x^2 - 3x + 2$

The simplified expression is $x^2 - 3x + 2$.

Evaluation:

Now, substitute $x = 1$ into the simplified expression $x^2 - 3x + 2$.

Value $= (1)^2 - 3(1) + 2$

$= 1 - 3 + 2$

$= -2 + 2$

$= 0$

Thus, for $x=1$, the value of $x(x-3)+2$ is 0.

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(ii) $3y (2y – 7) – 3 (y – 4) – 63$ for $y = –2$

Simplification:

First, simplify the expression $3y(2y – 7) – 3(y – 4) – 63$.

Apply the distributive property to $3y(2y-7)$:

$3y(2y-7) = 3y \cdot 2y - 3y \cdot 7 = 6y^2 - 21y$

Apply the distributive property to $-3(y-4)$:

$-3(y-4) = -3 \cdot y - 3 \cdot (-4) = -3y + 12$

Now substitute these back into the original expression:

$6y^2 - 21y - 3y + 12 - 63$

Combine like terms:

$6y^2 + (-21y - 3y) + (12 - 63)$

$= 6y^2 - 24y - 51$

The simplified expression is $6y^2 - 24y - 51$.

Evaluation:

Now, substitute $y = -2$ into the simplified expression $6y^2 - 24y - 51$.

Value $= 6(-2)^2 - 24(-2) - 51$

$= 6(4) - (-48) - 51$

$= 24 + 48 - 51$

$= 72 - 51$

$= 21$

Thus, for $y=-2$, the value of $3y(2y – 7) – 3(y – 4) – 63$ is 21.

(i) Add $5m (3 – m)$ and $6m^2 – 13m$

First, simplify the expression $5m(3-m)$:

$5m(3-m) = 5m \cdot 3 - 5m \cdot m$

$= 15m - 5m^2$

Now, add this simplified expression to $6m^2 - 13m$:

Sum $= (15m - 5m^2) + (6m^2 - 13m)$

Rearrange the terms to group like terms together:

Sum $= -5m^2 + 6m^2 + 15m - 13m$

Combine the like terms:

For $m^2$ terms: $(-5 + 6)m^2 = 1m^2 = m^2$

For $m$ terms: $(15 - 13)m = 2m$

So, the sum is $m^2 + 2m$.

Thus, $5m (3 – m) + (6m^2 – 13m) = \mathbf{m^2 + 2m}$.

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(ii) Add $4y (3y^2 + 5y – 7)$ and $2 (y^3 – 4y^2 + 5)$

First, simplify the expression $4y(3y^2 + 5y – 7)$:

$4y(3y^2 + 5y – 7) = 4y \cdot 3y^2 + 4y \cdot 5y - 4y \cdot 7$

$= 12y^3 + 20y^2 - 28y$

Next, simplify the expression $2(y^3 – 4y^2 + 5)$:

$2(y^3 – 4y^2 + 5) = 2 \cdot y^3 - 2 \cdot 4y^2 + 2 \cdot 5$

$= 2y^3 - 8y^2 + 10$

Now, add these two simplified expressions:

Sum $= (12y^3 + 20y^2 - 28y) + (2y^3 - 8y^2 + 10)$

Rearrange the terms to group like terms together:

Sum $= 12y^3 + 2y^3 + 20y^2 - 8y^2 - 28y + 10$

Combine the like terms:

For $y^3$ terms: $(12 + 2)y^3 = 14y^3$

For $y^2$ terms: $(20 - 8)y^2 = 12y^2$

For $y$ terms: $-28y$

Constant terms: $+10$

So, the sum is $14y^3 + 12y^2 - 28y + 10$.

Thus, $4y (3y^2 + 5y – 7) + 2 (y^3 – 4y^2 + 5) = \mathbf{14y^3 + 12y^2 - 28y + 10}$.

We need to subtract the expression $3pq (p – q)$ from the expression $2pq (p + q)$.

This can be written as: $2pq (p + q) - 3pq (p – q)$.

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First, let's simplify the expression $2pq (p + q)$:

$2pq (p + q) = 2pq \cdot p + 2pq \cdot q$

$= 2p^{(1+1)}q + 2pq^{(1+1)}$

$= 2p^2q + 2pq^2$

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Next, let's simplify the expression $3pq (p – q)$:

$3pq (p – q) = 3pq \cdot p - 3pq \cdot q$

$= 3p^{(1+1)}q - 3pq^{(1+1)}$

$= 3p^2q - 3pq^2$

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Now, perform the subtraction: $(2p^2q + 2pq^2) - (3p^2q - 3pq^2)$.

When subtracting an expression, we change the sign of each term in the expression being subtracted:

$2p^2q + 2pq^2 - 3p^2q - (-3pq^2)$

$= 2p^2q + 2pq^2 - 3p^2q + 3pq^2$

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Group the like terms together:

Terms with $p^2q$: $2p^2q - 3p^2q$

Terms with $pq^2$: $2pq^2 + 3pq^2$

So, the expression becomes:

$(2p^2q - 3p^2q) + (2pq^2 + 3pq^2)$

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Combine the coefficients of the like terms:

$(2 - 3)p^2q + (2 + 3)pq^2$

$= -1p^2q + 5pq^2$

$= -p^2q + 5pq^2$

This can also be written as $5pq^2 - p^2q$.

Thus, subtracting $3pq (p – q)$ from $2pq (p + q)$ results in $5pq^2 - p^2q$.

To multiply a monomial by a polynomial (a binomial, trinomial, etc.), we use the distributive property of multiplication over addition (and subtraction).

This property states that to multiply a monomial by a polynomial, we multiply the monomial by each term of the polynomial separately and then add the products.

