Chapter 2 Polynomials

The degree of a polynomial is the highest power (exponent) of the variable in the polynomial.

---

(i) In the polynomial x⁵ - x⁴ + 3:

• The term x⁵ has a power of 5.
• The term -x⁴ has a power of 4.
• The term 3 (which is 3x⁰) has a power of 0.

The highest power among 5, 4, and 0 is 5. Therefore, the degree of the polynomial x⁵ - x⁴ + 3 is 5.

---

(ii) In the polynomial 2 - y² - y³ + y⁸:

• The term 2 (which is 2y⁰) has a power of 0.
• The term -y² has a power of 2.
• The term -y³ has a power of 3.
• The term y⁸ has a power of 8.

The highest power among 0, 2, 3, and 8 is 8. Therefore, the degree of the polynomial 2 - y² - y³ + y⁸ is 8.

---

(iii) The polynomial 2 is a constant polynomial. It can be written as 2x⁰ (or using any variable to the power of 0).

The highest power of the variable is 0. Therefore, the degree of the polynomial 2 is 0.

A polynomial in one variable has only one variable, and the exponents of the variable in each term must be non-negative integers (0, 1, 2, 3, ...).

---

(i) Yes, 4x² − 3x + 7 is a polynomial in one variable (x). All exponents (2, 1, and 0 for the constant term) are non-negative integers.

---

(ii) Yes, y² + $\sqrt{2}$ is a polynomial in one variable (y). The exponents (2 and 0 for the constant term $\sqrt{2}$) are non-negative integers.

---

(iii) No, 3$\sqrt{t}$ + t$\sqrt{2}$ is not a polynomial. The term 3$\sqrt{t}$ can be written as $3t^{\frac{1}{2}}$. The exponent $\frac{1}{2}$ is not an integer.

---

(iv) No, y + $\frac{2}{y}$ is not a polynomial. The term $\frac{2}{y}$ can be written as 2y^-1. The exponent -1 is not a non-negative integer.

---

(v) No, x¹⁰ + y³ + t⁵⁰ is not a polynomial in one variable. It is a polynomial, but it involves three variables (x, y, and t).

The coefficient is the numerical factor multiplying the specified variable term.

---

(i) In 2 + x² + x, the term with x² is x², which is equivalent to 1x². The coefficient of x² is 1.

(ii) In 2 − x² + x³, the term with x² is -x², which is equivalent to -1x². The coefficient of x² is -1.

(iii) In $\frac{\pi}{2}$ x² + x, the term with x² is $\frac{\pi}{2}$x². The coefficient of x² is $\frac{\pi}{2}$.

(iv) In $\sqrt{2}$ x − 1, there is no term containing x². This is equivalent to having 0x². The coefficient of x² is 0.

A binomial has two terms. A monomial has one term. The degree is the highest power of the variable.

Example of a binomial of degree 35: x³⁵ + 5 (Other examples: 3y³⁵ - y², 7z³⁵ - 10)

Example of a monomial of degree 100: y¹⁰⁰ (Other examples: -8x¹⁰⁰, $\sqrt{3}$t¹⁰⁰)

The degree is the highest power of the variable in the polynomial.

---

(i) In 5x³ + 4x² + 7x (which is 7x¹), the powers are 3, 2, and 1. The highest power is 3. The degree is 3.

(ii) In 4 − y², the powers are 0 (for 4) and 2. The highest power is 2. The degree is 2.

(iii) In 5t − $\sqrt{7}$ (which is 5t¹ − $\sqrt{7}$t⁰), the powers are 1 and 0. The highest power is 1. The degree is 1.

(iv) The polynomial 3 is a constant polynomial (3x⁰). The highest power is 0. The degree is 0.

• A polynomial of degree 1 is called linear.
• A polynomial of degree 2 is called quadratic.
• A polynomial of degree 3 is called cubic.

---

(i) x² + x: The degree is 2. It is quadratic.

(ii) x − x³: The degree is 3. It is cubic.

(iii) y + y² + 4: The degree is 2. It is quadratic.

(iv) 1 + x: The degree is 1 (for x¹). It is linear.

(v) 3t: The degree is 1 (for t¹). It is linear.

(vi) r²: The degree is 2. It is quadratic.

(vii) 7x³: The degree is 3. It is cubic.

(i) We have p(x) = 5x² − 3x + 7.

Substitute x = 1 in the polynomial:

p(1) = 5(1)² − 3(1) + 7

p(1) = 5(1) − 3 + 7

p(1) = 5 − 3 + 7

p(1) = 9

So, the value of p(x) at x = 1 is 9.

---

(ii) We have q(y) = 3y³ − 4y + $\sqrt{11}$.

Substitute y = 2 in the polynomial:

q(2) = 3(2)³ − 4(2) + $\sqrt{11}$

q(2) = 3(8) − 8 + $\sqrt{11}$

q(2) = 24 − 8 + $\sqrt{11}$

q(2) = 16 + $\sqrt{11}$

So, the value of q(y) at y = 2 is 16 + $\sqrt{11}$.

---

(iii) We have p(t) = 4t⁴ + 5t³ − t² + 6.

Substitute t = a in the polynomial:

p(a) = 4(a)⁴ + 5(a)³ − (a)² + 6

p(a) = 4a⁴ + 5a³ − a² + 6

So, the value of p(t) at t = a is 4a⁴ + 5a³ − a² + 6.

Let the polynomial be p(x) = x + 2.

A number is a zero of the polynomial if the value of the polynomial at that number is 0.

---

Check for x = -2:

p(-2) = (-2) + 2 = 0

Since p(-2) = 0, -2 is a zero of the polynomial x + 2.

---

Check for x = 2:

p(2) = (2) + 2 = 4

Since p(2) ≠ 0, 2 is not a zero of the polynomial x + 2.

To find a zero of the polynomial p(x) = 2x + 1, we need to find the value of x for which we will take p(x) = 0. So,

2x + 1 = 0

2x = -1

x = $-\frac{1}{2}$

Therefore, $-\frac{1}{2}$ is a zero of the polynomial p(x) = 2x + 1.

Verification:

p$\left( -\frac{1}{2} \right)$ = 2$\left( -\frac{1}{2} \right)$ + 1 = -1 + 1 = 0

Let the polynomial be p(x) = x² – 2x.

---

Check for x = 2:

p(2) = (2)² – 2(2)

p(2) = 4 – 4 = 0

Since p(2) = 0, 2 is a zero of the polynomial x² – 2x.

---

Check for x = 0:

p(0) = (0)² – 2(0)

p(0) = 0 – 0 = 0

Since p(0) = 0, 0 is a zero of the polynomial x² – 2x.

Let the polynomial be p(x) = 5x – 4x² + 3.