In general, if $M$ is a monomial and $(A+B)$ is a binomial, then:

$M \times (A + B) = M \times A + M \times B = MA + MB$

If we have subtraction, like $(A-B)$, the rule is similar:

$M \times (A - B) = M \times A - M \times B = MA - MB$

For a trinomial $(A+B+C)$, the property extends:

$M \times (A + B + C) = M \times A + M \times B + M \times C = MA + MB + MC$

The same property applies if the monomial is after the polynomial, e.g., $(A+B) \times M = AM + BM$.

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(i) $4p, q + r$

We need to multiply $4p$ by $(q+r)$.

Product $= 4p \times (q+r)$

Using the distributive property:

$= 4p \cdot q + 4p \cdot r$

$= 4pq + 4pr$

Thus, the product of $4p$ and $q+r$ is $4pq + 4pr$.

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(ii) $ab, a – b$

We need to multiply $ab$ by $(a-b)$.

Product $= ab \times (a-b)$

Using the distributive property:

$= ab \cdot a - ab \cdot b$

Recall that $x^m \cdot x^n = x^{m+n}$.

$= a^{1+1}b - ab^{1+1}$

$= a^2b - ab^2$

Thus, the product of $ab$ and $a-b$ is $a^2b - ab^2$.

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(iii) $a + b, 7a^2b^2$

We need to multiply $(a+b)$ by $7a^2b^2$.

Product $= (a+b) \times 7a^2b^2$

Using the distributive property:

$= a \cdot 7a^2b^2 + b \cdot 7a^2b^2$

$= 7a^{1+2}b^2 + 7a^2b^{1+2}$

$= 7a^3b^2 + 7a^2b^3$

Thus, the product of $a+b$ and $7a^2b^2$ is $7a^3b^2 + 7a^2b^3$.

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(iv) $a^2 – 9, 4a$

We need to multiply $(a^2-9)$ by $4a$.

Product $= (a^2-9) \times 4a$

Using the distributive property:

$= a^2 \cdot 4a - 9 \cdot 4a$

$= 4a^{2+1} - 36a$

$= 4a^3 - 36a$

Thus, the product of $a^2-9$ and $4a$ is $4a^3 - 36a$.

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(v) $pq + qr + rp, 0$

We need to multiply $(pq+qr+rp)$ by $0$.

Product $= (pq + qr + rp) \times 0$

Any expression multiplied by zero is zero.

$= 0$

Thus, the product of $pq+qr+rp$ and $0$ is $0$.

To complete the table, we need to find the product of the "First Expression" and the "Second Expression" for each row. We will use the distributive property of multiplication over addition/subtraction.

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(i) First Expression: $a$, Second Expression: $b + c + d$

Product $= a \times (b + c + d)$

Using the distributive property $M(A+B+C) = MA + MB + MC$:

$= a \cdot b + a \cdot c + a \cdot d$

$= ab + ac + ad$

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(ii) First Expression: $x + y – 5$, Second Expression: $5xy$

Product $= (x + y – 5) \times 5xy$

Using the distributive property $(A+B-C)M = AM + BM - CM$:

$= x \cdot 5xy + y \cdot 5xy - 5 \cdot 5xy$

$= 5x^{1+1}y + 5xy^{1+1} - 25xy$

$= 5x^2y + 5xy^2 - 25xy$

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(iii) First Expression: $p$, Second Expression: $6p^2 – 7p + 5$

Product $= p \times (6p^2 – 7p + 5)$

Using the distributive property $M(A-B+C) = MA - MB + MC$:

$= p \cdot 6p^2 - p \cdot 7p + p \cdot 5$

$= 6p^{1+2} - 7p^{1+1} + 5p$

$= 6p^3 - 7p^2 + 5p$

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(iv) First Expression: $4p^2q^2$, Second Expression: $p^2 – q^2$

Product $= 4p^2q^2 \times (p^2 – q^2)$

Using the distributive property $M(A-B) = MA - MB$:

$= 4p^2q^2 \cdot p^2 - 4p^2q^2 \cdot q^2$

$= 4p^{(2+2)}q^2 - 4p^2q^{(2+2)}$

$= 4p^4q^2 - 4p^2q^4$

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(v) First Expression: $a + b + c$, Second Expression: $abc$

Product $= (a + b + c) \times abc$

Using the distributive property $(A+B+C)M = AM + BM + CM$:

$= a \cdot abc + b \cdot abc + c \cdot abc$

$= a^{1+1}bc + ab^{1+1}c + abc^{1+1}$

$= a^2bc + ab^2c + abc^2$

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Completed Table:

	First Expression	Second Expression	Product
(i)	$a$	$b + c + d$	$ab + ac + ad$
(ii)	$x + y – 5$	$5xy$	$5x^2y + 5xy^2 - 25xy$
(iii)	$p$	$6p^2 – 7p + 5$	$6p^3 - 7p^2 + 5p$
(iv)	$4p^2q^2$	$p^2 – q^2$	$4p^4q^2 - 4p^2q^4$
(v)	$a + b + c$	$abc$	$a^2bc + ab^2c + abc^2$

To find the product of monomials, we multiply their numerical coefficients and then multiply their variable parts. For variables with exponents, we use the rule $x^m \times x^n = x^{m+n}$.

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(i) $(a^2) \times (2a^{22}) \times (4a^{26})$

Product $= (1 \cdot a^2) \times (2 \cdot a^{22}) \times (4 \cdot a^{26})$

Multiply the numerical coefficients:

$1 \times 2 \times 4 = 8$

Multiply the variable parts:

$a^2 \times a^{22} \times a^{26} = a^{(2+22+26)}$

$= a^{(24+26)}$

$= a^{50}$

Combine these results:

Product $= 8 \times a^{50} = 8a^{50}$

Thus, the product is $8a^{50}$.

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(ii) $\left( \frac{2}{3}xy \right)\times\left( \frac{-9}{10}x^2y^2 \right)$

Product $= \left( \frac{2}{3} \right) \times \left( \frac{-9}{10} \right) \times (xy) \times (x^2y^2)$

Multiply the numerical coefficients:

$\frac{2}{3} \times \frac{-9}{10} = \frac{2 \times (-9)}{3 \times 10} = \frac{-18}{30}$

Simplify the fraction: $\frac{-18}{30} = \frac{-\cancel{18}^3}{\cancel{30}^5} = -\frac{3}{5}$ (dividing numerator and denominator by 6)

Multiply the variable parts:

For $x$: $x \times x^2 = x^1 \times x^2 = x^{(1+2)} = x^3$

For $y$: $y \times y^2 = y^1 \times y^2 = y^{(1+2)} = y^3$

So, $(xy) \times (x^2y^2) = x^3y^3$

Combine these results:

Product $= -\frac{3}{5} \times x^3y^3 = -\frac{3}{5}x^3y^3$

Thus, the product is $-\frac{3}{5}x^3y^3$.

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(iii) $\left( -\frac{10}{3}pq^3 \right)\times\left( \frac{6}{5}p^3q \right)$

Product $= \left( -\frac{10}{3} \right) \times \left( \frac{6}{5} \right) \times (pq^3) \times (p^3q)$

Multiply the numerical coefficients:

$-\frac{10}{3} \times \frac{6}{5} = \frac{-10 \times 6}{3 \times 5} = \frac{-60}{15}$

Simplify the fraction: $\frac{-60}{15} = -4$

Multiply the variable parts:

For $p$: $p \times p^3 = p^1 \times p^3 = p^{(1+3)} = p^4$

For $q$: $q^3 \times q = q^3 \times q^1 = q^{(3+1)} = q^4$

So, $(pq^3) \times (p^3q) = p^4q^4$

Combine these results:

Product $= -4 \times p^4q^4 = -4p^4q^4$

Thus, the product is $-4p^4q^4$.

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(iv) $x \times x^2 \times x^3 \times x^4$

Product $= x^1 \times x^2 \times x^3 \times x^4$

The numerical coefficient for each term is 1, so their product is $1 \times 1 \times 1 \times 1 = 1$.

Multiply the variable parts by adding the exponents:

$x^{(1+2+3+4)} = x^{(3+3+4)} = x^{(6+4)} = x^{10}$

Combine these results:

Product $= 1 \times x^{10} = x^{10}$

Thus, the product is $x^{10}$.

(a) Simplify $3x (4x – 5) + 3$ and find its values for (i) $x = 3$ (ii) $x = \frac{1}{2}$

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Simplification:

The given expression is $3x(4x-5) + 3$.

Using the distributive property, $A(B-C) = AB - AC$:

$3x(4x-5) = (3x)(4x) - (3x)(5)$

$= 12x^2 - 15x$

So, the expression becomes:

$12x^2 - 15x + 3$

The simplified expression is $12x^2 - 15x + 3$.

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Evaluation:

(i) For $x = 3$:

Substitute $x=3$ into the simplified expression $12x^2 - 15x + 3$:

Value $= 12(3)^2 - 15(3) + 3$

$= 12(9) - 45 + 3$

$= 108 - 45 + 3$

$= 63 + 3$

$= 66$

Thus, for $x=3$, the value is 66.

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(ii) For $x = \frac{1}{2}$:

Substitute $x=\frac{1}{2}$ into the simplified expression $12x^2 - 15x + 3$:

Value $= 12\left(\frac{1}{2}\right)^2 - 15\left(\frac{1}{2}\right) + 3$

$= 12\left(\frac{1}{4}\right) - \frac{15}{2} + 3$

$= \frac{12}{4} - \frac{15}{2} + 3$

$= 3 - \frac{15}{2} + 3$

$= 6 - \frac{15}{2}$

To subtract, find a common denominator:

$= \frac{6 \times 2}{2} - \frac{15}{2}$

$= \frac{12}{2} - \frac{15}{2}$

$= \frac{12 - 15}{2}$

$= -\frac{3}{2}$

Thus, for $x=\frac{1}{2}$, the value is $-\frac{3}{2}$.

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(b) Simplify $a (a^2 + a + 1) + 5$ and find its value for (i) $a = 0$ (ii) $a = 1$ (iii) $a = – 1$

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Simplification:

The given expression is $a(a^2 + a + 1) + 5$.

Using the distributive property, $X(A+B+C) = XA + XB + XC$:

$a(a^2 + a + 1) = (a)(a^2) + (a)(a) + (a)(1)$

$= a^3 + a^2 + a$

So, the expression becomes:

$a^3 + a^2 + a + 5$

The simplified expression is $a^3 + a^2 + a + 5$.

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Evaluation:

(i) For $a = 0$:

Substitute $a=0$ into the simplified expression $a^3 + a^2 + a + 5$:

Value $= (0)^3 + (0)^2 + (0) + 5$

$= 0 + 0 + 0 + 5$

$= 5$

Thus, for $a=0$, the value is 5.

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(ii) For $a = 1$:

Substitute $a=1$ into the simplified expression $a^3 + a^2 + a + 5$:

Value $= (1)^3 + (1)^2 + (1) + 5$

$= 1 + 1 + 1 + 5$

$= 3 + 5$

$= 8$

Thus, for $a=1$, the value is 8.

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(iii) For $a = – 1$:

Substitute $a=-1$ into the simplified expression $a^3 + a^2 + a + 5$:

Value $= (-1)^3 + (-1)^2 + (-1) + 5$

$= -1 + 1 - 1 + 5$

$= 0 - 1 + 5$

$= 4$

Thus, for $a=-1$, the value is 4.