---

(i) At x = 0:

p(0) = 5(0) – 4(0)² + 3

p(0) = 0 – 0 + 3

p(0) = 3

---

(ii) At x = −1:

p(−1) = 5(−1) – 4(−1)² + 3

p(−1) = −5 – 4(1) + 3

p(−1) = −5 – 4 + 3 = −9 + 3

p(−1) = −6

---

(iii) At x = 2:

p(2) = 5(2) – 4(2)² + 3

p(2) = 10 – 4(4) + 3

p(2) = 10 – 16 + 3 = −6 + 3

p(2) = −3

(i) p(y) = y² − y + 1

• p(0) = (0)² − (0) + 1 = 1
• p(1) = (1)² − (1) + 1 = 1 - 1 + 1 = 1
• p(2) = (2)² − (2) + 1 = 4 - 2 + 1 = 3

---

(ii) p(t) = 2 + t + 2t² − t³

• p(0) = 2 + (0) + 2(0)² − (0)³ = 2
• p(1) = 2 + (1) + 2(1)² − (1)³ = 2 + 1 + 2 - 1 = 4
• p(2) = 2 + (2) + 2(2)² − (2)³ = 2 + 2 + 2(4) - 8 = 4 + 8 - 8 = 4

---

(iii) p(x) = x³

• p(0) = (0)³ = 0
• p(1) = (1)³ = 1
• p(2) = (2)³ = 8

---

(iv) p(x) = (x − 1) (x + 1)

• p(0) = (0 − 1)(0 + 1) = (−1)(1) = −1
• p(1) = (1 − 1)(1 + 1) = (0)(2) = 0
• p(2) = (2 − 1)(2 + 1) = (1)(3) = 3

(Alternatively, p(x) = x² - 1)

A value is a zero if p(value) = 0.

(i) p(x) = 3x + 1, x = $−\frac{1}{3}$

p($−\frac{1}{3}$) = 3($−\frac{1}{3}$) + 1 = -1 + 1 = 0.

Yes, $−\frac{1}{3}$ is a zero.

---

(ii) p(x) = 5x − $\pi$, x = $\frac{4}{5}$

p($\frac{4}{5}$) = 5($\frac{4}{5}$) − $\pi$ = 4 − $\pi$.

Since 4 - $\pi$ ≠ 0, No, $\frac{4}{5}$ is not a zero.

---

(iii) p(x) = x² − 1, x = 1, −1

p(1) = (1)² − 1 = 1 − 1 = 0.

Yes, 1 is a zero.

p(−1) = (−1)² − 1 = 1 − 1 = 0.

Yes, −1 is a zero.

---

(iv) p(x) = (x + 1) (x − 2), x = −1, 2

p(−1) = (−1 + 1)(−1 − 2) = (0)(−3) = 0.

Yes, −1 is a zero.

p(2) = (2 + 1)(2 − 2) = (3)(0) = 0.

Yes, 2 is a zero.

---

(v) p(x) = x², x = 0

p(0) = (0)² = 0.

Yes, 0 is a zero.

---

(vi) p(x) = lx + m, x = $-\frac{m}{l}$

p($-\frac{m}{l}$) = l($-\frac{m}{l}$) + m = -m + m = 0.

Yes, $-\frac{m}{l}$ is a zero.

---

(vii) p(x) = 3x² − 1, x = $-\frac{1}{\sqrt{3}}$ , $\frac{2}{\sqrt{3}}$

p($-\frac{1}{\sqrt{3}}$) = 3($-\frac{1}{\sqrt{3}}$)² − 1 = 3($\frac{1}{3}$) − 1 = 1 − 1 = 0.

Yes, $-\frac{1}{\sqrt{3}}$ is a zero.

p($\frac{2}{\sqrt{3}}$) = 3($\frac{2}{\sqrt{3}}$)² − 1 = 3($\frac{4}{3}$) − 1 = 4 − 1 = 3.

Since 3 ≠ 0, No, $\frac{2}{\sqrt{3}}$ is not a zero.

---

(viii) p(x) = 2x + 1, x = $\frac{1}{2}$

p($\frac{1}{2}$) = 2($\frac{1}{2}$) + 1 = 1 + 1 = 2.

Since 2 ≠ 0, No, $\frac{1}{2}$ is not a zero.

To find the zero, set p(x) = 0 and solve for x.

(i) p(x) = x + 5

x + 5 = 0

⇒ x = -5

---

(ii) p(x) = x − 5

x − 5 = 0

⇒ x = 5

---

(iii) p(x) = 2x + 5

2x + 5 = 0

⇒ 2x = -5

⇒ x = $-\frac{5}{2}$

---

(iv) p(x) = 3x − 2

3x − 2 = 0

⇒ 3x = 2

⇒ x = $\frac{2}{3}$

---

(v) p(x) = 3x

3x = 0

⇒ x = 0

---

(vi) p(x) = ax, a ≠ 0

ax = 0.

Since a ≠ 0, we must have x = 0.

---

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

cx + d = 0

⇒ cx = -d.

Since c ≠ 0, x = $-\frac{d}{c}$.

Solution:

We need to divide the polynomial $p(x) = x + 3x^2 - 1$ by $g(x) = 1 + x$.

First, we write both the dividend and the divisor in the standard form, which means arranging the terms in descending order of their powers.

Dividend: $p(x) = 3x^2 + x - 1$

Divisor: $g(x) = x + 1$

Now we perform the long division:

$\begin{array}{r} 3x-2\phantom{2x+3)} \\ x+1{\overline{\smash{\big)}\,3x^2+x-1\phantom{)}}} \\ \underline{-~\phantom{(}(3x^2+3x)\phantom{-b)}} \\ 0-2x-1\phantom{)} \\ \underline{-~\phantom{()}(-2x-2)} \\ 0+1\phantom{)} \end{array}$

Explanation of the steps:

Step 1: Divide the first term of the dividend ($3x^2$) by the first term of the divisor ($x$).

$\frac{3x^2}{x} = 3x$. This is the first term of the quotient.

Step 2: Multiply the divisor ($x+1$) by this term ($3x$).

$3x(x+1) = 3x^2 + 3x$.

Step 3: Subtract this result from the dividend.

$(3x^2 + x - 1) - (3x^2 + 3x) = -2x - 1$.

Step 4: Bring down the next term. The new dividend is $-2x - 1$.

Step 5: Repeat the process. Divide the first term of the new dividend ($-2x$) by the first term of the divisor ($x$).

$\frac{-2x}{x} = -2$. This is the second term of the quotient.

Step 6: Multiply the divisor ($x+1$) by this term ($-2$).

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

Step 7: Subtract this result from the new dividend.

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

The degree of the remainder (1) is 0, which is less than the degree of the divisor ($x+1$), which is 1. So, we stop here.

From the division, we have:

Quotient = $3x - 2$

Remainder = 1

Solution:

We need to divide $3x^4 - 4x^3 - 3x - 1$ by $x - 1$.

The dividend is $3x^4 - 4x^3 - 3x - 1$. Notice that the term with $x^2$ is missing. To make the long division process easier, we can write the dividend as $3x^4 - 4x^3 + 0x^2 - 3x - 1$.

The divisor is $x - 1$.