(a) Add: $p (p – q)$, $q (q – r)$ and $r (r – p)$

First, simplify each expression:

$p(p-q) = p \cdot p - p \cdot q = p^2 - pq$

$q(q-r) = q \cdot q - q \cdot r = q^2 - qr$

$r(r-p) = r \cdot r - r \cdot p = r^2 - rp$

Now, add these simplified expressions:

Sum $= (p^2 - pq) + (q^2 - qr) + (r^2 - rp)$

$= p^2 - pq + q^2 - qr + r^2 - rp$

Rearranging the terms (optional, for conventional order):

$= p^2 + q^2 + r^2 - pq - qr - rp$

Thus, the sum is $p^2 + q^2 + r^2 - pq - qr - rp$.

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(b) Add: $2x (z – x – y)$ and $2y (z – y – x)$

First, simplify each expression:

$2x(z - x - y) = 2x \cdot z - 2x \cdot x - 2x \cdot y = 2xz - 2x^2 - 2xy$

$2y(z - y - x) = 2y \cdot z - 2y \cdot y - 2y \cdot x = 2yz - 2y^2 - 2yx$

Since $yx = xy$, the second expression is $2yz - 2y^2 - 2xy$.

Now, add these simplified expressions:

Sum $= (2xz - 2x^2 - 2xy) + (2yz - 2y^2 - 2xy)$

$= 2xz - 2x^2 - 2xy + 2yz - 2y^2 - 2xy$

Combine like terms:

Group $x^2$ terms: $-2x^2$

Group $y^2$ terms: $-2y^2$

Group $xy$ terms: $-2xy - 2xy = -4xy$

Group $xz$ terms: $2xz$

Group $yz$ terms: $2yz$

So, Sum $= -2x^2 - 2y^2 - 4xy + 2xz + 2yz$

Thus, the sum is $-2x^2 - 2y^2 - 4xy + 2xz + 2yz$. (The order of terms can vary).

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(c) Subtract: $3l (l – 4m + 5n)$ from $4l (10n – 3m + 2l)$

This means we need to calculate: $4l (10n – 3m + 2l) - 3l (l – 4m + 5n)$.

First, simplify the expression $4l (10n – 3m + 2l)$:

$4l(10n - 3m + 2l) = 4l \cdot 10n - 4l \cdot 3m + 4l \cdot 2l$

$= 40ln - 12lm + 8l^2$

Next, simplify the expression $3l (l – 4m + 5n)$:

$3l(l - 4m + 5n) = 3l \cdot l - 3l \cdot 4m + 3l \cdot 5n$

$= 3l^2 - 12lm + 15ln$

Now, perform the subtraction:

$(40ln - 12lm + 8l^2) - (3l^2 - 12lm + 15ln)$

$= 40ln - 12lm + 8l^2 - 3l^2 + 12lm - 15ln$

Combine like terms:

Group $l^2$ terms: $8l^2 - 3l^2 = 5l^2$

Group $lm$ terms: $-12lm + 12lm = 0$

Group $ln$ terms: $40ln - 15ln = 25ln$

So, the result is $5l^2 + 0 + 25ln = 5l^2 + 25ln$.

Thus, the result of the subtraction is $5l^2 + 25ln$.

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(d) Subtract: $3a (a + b + c) – 2b (a – b + c)$ from $4c (– a + b + c )$

Let the first expression (to be subtracted) be $E_1 = 3a (a + b + c) – 2b (a – b + c)$.

Let the second expression (from which to subtract) be $E_2 = 4c (– a + b + c )$.

We need to calculate $E_2 - E_1$.

First, simplify $E_1$:

$3a(a+b+c) = 3a^2 + 3ab + 3ac$

$2b(a-b+c) = 2ab - 2b^2 + 2bc$

$E_1 = (3a^2 + 3ab + 3ac) - (2ab - 2b^2 + 2bc)$

$E_1 = 3a^2 + 3ab + 3ac - 2ab + 2b^2 - 2bc$

Combine like terms in $E_1$:

$E_1 = 3a^2 + 2b^2 + (3ab - 2ab) + 3ac - 2bc$

$E_1 = 3a^2 + 2b^2 + ab + 3ac - 2bc$

Next, simplify $E_2$:

$E_2 = 4c(– a + b + c) = 4c(-a) + 4c(b) + 4c(c)$

$E_2 = -4ac + 4bc + 4c^2$

Now, perform the subtraction $E_2 - E_1$:

$(-4ac + 4bc + 4c^2) - (3a^2 + 2b^2 + ab + 3ac - 2bc)$

$= -4ac + 4bc + 4c^2 - 3a^2 - 2b^2 - ab - 3ac + 2bc$

Combine like terms:

Group $a^2$ terms: $-3a^2$

Group $b^2$ terms: $-2b^2$

Group $c^2$ terms: $4c^2$

Group $ab$ terms: $-ab$

Group $ac$ terms: $-4ac - 3ac = -7ac$

Group $bc$ terms: $4bc + 2bc = 6bc$

So, the result is $-3a^2 - 2b^2 + 4c^2 - ab - 7ac + 6bc$.

Arranging in a standard order (optional):

$-3a^2 - 2b^2 + 4c^2 - ab + 6bc - 7ac$

Thus, the result of the subtraction is $-3a^2 - 2b^2 + 4c^2 - ab + 6bc - 7ac$.

To multiply two binomials, we use the distributive property. If we have two binomials $(a+b)$ and $(c+d)$, their product is given by:

$(a+b)(c+d) = a(c+d) + b(c+d)$

$= ac + ad + bc + bd$

This is sometimes remembered by the acronym FOIL (First, Outer, Inner, Last terms).

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(i) Multiply $(x – 4)$ and $(2x + 3)$

Let the first binomial be $(x-4)$ and the second binomial be $(2x+3)$.

Product $= (x-4)(2x+3)$

Using the distributive property, we multiply each term in the first binomial by each term in the second binomial:

$= x(2x+3) - 4(2x+3)$

Now, distribute $x$ into $(2x+3)$ and $-4$ into $(2x+3)$:

$= (x \cdot 2x + x \cdot 3) - (4 \cdot 2x + 4 \cdot 3)$

$= (2x^2 + 3x) - (8x + 12)$

$= 2x^2 + 3x - 8x - 12$

Combine the like terms (the terms with $x$):

$3x - 8x = -5x$

So, the product is:

$= 2x^2 - 5x - 12$

Thus, $(x – 4) \times (2x + 3) = \mathbf{2x^2 - 5x - 12}$.

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(ii) Multiply $(x – y)$ and $(3x + 5y)$

Let the first binomial be $(x-y)$ and the second binomial be $(3x+5y)$.

Product $= (x-y)(3x+5y)$

Using the distributive property:

$= x(3x+5y) - y(3x+5y)$

Now, distribute $x$ into $(3x+5y)$ and $-y$ into $(3x+5y)$:

$= (x \cdot 3x + x \cdot 5y) - (y \cdot 3x + y \cdot 5y)$

$= (3x^2 + 5xy) - (3yx + 5y^2)$

Since $yx = xy$, we can rewrite the expression as:

$= 3x^2 + 5xy - 3xy - 5y^2$

Combine the like terms (the terms with $xy$):

$5xy - 3xy = 2xy$

So, the product is:

$= 3x^2 + 2xy - 5y^2$

Thus, $(x – y) \times (3x + 5y) = \mathbf{3x^2 + 2xy - 5y^2}$.

To multiply two binomials, we use the distributive property: $(X+Y)(U+V) = X(U+V) + Y(U+V) = XU + XV + YU + YV$.

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(i) Multiply $(a + 7)$ and $(b – 5)$

Let the first binomial be $(a+7)$ and the second binomial be $(b-5)$.

Product $= (a+7)(b-5)$

Using the distributive property:

$= a(b-5) + 7(b-5)$

Now, distribute $a$ into $(b-5)$ and $7$ into $(b-5)$:

$= (a \cdot b + a \cdot (-5)) + (7 \cdot b + 7 \cdot (-5))$

$= (ab - 5a) + (7b - 35)$

$= ab - 5a + 7b - 35$

In this case, there are no like terms to combine.

Thus, $(a + 7) \times (b – 5) = \mathbf{ab - 5a + 7b - 35}$.

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(ii) Multiply $(a^2 + 2b^2)$ and $(5a – 3b)$

Let the first binomial be $(a^2 + 2b^2)$ and the second binomial be $(5a - 3b)$.

Product $= (a^2 + 2b^2)(5a - 3b)$

Using the distributive property:

$= a^2(5a - 3b) + 2b^2(5a - 3b)$

Now, distribute $a^2$ into $(5a - 3b)$ and $2b^2$ into $(5a - 3b)$:

$= (a^2 \cdot 5a + a^2 \cdot (-3b)) + (2b^2 \cdot 5a + 2b^2 \cdot (-3b))$

$= (5a^{2+1} - 3a^2b) + (10b^2a - 6b^{2+1})$

$= (5a^3 - 3a^2b) + (10ab^2 - 6b^3)$

$= 5a^3 - 3a^2b + 10ab^2 - 6b^3$

In this case, there are no like terms to combine (since $a^2b$ is different from $ab^2$).

Thus, $(a^2 + 2b^2) \times (5a – 3b) = \mathbf{5a^3 - 3a^2b + 10ab^2 - 6b^3}$.

We need to simplify the expression $(a + b) (2a – 3b + c) – (2a – 3b) c$.

Let's break this down into two parts.

Part 1: Simplify $(a + b) (2a – 3b + c)$

Using the distributive property, we multiply each term in the first expression $(a+b)$ by each term in the second expression $(2a - 3b + c)$:

$(a + b) (2a – 3b + c) = a(2a – 3b + c) + b(2a – 3b + c)$

Now, distribute $a$ and $b$ into the trinomial:

$= (a \cdot 2a - a \cdot 3b + a \cdot c) + (b \cdot 2a - b \cdot 3b + b \cdot c)$

$= (2a^2 - 3ab + ac) + (2ab - 3b^2 + bc)$

Remove the parentheses and combine like terms:

$= 2a^2 - 3ab + ac + 2ab - 3b^2 + bc$

Combine the $ab$ terms: $-3ab + 2ab = -ab$.

So, $(a + b) (2a – 3b + c) = 2a^2 - ab + ac - 3b^2 + bc$.

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Part 2: Simplify $(2a – 3b) c$

Using the distributive property, we multiply $c$ by each term in the binomial $(2a - 3b)$:

$(2a – 3b) c = 2a \cdot c - 3b \cdot c$

$= 2ac - 3bc$.

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Part 3: Substitute the simplified parts back into the original expression and simplify.

The original expression is $(a + b) (2a – 3b + c) – (2a – 3b) c$.

Substituting the results from Part 1 and Part 2:

$= (2a^2 - ab + ac - 3b^2 + bc) - (2ac - 3bc)$

When subtracting an expression in parentheses, change the sign of each term inside the parentheses:

$= 2a^2 - ab + ac - 3b^2 + bc - 2ac + 3bc$

Now, combine like terms:

Terms with $a^2$: $2a^2$

Terms with $b^2$: $-3b^2$

Terms with $ab$: $-ab$

Terms with $ac$: $ac - 2ac = (1-2)ac = -ac$

Terms with $bc$: $bc + 3bc = (1+3)bc = 4bc$

Combining these terms, we get:

$2a^2 - 3b^2 - ab - ac + 4bc$

Thus, the simplified expression is $2a^2 - 3b^2 - ab - ac + 4bc$. (The order of terms may vary, e.g., $2a^2 - ab - ac - 3b^2 + 4bc$).