Now we perform the long division:

$\begin{array}{r} 3x^3-x^2-x-4\phantom{2x+3)} \\ x-1{\overline{\smash{\big)}\,3x^4-4x^3+0x^2-3x-1\phantom{)}}} \\ \underline{-~\phantom{(}(3x^4-3x^3)\phantom{-.....-b)}} \\ 0-x^3+0x^2\phantom{...-1} \\ \underline{-~\phantom{()}(-x^3+x^2)\phantom{-..-1}} \\ 0-x^2-3x\phantom{..} \\ \underline{-~\phantom{()}(-x^2+x)\phantom{...}} \\ 0-4x-1\phantom{)} \\ \underline{-~\phantom{()}(-4x+4)\phantom{}} \\ 0-5\phantom{)} \end{array}$

From the division, we find:

Quotient = $3x^3 - x^2 - x - 4$

Remainder = -5

We can solve this problem in two ways: by long division and by using the Remainder Theorem.

---

Method 1: Long Division

We need to divide $p(x) = x^3 + 1$ by $x + 1$. We can write the dividend as $x^3 + 0x^2 + 0x + 1$ to include the missing terms.

$\begin{array}{r} x^2-x+1\phantom{2x+3)} \\ x+1{\overline{\smash{\big)}\,x^3+0x^2+0x+1\phantom{)}}} \\ \underline{-~\phantom{(}(x^3+x^2)\phantom{-...-b)}} \\ 0-x^2+0x\phantom{-..1} \\ \underline{-~\phantom{()}(-x^2-x)\phantom{-..1}} \\ 0+x+1\phantom{)} \\ \underline{-~\phantom{()}(x+1)\phantom{}} \\ 0\phantom{)} \end{array}$

The remainder from the long division is 0.

---

Method 2: Using the Remainder Theorem

The Remainder Theorem states that if a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, the remainder is equal to $p(a)$.

In this problem:

The polynomial is $p(x) = x^3 + 1$.

The divisor is $x + 1$. To find the value of 'a', we set the divisor to zero:

$x + 1 = 0 \implies x = -1$

So, $a = -1$.

According to the theorem, the remainder is $p(-1)$. We substitute $x = -1$ into the polynomial $p(x)$:

$p(-1) = (-1)^3 + 1$

$p(-1) = -1 + 1$

$p(-1) = 0$

The remainder is 0.

Both methods give the same result.

Solution:

We will use the Remainder Theorem to find the remainder without performing long division.

The Remainder Theorem states that when a polynomial $p(x)$ is divided by $(x - a)$, the remainder is $p(a)$.

Here, the polynomial (dividend) is:

$p(x) = x^4 + x^3 - 2x^2 + x + 1$

The divisor is:

$x - 1$

To find the value of 'a', we set the divisor to zero:

$x - 1 = 0 \implies x = 1$

So, $a = 1$.

Now, we need to calculate the value of the polynomial at $x = 1$, which is $p(1)$.

$p(1) = (1)^4 + (1)^3 - 2(1)^2 + (1) + 1$

$p(1) = 1 + 1 - 2(1) + 1 + 1$

$p(1) = 1 + 1 - 2 + 1 + 1$

$p(1) = 2 - 2 + 2$

$p(1) = 2$

Therefore, the remainder when $x^4 + x^3 - 2x^2 + x + 1$ is divided by $x - 1$ is 2.

Solution:

To check if the polynomial $q(t)$ is a multiple of $2t+1$, we need to determine if $2t+1$ is a factor of $q(t)$.

According to the Factor Theorem (a special case of the Remainder Theorem), a polynomial $p(x)$ has a factor $(x-a)$ if and only if the remainder of the division is zero, which means $p(a) = 0$.

Here, the polynomial is $q(t) = 4t^3 + 4t^2 - t - 1$.

The potential factor is $2t + 1$.

First, we find the zero of the linear polynomial $2t + 1$ by setting it to zero:

$2t + 1 = 0$

$2t = -1$

$t = -\frac{1}{2}$

Now, we substitute this value of $t$ into the polynomial $q(t)$. If the result is 0, then $2t+1$ is a factor. The result we get is the remainder.

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

$q\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{8}\right) + 4\left(\frac{1}{4}\right) + \frac{1}{2} - 1$

$q\left(-\frac{1}{2}\right) = -\frac{4}{8} + \frac{4}{4} + \frac{1}{2} - 1$

$q\left(-\frac{1}{2}\right) = -\frac{1}{2} + 1 + \frac{1}{2} - 1$

$q\left(-\frac{1}{2}\right) = \left(-\frac{1}{2} + \frac{1}{2}\right) + (1 - 1)$

$q\left(-\frac{1}{2}\right) = 0 + 0 = 0$

Since the remainder $q\left(-\frac{1}{2}\right)$ is 0, this means that $(2t+1)$ is a factor of $q(t)$.

Therefore, yes, the polynomial $q(t) = 4t^3 + 4t^2 - t - 1$ is a multiple of $2t + 1$.

To find the remainder in each case, we will use the Remainder Theorem. The theorem states that if a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, the remainder is $p(a)$.

The given polynomial is $p(x) = x^3 + 3x^2 + 3x + 1$.

---

(i) Divided by (x + 1)

First, we find the zero of the divisor $(x+1)$ by setting it to zero:

$x + 1 = 0 \implies x = -1$

According to the Remainder Theorem, the remainder is $p(-1)$.

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

$p(-1) = -1 + 3(1) - 3 + 1$

$p(-1) = -1 + 3 - 3 + 1$

$p(-1) = 0$

Therefore, the remainder is 0.

---

(ii) Divided by $x - \frac{1}{2}$

First, we find the zero of the divisor $(x - \frac{1}{2})$:

$x - \frac{1}{2} = 0 \implies x = \frac{1}{2}$

The remainder is $p(\frac{1}{2})$.

$p\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) + 1$

$p\left(\frac{1}{2}\right) = \frac{1}{8} + 3\left(\frac{1}{4}\right) + \frac{3}{2} + 1$

$p\left(\frac{1}{2}\right) = \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1$

$p\left(\frac{1}{2}\right) = \frac{1 + 6 + 12 + 8}{8} = \frac{27}{8}$

Therefore, the remainder is $\frac{27}{8}$.

---

(iii) Divided by x

First, we find the zero of the divisor $x$:

$x = 0$

The remainder is $p(0)$.

$p(0) = (0)^3 + 3(0)^2 + 3(0) + 1$

$p(0) = 0 + 0 + 0 + 1$

$p(0) = 1$

Therefore, the remainder is 1.

---

(iv) Divided by x + π

First, we find the zero of the divisor $(x + \pi)$:

$x + \pi = 0 \implies x = -\pi$

The remainder is $p(-\pi)$.

$p(-\pi) = (-\pi)^3 + 3(-\pi)^2 + 3(-\pi) + 1$

$p(-\pi) = -\pi^3 + 3\pi^2 - 3\pi + 1$

Therefore, the remainder is $-\pi^3 + 3\pi^2 - 3\pi + 1$.

---

(v) Divided by 5 + 2x

First, we find the zero of the divisor $(5 + 2x)$:

$5 + 2x = 0 \implies 2x = -5 \implies x = -\frac{5}{2}$

The remainder is $p(-\frac{5}{2})$.