To multiply two binomials $(A+B)$ and $(C+D)$, we use the distributive property (often remembered by FOIL):

$(A+B)(C+D) = A(C+D) + B(C+D) = AC + AD + BC + BD$

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(i) $(2x + 5)$ and $(4x – 3)$

Product $= (2x + 5)(4x - 3)$

$= 2x(4x - 3) + 5(4x - 3)$

$= (2x \cdot 4x - 2x \cdot 3) + (5 \cdot 4x - 5 \cdot 3)$

$= (8x^2 - 6x) + (20x - 15)$

$= 8x^2 - 6x + 20x - 15$

Combine like terms ($-6x + 20x = 14x$):

$= 8x^2 + 14x - 15$

Thus, the product is $8x^2 + 14x - 15$.

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(ii) $(y – 8)$ and $(3y – 4)$

Product $= (y - 8)(3y - 4)$

$= y(3y - 4) - 8(3y - 4)$

$= (y \cdot 3y - y \cdot 4) - (8 \cdot 3y - 8 \cdot 4)$

$= (3y^2 - 4y) - (24y - 32)$

$= 3y^2 - 4y - 24y + 32$

Combine like terms ($-4y - 24y = -28y$):

$= 3y^2 - 28y + 32$

Thus, the product is $3y^2 - 28y + 32$.

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(iii) $(2.5l – 0.5m)$ and $(2.5l + 0.5m)$

This is in the form $(A-B)(A+B)$, which equals $A^2 - B^2$.

Here, $A = 2.5l$ and $B = 0.5m$.

Product $= (2.5l)^2 - (0.5m)^2$

$= (2.5)^2 l^2 - (0.5)^2 m^2$

Calculate the squares of the coefficients:

$(2.5)^2 = 2.5 \times 2.5 = 6.25$

$(0.5)^2 = 0.5 \times 0.5 = 0.25$

So, Product $= 6.25l^2 - 0.25m^2$

Alternate method (FOIL):

Product $= (2.5l - 0.5m)(2.5l + 0.5m)$

$= 2.5l(2.5l + 0.5m) - 0.5m(2.5l + 0.5m)$

$= (2.5l \cdot 2.5l + 2.5l \cdot 0.5m) - (0.5m \cdot 2.5l + 0.5m \cdot 0.5m)$

$= (6.25l^2 + 1.25lm) - (1.25ml + 0.25m^2)$

$= 6.25l^2 + 1.25lm - 1.25lm - 0.25m^2$

Combine like terms ($1.25lm - 1.25lm = 0$):

$= 6.25l^2 - 0.25m^2$

Thus, the product is $6.25l^2 - 0.25m^2$.

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(iv) $(a + 3b)$ and $(x + 5)$

Product $= (a + 3b)(x + 5)$

$= a(x + 5) + 3b(x + 5)$

$= (a \cdot x + a \cdot 5) + (3b \cdot x + 3b \cdot 5)$

$= (ax + 5a) + (3bx + 15b)$

$= ax + 5a + 3bx + 15b$

There are no like terms to combine.

Thus, the product is $ax + 5a + 3bx + 15b$.

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(v) $(2pq + 3q^2)$ and $(3pq – 2q^2)$

Product $= (2pq + 3q^2)(3pq - 2q^2)$

$= 2pq(3pq - 2q^2) + 3q^2(3pq - 2q^2)$

$= (2pq \cdot 3pq - 2pq \cdot 2q^2) + (3q^2 \cdot 3pq - 3q^2 \cdot 2q^2)$

$= (6p^{1+1}q^{1+1} - 4pq^{1+2}) + (9pq^{2+1} - 6q^{2+2})$

$= (6p^2q^2 - 4pq^3) + (9pq^3 - 6q^4)$

$= 6p^2q^2 - 4pq^3 + 9pq^3 - 6q^4$

Combine like terms ($-4pq^3 + 9pq^3 = 5pq^3$):

$= 6p^2q^2 + 5pq^3 - 6q^4$

Thus, the product is $6p^2q^2 + 5pq^3 - 6q^4$.

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(vi) $\left( \frac{3}{4}a^2 + 3b^2 \right)$ and $4\left( a^2 - \frac{2}{3}b^2 \right)$

First, distribute the $4$ into the second binomial:

$4\left( a^2 - \frac{2}{3}b^2 \right) = 4 \cdot a^2 - 4 \cdot \frac{2}{3}b^2 = 4a^2 - \frac{8}{3}b^2$

Now multiply $\left( \frac{3}{4}a^2 + 3b^2 \right)$ by $\left( 4a^2 - \frac{8}{3}b^2 \right)$:

Product $= \left( \frac{3}{4}a^2 + 3b^2 \right) \left( 4a^2 - \frac{8}{3}b^2 \right)$

$= \frac{3}{4}a^2 \left( 4a^2 - \frac{8}{3}b^2 \right) + 3b^2 \left( 4a^2 - \frac{8}{3}b^2 \right)$

$= \left( \frac{3}{4}a^2 \cdot 4a^2 - \frac{3}{4}a^2 \cdot \frac{8}{3}b^2 \right) + \left( 3b^2 \cdot 4a^2 - 3b^2 \cdot \frac{8}{3}b^2 \right)$

$= \left( \frac{3 \cdot 4}{4}a^{2+2} - \frac{3 \cdot 8}{4 \cdot 3}a^2b^2 \right) + \left( 12b^2a^2 - \frac{3 \cdot 8}{3}b^{2+2} \right)$

$= \left( 3a^4 - \frac{\cancel{3} \cdot \cancel{8}^2}{\cancel{4} \cdot \cancel{3}}a^2b^2 \right) + \left( 12a^2b^2 - \frac{\cancel{3} \cdot 8}{\cancel{3}}b^4 \right)$

$= (3a^4 - 2a^2b^2) + (12a^2b^2 - 8b^4)$

$= 3a^4 - 2a^2b^2 + 12a^2b^2 - 8b^4$

Combine like terms ($-2a^2b^2 + 12a^2b^2 = 10a^2b^2$):

$= 3a^4 + 10a^2b^2 - 8b^4$

Thus, the product is $3a^4 + 10a^2b^2 - 8b^4$.

To find the product of two binomials, we use the distributive property, often remembered by the acronym FOIL (First, Outer, Inner, Last). For binomials $(A+B)$ and $(C+D)$, the product is $A(C+D) + B(C+D) = AC + AD + BC + BD$.

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(i) $(5 – 2x) (3 + x)$

Product $= (5 - 2x)(3 + x)$

Using the distributive property:

$= 5(3 + x) - 2x(3 + x)$

$= (5 \cdot 3 + 5 \cdot x) - (2x \cdot 3 + 2x \cdot x)$

$= (15 + 5x) - (6x + 2x^2)$

$= 15 + 5x - 6x - 2x^2$

Combine like terms ($5x - 6x = -x$):

$= 15 - x - 2x^2$

Writing in standard form (decreasing powers of x):

$= -2x^2 - x + 15$

Thus, the product is $-2x^2 - x + 15$.

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(ii) $(x + 7y) (7x – y)$

Product $= (x + 7y)(7x - y)$

Using the distributive property:

$= x(7x - y) + 7y(7x - y)$

$= (x \cdot 7x - x \cdot y) + (7y \cdot 7x - 7y \cdot y)$

$= (7x^2 - xy) + (49yx - 7y^2)$

Since $yx = xy$, we have:

$= 7x^2 - xy + 49xy - 7y^2$

Combine like terms ($-xy + 49xy = 48xy$):

$= 7x^2 + 48xy - 7y^2$

Thus, the product is $7x^2 + 48xy - 7y^2$.

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(iii) $(a^2 + b) (a + b^2)$

Product $= (a^2 + b)(a + b^2)$

Using the distributive property:

$= a^2(a + b^2) + b(a + b^2)$

$= (a^2 \cdot a + a^2 \cdot b^2) + (b \cdot a + b \cdot b^2)$

$= (a^{2+1} + a^2b^2) + (ab + b^{1+2})$

$= a^3 + a^2b^2 + ab + b^3$

There are no like terms to combine.

Thus, the product is $a^3 + a^2b^2 + ab + b^3$.

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(iv) $(p^2 – q^2) (2p + q)$

Product $= (p^2 - q^2)(2p + q)$

Using the distributive property:

$= p^2(2p + q) - q^2(2p + q)$

$= (p^2 \cdot 2p + p^2 \cdot q) - (q^2 \cdot 2p + q^2 \cdot q)$

$= (2p^{2+1} + p^2q) - (2q^2p + q^{2+1})$

$= (2p^3 + p^2q) - (2pq^2 + q^3)$

$= 2p^3 + p^2q - 2pq^2 - q^3$

There are no like terms to combine.

Thus, the product is $2p^3 + p^2q - 2pq^2 - q^3$.

To simplify the given expressions, we will perform the multiplications (distributive property or FOIL method for binomials) and then combine any like terms.

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(i) $(x^2 – 5) (x + 5) + 25$

First, multiply $(x^2 – 5)$ by $(x + 5)$:

$(x^2 – 5)(x + 5) = x^2(x + 5) – 5(x + 5)$

$= (x^2 \cdot x + x^2 \cdot 5) – (5 \cdot x + 5 \cdot 5)$

$= (x^3 + 5x^2) – (5x + 25)$

$= x^3 + 5x^2 – 5x – 25$

Now, add 25 to this result:

Expression $= (x^3 + 5x^2 – 5x – 25) + 25$

$= x^3 + 5x^2 – 5x – 25 + 25$

Combine the constant terms ($-25 + 25 = 0$):

$= x^3 + 5x^2 – 5x$

Thus, the simplified expression is $x^3 + 5x^2 – 5x$.

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(ii) $(a^2 + 5) (b^3 + 3) + 5$

First, multiply $(a^2 + 5)$ by $(b^3 + 3)$:

$(a^2 + 5)(b^3 + 3) = a^2(b^3 + 3) + 5(b^3 + 3)$

$= (a^2 \cdot b^3 + a^2 \cdot 3) + (5 \cdot b^3 + 5 \cdot 3)$

$= (a^2b^3 + 3a^2) + (5b^3 + 15)$

$= a^2b^3 + 3a^2 + 5b^3 + 15$

Now, add 5 to this result:

Expression $= (a^2b^3 + 3a^2 + 5b^3 + 15) + 5$

$= a^2b^3 + 3a^2 + 5b^3 + 15 + 5$

Combine the constant terms ($15 + 5 = 20$):

$= a^2b^3 + 3a^2 + 5b^3 + 20$

Thus, the simplified expression is $a^2b^3 + 3a^2 + 5b^3 + 20$.

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(iii) $(t + s^2) (t^2 – s)$

Multiply $(t + s^2)$ by $(t^2 – s)$ using the distributive property:

$(t + s^2)(t^2 – s) = t(t^2 – s) + s^2(t^2 – s)$

$= (t \cdot t^2 - t \cdot s) + (s^2 \cdot t^2 - s^2 \cdot s)$

$= (t^3 - ts) + (s^2t^2 - s^3)$

$= t^3 - ts + s^2t^2 - s^3$

Rearranging terms (optional, for conventional order):

$= t^3 + s^2t^2 - ts - s^3$

Thus, the simplified expression is $t^3 + s^2t^2 - ts - s^3$.