$p\left(-\frac{5}{2}\right) = \left(-\frac{5}{2}\right)^3 + 3\left(-\frac{5}{2}\right)^2 + 3\left(-\frac{5}{2}\right) + 1$

$p\left(-\frac{5}{2}\right) = -\frac{125}{8} + 3\left(\frac{25}{4}\right) - \frac{15}{2} + 1$

$p\left(-\frac{5}{2}\right) = -\frac{125}{8} + \frac{75}{4} - \frac{15}{2} + 1$

$p\left(-\frac{5}{2}\right) = \frac{-125 + 150 - 60 + 8}{8}$

$p\left(-\frac{5}{2}\right) = \frac{25 - 60 + 8}{8} = \frac{-27}{8}$

Therefore, the remainder is $-\frac{27}{8}$.

Solution:

We need to find the remainder when the polynomial $p(x) = x^3 - ax^2 + 6x - a$ is divided by the linear polynomial $g(x) = x - a$.

We can use the Remainder Theorem for this purpose. The theorem states that if a polynomial $p(x)$ is divided by $(x - a)$, the remainder is equal to $p(a)$.

In this case, the polynomial is:

$p(x) = x^3 - ax^2 + 6x - a$

The divisor is already in the form $(x - a)$.

To find the remainder, we need to calculate the value of the polynomial $p(x)$ at $x=a$.

Substitute $x=a$ into the polynomial $p(x)$:

$p(a) = (a)^3 - a(a)^2 + 6(a) - a$

Now, we simplify the expression:

$p(a) = a^3 - a(a^2) + 6a - a$

$p(a) = a^3 - a^3 + 6a - a$

$p(a) = 0 + 5a$

$p(a) = 5a$

Therefore, the remainder when $x^3 - ax^2 + 6x - a$ is divided by $x - a$ is 5a.

Solution:

To check whether $7 + 3x$ is a factor of the polynomial $3x^3 + 7x$, we will use the Remainder Theorem.

The Remainder Theorem helps us find the remainder when a polynomial is divided by a linear expression. A direct consequence of this is the Factor Theorem, which states that a linear expression like $(ax+b)$ is a factor of a polynomial $p(x)$ if and only if the remainder of the division is zero.

Let the polynomial be $p(x) = 3x^3 + 7x$.

The potential factor is the divisor, $g(x) = 7 + 3x$.

---

Step 1: Find the zero of the divisor.

We set the divisor equal to zero and solve for $x$:

$7 + 3x = 0$

$3x = -7$

$x = -\frac{7}{3}$

---

Step 2: Apply the Remainder Theorem.

The remainder when $p(x)$ is divided by $7 + 3x$ is the value of the polynomial $p(x)$ at $x = -\frac{7}{3}$. We need to calculate $p\left(-\frac{7}{3}\right)$.

Substitute $x = -\frac{7}{3}$ into the polynomial $p(x)$:

$p\left(-\frac{7}{3}\right) = 3\left(-\frac{7}{3}\right)^3 + 7\left(-\frac{7}{3}\right)$

Now, we simplify the expression:

$p\left(-\frac{7}{3}\right) = 3\left(\frac{(-7)^3}{3^3}\right) + 7\left(-\frac{7}{3}\right)$

$p\left(-\frac{7}{3}\right) = 3\left(\frac{-343}{27}\right) - \frac{49}{3}$

$p\left(-\frac{7}{3}\right) = \frac{\cancel{3} \times (-343)}{\cancel{27}_9} - \frac{49}{3}$

$p\left(-\frac{7}{3}\right) = -\frac{343}{9} - \frac{49}{3}$

To subtract the fractions, we use a common denominator, which is 9.

$p\left(-\frac{7}{3}\right) = -\frac{343}{9} - \frac{49 \times 3}{3 \times 3}$

$p\left(-\frac{7}{3}\right) = -\frac{343}{9} - \frac{147}{9}$

$p\left(-\frac{7}{3}\right) = \frac{-343 - 147}{9}$

$p\left(-\frac{7}{3}\right) = -\frac{490}{9}$

---

Step 3: Conclusion.

The remainder is $-\frac{490}{9}$.

Since the remainder is not equal to 0, $7 + 3x$ is not a factor of the polynomial $3x^3 + 7x$.

Therefore, no, $7 + 3x$ is not a factor of $3x^3 + 7x$.

Solution:

To determine if $(x+2)$ is a factor of the given polynomials, we will use the Factor Theorem. This theorem states that $(x-a)$ is a factor of a polynomial $p(x)$ if and only if $p(a)=0$.

The potential factor is $x+2$. First, we find its zero:

$x+2=0 \implies x = -2$.

So, we need to check if the value of each polynomial is zero when $x=-2$.

---

Part 1: For the polynomial $p(x) = x^3 + 3x^2 + 5x + 6$

We substitute $x = -2$ into $p(x)$:

$p(-2) = (-2)^3 + 3(-2)^2 + 5(-2) + 6$

$p(-2) = -8 + 3(4) - 10 + 6$

$p(-2) = -8 + 12 - 10 + 6$

$p(-2) = 4 - 4$

$p(-2) = 0$

Since the remainder $p(-2)$ is 0, yes, $(x+2)$ is a factor of $x^3 + 3x^2 + 5x + 6$.

---

Part 2: For the polynomial $g(x) = 2x + 4$

We substitute $x = -2$ into $g(x)$:

$g(-2) = 2(-2) + 4$

$g(-2) = -4 + 4$

$g(-2) = 0$

Since the remainder $g(-2)$ is 0, yes, $(x+2)$ is a factor of $2x + 4$.

Alternatively for Part 2: We can factor the polynomial $2x+4$ directly by taking out the common factor, which is 2:

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

From this factored form, it is clear that $(x+2)$ is a factor.

Solution:

We are given that $(x - 1)$ is a factor of the polynomial $p(x) = 4x^3 + 3x^2 - 4x + k$.

According to the Factor Theorem, if $(x - a)$ is a factor of a polynomial $p(x)$, then the value of the polynomial at $x=a$ must be zero, i.e., $p(a) = 0$.

In this problem, the factor is $(x - 1)$. We find its zero by setting it to zero:

$x - 1 = 0 \implies x = 1$.

So, we have $a=1$.

Since $(x - 1)$ is a factor of $p(x)$, we must have $p(1) = 0$.

Now, let's substitute $x=1$ into the polynomial $p(x)$ and set the expression equal to zero:

$p(1) = 4(1)^3 + 3(1)^2 - 4(1) + k = 0$

Now we solve this equation for $k$:

$4(1) + 3(1) - 4 + k = 0$

$4 + 3 - 4 + k = 0$

$3 + k = 0$

$k = -3$

Therefore, the value of $k$ for which $(x-1)$ is a factor of the given polynomial is -3.

Method 1: By Splitting the Middle Term

The given quadratic polynomial is $p(x) = 6x^2 + 17x + 5$.

To factorize by splitting the middle term, we need to find two numbers whose sum is the coefficient of the middle term (17) and whose product is the product of the first and last coefficients ($6 \times 5 = 30$).

We are looking for two numbers, say 'a' and 'b', such that:

$a + b = 17$

$a \times b = 30$

Let's list the pairs of factors of 30: (1, 30), (2, 15), (3, 10), (5, 6).

The pair (2, 15) satisfies the condition, as $2 + 15 = 17$.

Now, we split the middle term $17x$ into $2x + 15x$:

$6x^2 + 17x + 5 = 6x^2 + 2x + 15x + 5$

Next, we group the terms and take out the common factors:

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

$= 2x(3x + 1) + 5(3x + 1)$

Now, we take out the common binomial factor $(3x+1)$:

$= (3x + 1)(2x + 5)$

So, the factors are $(3x + 1)$ and $(2x + 5)$.

---

Method 2: By Using the Factor Theorem

Let $p(x) = 6x^2 + 17x + 5$.

We can write $p(x)$ as $6(x^2 + \frac{17}{6}x + \frac{5}{6})$.

If $(x-a)$ and $(x-b)$ are the factors of $p(x)$, then 'a' and 'b' are the zeroes of the polynomial. The possible rational zeroes are of the form $\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}$.

Factors of the constant term (5) are $\pm 1, \pm 5$.

Factors of the leading coefficient (6) are $\pm 1, \pm 2, \pm 3, \pm 6$.

Possible zeroes are $\pm 1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{1}{6}, \pm \frac{5}{6}$.

Let's try some of these values in $p(x)$:

Try $x = -1/3$:

$p(-\frac{1}{3}) = 6(-\frac{1}{3})^2 + 17(-\frac{1}{3}) + 5 = 6(\frac{1}{9}) - \frac{17}{3} + 5 \ $$ = \frac{2}{3} - \frac{17}{3} + \frac{15}{3} = \frac{2-17+15}{3} = 0$.

Since $p(-\frac{1}{3}) = 0$, $(x - (-\frac{1}{3})) = (x + \frac{1}{3})$ is a factor. This implies $(3x+1)$ is a factor.

Try $x = -5/2$:

$p(-\frac{5}{2}) = 6(-\frac{5}{2})^2 + 17(-\frac{5}{2}) + 5 = 6(\frac{25}{4}) - \frac{85}{2} + 5 = \frac{75}{2} - \frac{85}{2} + \frac{10}{2} \ $$ = \frac{75-85+10}{2} = 0$.

Since $p(-\frac{5}{2}) = 0$, $(x - (-\frac{5}{2})) = (x + \frac{5}{2})$ is a factor. This implies $(2x+5)$ is a factor.

Since we have found two linear factors for a quadratic polynomial, the factorization is complete.

$6x^2 + 17x + 5 = (3x + 1)(2x + 5)$

Solution:

Let the given polynomial be $p(y) = y^2 - 5y + 6$.

We will use the Factor Theorem to find the factors of this polynomial. According to the theorem, if $(y-a)$ is a factor of $p(y)$, then $p(a) = 0$.

The possible values for 'a' (the zeroes of the polynomial) are the integer factors of the constant term, which is 6.

The factors of 6 are $\pm 1, \pm 2, \pm 3, \pm 6$.

We will test these values by substituting them into the polynomial $p(y)$ until we find a value that makes $p(y)=0$.

Step 1: Test y = 1

$p(1) = (1)^2 - 5(1) + 6 = 1 - 5 + 6 = 2$. Since $p(1) \neq 0$, $(y-1)$ is not a factor.

Step 2: Test y = 2

$p(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = 0$. Since $p(2) = 0$, $(y-2)$ is a factor of $p(y)$.

Step 3: Test y = 3

$p(3) = (3)^2 - 5(3) + 6 = 9 - 15 + 6 = 0$. Since $p(3) = 0$, $(y-3)$ is a factor of $p(y)$.

Since the given polynomial is a quadratic (degree 2), it can have at most two linear factors. We have found both factors.

Therefore, the factorization of $y^2 - 5y + 6$ is:

$(y-2)(y-3)$

We can verify this by multiplying the factors: $(y-2)(y-3) = y^2 - 3y - 2y + 6 = y^2 - 5y + 6$, which is the original polynomial.

Solution:

Let the given polynomial be $p(x) = x^3 - 23x^2 + 142x - 120$.

This is a cubic polynomial, so we expect to find three linear factors. We will start by using the Factor Theorem to find one factor, and then use polynomial division to find the remaining quadratic factor.

Step 1: Find one factor using the Factor Theorem.

The possible integer zeroes of $p(x)$ are the factors of the constant term, -120. The factors are numerous, so we'll start with the simplest ones: $\pm 1, \pm 2, \pm 3, \dots$.

Let's try $x=1$:

$p(1) = (1)^3 - 23(1)^2 + 142(1) - 120 = 1 - 23 + 142 - 120 \ $$ = 143 - 143 = 0$.

Since $p(1) = 0$, we know that $(x - 1)$ is a factor of $p(x)$.

Step 2: Divide the polynomial by the known factor.

Now we divide $p(x)$ by $(x - 1)$ using long division to find the other factors.

$\begin{array}{r} x^2-22x+120\phantom{2x+3)} \\ x-1{\overline{\smash{\big)}\,x^3-23x^2+142x-120\phantom{)}}} \\ \underline{-~\phantom{(}(x^3-x^2)\phantom{-......-b)}} \\ 0-22x^2+142x\phantom{...-1} \\ \underline{-~\phantom{()}(-22x^2+22x)\phantom{-..-1}} \\ 0+120x-120\phantom{)} \\ \underline{-~\phantom{()}(120x-120)\phantom{}} \\ 0\phantom{)} \end{array}$

The division gives us a quotient of $x^2 - 22x + 120$ and a remainder of 0, as expected.

So, we can now write the polynomial as:

$p(x) = (x - 1)(x^2 - 22x + 120)$

Step 3: Factorize the resulting quadratic expression.

Now we need to factorize the quadratic $x^2 - 22x + 120$. We can do this by splitting the middle term.

We need two numbers that add up to -22 and multiply to 120. Let's think of factors of 120:

• $10 \times 12 = 120$, and $10 + 12 = 22$.

Since we need the sum to be -22, we can use -10 and -12. Their sum is $-10 + (-12) = -22$, and their product is $(-10) \times (-12) = 120$.

Now, we split the middle term:

$x^2 - 22x + 120 = x^2 - 10x - 12x + 120$

$= x(x - 10) - 12(x - 10)$

$= (x - 10)(x - 12)$

Step 4: Combine all the factors.

We combine the factor from Step 1 with the factors from Step 3 to get the complete factorization of the cubic polynomial.

$x^3 - 23x^2 + 142x - 120 = \bf{(x - 1)(x - 10)(x - 12)}$

To determine if $(x+1)$ is a factor of a polynomial $p(x)$, we use the Factor Theorem. It states that $(x+1)$ is a factor if and only if $p(-1)=0$.