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(iv) $(a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)$

First, expand each product:

1. $(a + b)(c – d) = a(c-d) + b(c-d) = ac - ad + bc - bd$

2. $(a – b)(c + d) = a(c+d) - b(c+d) = ac + ad - bc - bd$

3. $2(ac + bd) = 2ac + 2bd$

Now, add these three results:

Expression $= (ac - ad + bc - bd) + (ac + ad - bc - bd) + (2ac + 2bd)$

$= ac - ad + bc - bd + ac + ad - bc - bd + 2ac + 2bd$

Combine like terms:

Terms with $ac$: $ac + ac + 2ac = 4ac$

Terms with $ad$: $-ad + ad = 0$

Terms with $bc$: $bc - bc = 0$

Terms with $bd$: $-bd - bd + 2bd = -2bd + 2bd = 0$

So, the expression simplifies to:

$= 4ac + 0 + 0 + 0 = 4ac$

Thus, the simplified expression is $4ac$.

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(v) $(x + y)(2x + y) + (x + 2y)(x – y)$

First, expand each product:

1. $(x + y)(2x + y) = x(2x+y) + y(2x+y)$

$= 2x^2 + xy + 2xy + y^2 = 2x^2 + 3xy + y^2$

2. $(x + 2y)(x – y) = x(x-y) + 2y(x-y)$

$= x^2 - xy + 2xy - 2y^2 = x^2 + xy - 2y^2$

Now, add these two results:

Expression $= (2x^2 + 3xy + y^2) + (x^2 + xy - 2y^2)$

$= 2x^2 + 3xy + y^2 + x^2 + xy - 2y^2$

Combine like terms:

Terms with $x^2$: $2x^2 + x^2 = 3x^2$

Terms with $xy$: $3xy + xy = 4xy$

Terms with $y^2$: $y^2 - 2y^2 = -y^2$

So, the expression simplifies to:

$= 3x^2 + 4xy - y^2$

Thus, the simplified expression is $3x^2 + 4xy - y^2$.

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(vi) $(x + y)(x^2 – xy + y^2)$

This expression is in the form of the sum of cubes identity: $(A+B)(A^2 - AB + B^2) = A^3 + B^3$.

Here, $A=x$ and $B=y$.

So, $(x + y)(x^2 – xy + y^2) = x^3 + y^3$.

Alternate method (Direct multiplication):

$(x + y)(x^2 – xy + y^2) = x(x^2 – xy + y^2) + y(x^2 – xy + y^2)$

$= (x \cdot x^2 - x \cdot xy + x \cdot y^2) + (y \cdot x^2 - y \cdot xy + y \cdot y^2)$

$= (x^3 - x^2y + xy^2) + (x^2y - xy^2 + y^3)$

$= x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3$

Combine like terms:

$-x^2y + x^2y = 0$

$xy^2 - xy^2 = 0$

So, the expression simplifies to $x^3 + y^3$.

Thus, the simplified expression is $x^3 + y^3$.

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(vii) $(1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y$

First, multiply $(1.5x – 4y)$ by $(1.5x + 4y + 3)$:

Let $A = 1.5x$ and $B = 4y$. The first part is $(A-B)(A+B+3)$.

$(1.5x – 4y)(1.5x + 4y + 3) = (1.5x – 4y)((1.5x + 4y) + 3)$

$= (1.5x – 4y)(1.5x + 4y) + (1.5x – 4y)(3)$

The term $(1.5x – 4y)(1.5x + 4y)$ is a difference of squares, $(U-V)(U+V) = U^2 - V^2$:

$(1.5x)^2 – (4y)^2 = (1.5)^2x^2 – (4)^2y^2 = 2.25x^2 – 16y^2$.

The term $(1.5x – 4y)(3)$ is:

$3(1.5x) – 3(4y) = 4.5x – 12y$.

So, $(1.5x – 4y)(1.5x + 4y + 3) = 2.25x^2 – 16y^2 + 4.5x – 12y$.

Now, substitute this back into the original expression:

Expression $= (2.25x^2 – 16y^2 + 4.5x – 12y) – 4.5x + 12y$

$= 2.25x^2 – 16y^2 + 4.5x – 12y – 4.5x + 12y$

Combine like terms:

Terms with $x$: $4.5x - 4.5x = 0$

Terms with $y$: $-12y + 12y = 0$

So, the expression simplifies to:

$= 2.25x^2 – 16y^2$

Thus, the simplified expression is $2.25x^2 – 16y^2$.

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(viii) $(a + b + c)(a + b – c)$

Let $X = (a+b)$. Then the expression becomes $(X+c)(X-c)$.

This is in the form of the difference of squares identity: $(U+V)(U-V) = U^2 - V^2$.

Here, $U = X = (a+b)$ and $V = c$.

So, $(X+c)(X-c) = X^2 - c^2 = (a+b)^2 - c^2$.

Now, expand $(a+b)^2$ using the identity $(P+Q)^2 = P^2 + 2PQ + Q^2$:

$(a+b)^2 = a^2 + 2ab + b^2$.

Substitute this back:

Expression $= (a^2 + 2ab + b^2) - c^2$

$= a^2 + 2ab + b^2 - c^2$

Alternate method (Direct multiplication):

$(a + b + c)(a + b – c) = a(a+b-c) + b(a+b-c) + c(a+b-c)$

$= (a \cdot a + a \cdot b - a \cdot c) + (b \cdot a + b \cdot b - b \cdot c) + (c \cdot a + c \cdot b - c \cdot c)$

$= (a^2 + ab - ac) + (ab + b^2 - bc) + (ac + bc - c^2)$

$= a^2 + ab - ac + ab + b^2 - bc + ac + bc - c^2$

Combine like terms:

$a^2$

$b^2$

$-c^2$

$ab + ab = 2ab$

$-ac + ac = 0$

$-bc + bc = 0$

So, the expression simplifies to $a^2 + b^2 - c^2 + 2ab$.

Rearranging, this is $a^2 + 2ab + b^2 - c^2$.

Thus, the simplified expression is $a^2 + 2ab + b^2 - c^2$.