The zero of the factor $(x+1)$ is found by setting $x+1=0$, which gives $x=-1$.

---

(i) $p(x) = x^3 + x^2 + x + 1$

$p(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0$.

Since the remainder is 0, $(x+1)$ is a factor of this polynomial.

---

(ii) $p(x) = x^4 + x^3 + x^2 + x + 1$

$p(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1 - 1 + 1 - 1 + 1 = 1$.

Since the remainder is 1 (not 0), $(x+1)$ is not a factor of this polynomial.

---

(iii) $p(x) = x^4 + 3x^3 + 3x^2 + x + 1$

$p(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1 \ $$ = 1 + 3(-1) + 3(1) - 1 + 1 = 1 - 3 + 3 - 1 + 1 = 1$.

Since the remainder is 1 (not 0), $(x+1)$ is not a factor of this polynomial.

---

(iv) $p(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$

$p(-1) = (-1)^3 - (-1)^2 - (2 + \sqrt{2})(-1) + \sqrt{2}$

$= -1 - 1 + (2 + \sqrt{2}) + \sqrt{2}$

$= -2 + 2 + \sqrt{2} + \sqrt{2} = 2\sqrt{2}$.

Since the remainder is $2\sqrt{2}$ (not 0), $(x+1)$ is not a factor of this polynomial.

We will use the Factor Theorem, which states that $g(x)$ is a factor of $p(x)$ if the value of $p(x)$ at the zero of $g(x)$ is 0.

---

(i) $p(x) = 2x^3 + x^2 - 2x - 1$, $g(x) = x + 1$

The zero of $g(x)$ is found by setting $x+1=0$, which gives $x=-1$.

Now, we evaluate $p(-1)$:

$p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1 = 2(-1) + 1 + 2 - 1 \ $$ = -2 + 1 + 2 - 1 = 0$.

Since $p(-1)=0$, yes, $g(x)$ is a factor of $p(x)$.

---

(ii) $p(x) = x^3 + 3x^2 + 3x + 1$, $g(x) = x + 2$

The zero of $g(x)$ is found by setting $x+2=0$, which gives $x=-2$.

Now, we evaluate $p(-2)$:

$p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1 = -8 + 3(4) - 6 + 1 \ $$ = -8 + 12 - 6 + 1 = 4 - 5 = -1$.

Since $p(-2) \neq 0$, no, $g(x)$ is not a factor of $p(x)$.

---

(iii) $p(x) = x^3 - 4x^2 + x + 6$, $g(x) = x - 3$

The zero of $g(x)$ is found by setting $x-3=0$, which gives $x=3$.

Now, we evaluate $p(3)$:

$p(3) = (3)^3 - 4(3)^2 + (3) + 6 = 27 - 4(9) + 3 + 6 = 27 - 36 + 9 \ $$ = 36 - 36 = 0$.

Since $p(3)=0$, yes, $g(x)$ is a factor of $p(x)$.

If $(x-1)$ is a factor of a polynomial $p(x)$, then according to the Factor Theorem, $p(1)=0$. We will use this condition to find the value of $k$ in each case.

---

(i) $p(x) = x^2 + x + k$

Set $p(1)=0$:

$(1)^2 + (1) + k = 0$

$1 + 1 + k = 0$

$2 + k = 0$

$k = -2$

---

(ii) $p(x) = 2x^2 + kx + \sqrt{2}$

Set $p(1)=0$:

$2(1)^2 + k(1) + \sqrt{2} = 0$

$2(1) + k + \sqrt{2} = 0$

$2 + k + \sqrt{2} = 0$

$k = -(2 + \sqrt{2})$

---

(iii) $p(x) = kx^2 - \sqrt{2}x + 1$

Set $p(1)=0$:

$k(1)^2 - \sqrt{2}(1) + 1 = 0$

$k - \sqrt{2} + 1 = 0$

$k = \sqrt{2} - 1$

---

(iv) $p(x) = kx^2 - 3x + k$

Set $p(1)=0$:

$k(1)^2 - 3(1) + k = 0$

$k - 3 + k = 0$

$2k - 3 = 0$

$2k = 3$

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

We will factorize the following quadratic polynomials by splitting the middle term.

---

(i) 12x² − 7x + 1

We need two numbers that add up to -7 and multiply to $12 \times 1 = 12$. The numbers are -3 and -4.

$12x^2 - 7x + 1 = 12x^2 - 3x - 4x + 1$

$= 3x(4x - 1) - 1(4x - 1)$

$= \bf{(4x - 1)(3x - 1)}$

---

(ii) 2x² + 7x + 3

We need two numbers that add up to 7 and multiply to $2 \times 3 = 6$. The numbers are 6 and 1.

$2x^2 + 7x + 3 = 2x^2 + 6x + x + 3$

$= 2x(x + 3) + 1(x + 3)$

$= \bf{(x + 3)(2x + 1)}$

---

(iii) 6x² + 5x − 6

We need two numbers that add up to 5 and multiply to $6 \times (-6) = -36$. The numbers are 9 and -4.

$6x^2 + 5x - 6 = 6x^2 + 9x - 4x - 6$

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

$= \bf{(2x + 3)(3x - 2)}$

---

(iv) 3x² − x − 4

We need two numbers that add up to -1 and multiply to $3 \times (-4) = -12$. The numbers are -4 and 3.

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

$= x(3x - 4) + 1(3x - 4)$

$= \bf{(3x - 4)(x + 1)}$

We will factorize the following cubic polynomials by first finding one linear factor using the Factor Theorem, then using long division to find the remaining quadratic factor, and finally factorizing the quadratic.

---

(i) $p(x) = x^3 - 2x^2 - x + 2$

Step 1: Use the Factor Theorem to find one factor.

The possible integer zeroes of $p(x)$ are the factors of the constant term, 2. The factors are $\pm1, \pm2$.

Let's test $x=1$:

$p(1) = (1)^3 - 2(1)^2 - (1) + 2 = 1 - 2 - 1 + 2 = 0$.

Since $p(1) = 0$, by the Factor Theorem, $(x-1)$ is a factor of $p(x)$.

Step 2: Use long division to find the quadratic factor.

We divide $p(x)$ by $(x-1)$:

$\begin{array}{r} x^2-x-2\phantom{2x+3)} \\ x-1{\overline{\smash{\big)}\,x^3-2x^2-x+2\phantom{)}}} \\ \underline{-~\phantom{(}(x^3-x^2)\phantom{-...-b)}} \\ 0-x^2-x\phantom{...2} \\ \underline{-~\phantom{()}(-x^2+x)\phantom{-..2}} \\ 0-2x+2\phantom{)} \\ \underline{-~\phantom{()}(-2x+2)\phantom{}} \\ 0\phantom{)} \end{array}$

The quotient is $x^2 - x - 2$.

Step 3: Factorize the quadratic quotient.

$x^2 - x - 2 = x^2 - 2x + x - 2 = x(x-2) + 1(x-2) \ $$ = (x-2)(x+1)$.

Step 4: Combine the factors.

The complete factorization is: $\bf{(x-1)(x+1)(x-2)}$.

---

(ii) $p(x) = x^3 - 3x^2 - 9x - 5$

Step 1: Use the Factor Theorem to find one factor.

The possible integer zeroes are the factors of -5: $\pm1, \pm5$.

Let's test $x=-1$:

$p(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3 + 9 - 5 = 0$.

Since $p(-1) = 0$, $(x+1)$ is a factor.

Step 2: Use long division.

We divide $p(x)$ by $(x+1)$:

$\begin{array}{r} x^2-4x-5\phantom{2x+3)} \\ x+1{\overline{\smash{\big)}\,x^3-3x^2-9x-5\phantom{)}}} \\ \underline{-~\phantom{(}(x^3+x^2)\phantom{-...-b)}} \\ 0-4x^2-9x\phantom{-..5} \\ \underline{-~\phantom{()}(-4x^2-4x)\phantom{-..5}} \\ 0-5x-5\phantom{)} \\ \underline{-~\phantom{()}(-5x-5)\phantom{}} \\ 0\phantom{)} \end{array}$

The quotient is $x^2 - 4x - 5$.

Step 3: Factorize the quadratic quotient.

$x^2 - 4x - 5 = x^2 - 5x + x - 5 = x(x-5) + 1(x-5) \ $$ = (x-5)(x+1)$.

Step 4: Combine the factors.

The complete factorization is: $\bf{(x+1)(x+1)(x-5)}$ or $\bf{(x+1)^2(x-5)}$.

---

(iii) $p(x) = x^3 + 13x^2 + 32x + 20$

Step 1: Use the Factor Theorem.

Possible integer zeroes are factors of 20. Since all coefficients are positive, we only need to test negative factors.

Let's test $x=-1$:

$p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20 = -1 + 13 - 32 + 20 = 0$.

Since $p(-1) = 0$, $(x+1)$ is a factor.

Step 2: Use long division.

We divide $p(x)$ by $(x+1)$:

$\begin{array}{r} x^2+12x+20\phantom{2x+3)} \\ x+1{\overline{\smash{\big)}\,x^3+13x^2+32x+20\phantom{)}}} \\ \underline{-~\phantom{(}(x^3+x^2)\phantom{-......-b)}} \\ 0+12x^2+32x\phantom{...-1} \\ \underline{-~\phantom{()}(12x^2+12x)\phantom{-..-1}} \\ 0+20x+20\phantom{)} \\ \underline{-~\phantom{()}(20x+20)\phantom{}} \\ 0\phantom{)} \end{array}$

The quotient is $x^2 + 12x + 20$.

Step 3: Factorize the quadratic quotient.

$x^2 + 12x + 20 = x^2 + 10x + 2x + 20 = x(x+10) + 2(x+10) \ $$ = (x+10)(x+2)$.

Step 4: Combine the factors.

The complete factorization is: $\bf{(x+1)(x+2)(x+10)}$.

---

(iv) $p(y) = 2y^3 + y^2 - 2y - 1$

Step 1: Use the Factor Theorem.

Possible rational zeroes are factors of -1 divided by factors of 2: $\pm1, \pm\frac{1}{2}$.

Let's test $y=1$:

$p(1) = 2(1)^3 + (1)^2 - 2(1) - 1 = 2 + 1 - 2 - 1 = 0$.

Since $p(1) = 0$, $(y-1)$ is a factor.

Step 2: Use long division.

We divide $p(y)$ by $(y-1)$:

$\begin{array}{r} 2y^2+3y+1\phantom{2x+3)} \\ y-1{\overline{\smash{\big)}\,2y^3+y^2-2y-1\phantom{)}}} \\ \underline{-~\phantom{(}(2y^3-2y^2)\phantom{-...-b)}} \\ 0+3y^2-2y\phantom{-..1} \\ \underline{-~\phantom{()}(3y^2-3y)\phantom{-..1}} \\ 0+y-1\phantom{)} \\ \underline{-~\phantom{()}(y-1)\phantom{}} \\ 0\phantom{)} \end{array}$

The quotient is $2y^2 + 3y + 1$.

Step 3: Factorize the quadratic quotient.

$2y^2 + 3y + 1 = 2y^2 + 2y + y + 1 = 2y(y+1) + 1(y+1) \ $$ = (y+1)(2y+1)$.

Step 4: Combine the factors.

The complete factorization is: $\bf{(y-1)(y+1)(2y+1)}$.

(i) (x + 3) (x + 3)

This is of the form $(a+b)(a+b) = (a+b)^2$.

We use the identity: $(a+b)^2 = a^2 + 2ab + b^2$.

Here, $a=x$ and $b=3$.

$(x+3)^2 = (x)^2 + 2(x)(3) + (3)^2$

$= x^2 + 6x + 9$

So, $(x+3)(x+3) = x^2 + 6x + 9$.

---

(ii) (x – 3) (x + 5)

This is of the form $(x+a)(x+b)$.

We use the identity: $(x+a)(x+b) = x^2 + (a+b)x + ab$.

Here, $x=x$, $a=-3$, and $b=5$.

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

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

$= x^2 + 2x - 15$

So, $(x-3)(x+5) = x^2 + 2x - 15$.

Solution:

We need to evaluate $105 \times 106$ using an algebraic identity.

We can write the numbers 105 and 106 in a more convenient form:

$105 = 100 + 5$

$106 = 100 + 6$

So, the product becomes $(100 + 5)(100 + 6)$.

This expression is in the form of the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.

Here, $x=100$, $a=5$, and $b=6$.

Applying the identity:

$(100+5)(100+6) = (100)^2 + (5+6)(100) + (5)(6)$

$= 10000 + (11)(100) + 30$

$= 10000 + 1100 + 30$

$= 11130$

Therefore, $105 \times 106 = 11130$.

(i) 49a² + 70ab + 25b²

We can see that the first and last terms are perfect squares:

$49a^2 = (7a)^2$

$25b^2 = (5b)^2$

The middle term is $70ab = 2 \times 35ab = 2 \times (7a) \times (5b)$.

The expression is in the form $x^2 + 2xy + y^2$, which is the expansion of the identity $(x+y)^2$.

Here, $x=7a$ and $y=5b$.

$49a^2 + 70ab + 25b^2 = (7a)^2 + 2(7a)(5b) + (5b)^2 = \bf{(7a + 5b)^2}$.

The factors are $(7a+5b)$ and $(7a+5b)$.

---

(ii) $\frac{25}{4}x^2 - \frac{y^2}{9}$

This expression is a difference of two squares. We can write each term as a square:

$\frac{25}{4}x^2 = \left(\frac{5}{2}x\right)^2$

$\frac{y^2}{9} = \left(\frac{y}{3}\right)^2$

The expression is in the form $a^2 - b^2$, which corresponds to the identity $(a-b)(a+b)$.

Here, $a = \frac{5}{2}x$ and $b = \frac{y}{3}$.

$\frac{25}{4}x^2 - \frac{y^2}{9} = \left(\frac{5}{2}x\right)^2 - \left(\frac{y}{3}\right)^2 = \bf{\left(\frac{5}{2}x - \frac{y}{3}\right)\left(\frac{5}{2}x + \frac{y}{3}\right)}$.

Solution:

We need to expand $(3a + 4b + 5c)^2$.

This expression is in the form of the identity $(x+y+z)^2$.

The identity for the square of a trinomial is:

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$

In this problem, we have:

$x = 3a$

$y = 4b$

$z = 5c$

Now, we substitute these values into the identity:

$(3a+4b+5c)^2 = (3a)^2 + (4b)^2 + (5c)^2 + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)$

Now, we simplify each term:

$= 9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ca$

Therefore, the expanded form is $9a^2 + 16b^2 + 25c^2 + 24ab + 40bc + 30ca$.

Solution:

We need to expand $(4a - 2b - 3c)^2$.

We will use the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$.

To fit our expression to this identity, we can write it as $(4a + (-2b) + (-3c))^2$.

Here, we have:

$x = 4a$

$y = -2b$

$z = -3c$

Now, substitute these values into the identity:

$(4a - 2b - 3c)^2 = (4a)^2 + (-2b)^2 + (-3c)^2 + 2(4a)(-2b) \ $$ + 2(-2b)(-3c) \ $$ + 2(-3c)(4a)$

Simplify each term:

$= 16a^2 + 4b^2 + 9c^2 - 16ab + 12bc - 24ca$

Therefore, the expanded form is $16a^2 + 4b^2 + 9c^2 - 16ab + 12bc - 24ca$.

Solution:

The given expression is $4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz$.

This expression looks like the expansion of the identity $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.

First, let's write the first three terms as squares:

$4x^2 = (2x)^2$

$y^2 = (y)^2$

$z^2 = (z)^2$

So, our terms $a, b, c$ could be $2x, y, z$. Let's check the remaining terms. Notice that the terms with 'y' in them ($-4xy$ and $-2yz$) are negative. This suggests that the 'y' term in our identity might be negative.

Let's try $a = 2x$, $b = -y$, and $c = z$.

Now, let's check the cross-product terms:

$2ab = 2(2x)(-y) = -4xy$ (Matches)

$2bc = 2(-y)(z) = -2yz$ (Matches)

$2ca = 2(z)(2x) = 4xz$ (Matches)

Since all the terms match the expansion of $(2x - y + z)^2$, this is the correct factorization.

Therefore, $4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz = \bf{(2x - y + z)^2}$.

The factors are $(2x-y+z)$ and $(2x-y+z)$.

(i) (3a + 4b)³

We use the identity: $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$.

Here, $x = 3a$ and $y = 4b$.

$(3a+4b)^3 = (3a)^3 + (4b)^3 + 3(3a)(4b)(3a+4b)$

$= 27a^3 + 64b^3 + 36ab(3a+4b)$

$= 27a^3 + 64b^3 + (36ab)(3a) + (36ab)(4b)$

$= \bf{27a^3 + 64b^3 + 108a^2b + 144ab^2}$

---

(ii) (5p – 3q)³

We use the identity: $(x-y)^3 = x^3 - y^3 - 3xy(x-y)$.

Here, $x = 5p$ and $y = 3q$.

$(5p-3q)^3 = (5p)^3 - (3q)^3 - 3(5p)(3q)(5p-3q)$

$= 125p^3 - 27q^3 - 45pq(5p-3q)$

$= 125p^3 - 27q^3 - (45pq)(5p) - (45pq)(-3q)$

$= \bf{125p^3 - 27q^3 - 225p^2q + 135pq^2}$

(i) (104)³

We can write 104 as $(100 + 4)$. So we need to calculate $(100+4)^3$.

Using the identity $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$:

Here, $x=100$ and $y=4$.

$(100+4)^3 = (100)^3 + (4)^3 + 3(100)(4)(100+4)$

$= 1,000,000 + 64 + 1200(104)$

$= 1,000,000 + 64 + 124,800$

$= \bf{1,124,864}$

---

(ii) (999)³

We can write 999 as $(1000 - 1)$. So we need to calculate $(1000-1)^3$.

Using the identity $(x-y)^3 = x^3 - y^3 - 3xy(x-y)$:

Here, $x=1000$ and $y=1$.

$(1000-1)^3 = (1000)^3 - (1)^3 - 3(1000)(1)(1000-1)$

$= 1,000,000,000 - 1 - 3000(999)$

$= 1,000,000,000 - 1 - 2,997,000$

$= 997,003,000 - 1$

$= \bf{997,002,999}$

Solution:

The given expression is $8x^3 + 27y^3 + 36x^2y + 54xy^2$.

This expression has four terms, with the first two terms being perfect cubes. This suggests it might be the expansion of a cubic identity like $(a+b)^3$.

The identity is: $(a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2$.

Let's check if our expression fits this form.

First, write the cubic terms as cubes:

$8x^3 = (2x)^3$

$27y^3 = (3y)^3$

So, it's possible that $a = 2x$ and $b = 3y$.

Now, let's check if the other two terms match the identity's cross-product terms.

The $3a^2b$ term should be:

$3(2x)^2(3y) = 3(4x^2)(3y) = 36x^2y$. This matches the third term of our expression.

The $3ab^2$ term should be:

$3(2x)(3y)^2 = 3(2x)(9y^2) = 54xy^2$. This matches the fourth term of our expression.

Since all the terms match the expansion of $(2x+3y)^3$, this is the correct factorization.

Therefore, $8x^3 + 27y^3 + 36x^2y + 54xy^2 = \bf{(2x + 3y)^3}$.

The factors are $(2x+3y)$, $(2x+3y)$, and $(2x+3y)$.

Solution:

The given expression is $8x^3 + y^3 + 27z^3 - 18xyz$.

This expression matches the form of the identity:

$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

First, we need to express the cubic terms in the given expression as cubes to identify $a, b,$ and $c$.

$8x^3 = (2x)^3$

$y^3 = (y)^3$

$27z^3 = (3z)^3$

So, we can set:

$a = 2x$

$b = y$

$c = 3z$

Now, let's check if the term $-18xyz$ matches the $-3abc$ part of the identity.

$-3abc = -3(2x)(y)(3z) = -18xyz$. This matches perfectly.

Now that we have identified $a, b,$ and $c$, we can substitute them into the factored form of the identity.

The factored form is $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

Substituting the values:

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

Now, we simplify the second bracket:

$= \bf{(2x + y + 3z)(4x^2 + y^2 + 9z^2 - 2xy - 3yz - 6zx)}$

(i) (x + 4) (x + 10)

Using the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$:

$(x+4)(x+10) = x^2 + (4+10)x + (4)(10) = \bf{x^2 + 14x + 40}$.

---

(ii) (x + 8) (x – 10)

Using the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$:

$(x+8)(x-10) = x^2 + (8-10)x + (8)(-10) = \bf{x^2 - 2x - 80}$.

---

(iii) (3x + 4) (3x – 5)

This is of the form $(Y+a)(Y+b)$ where $Y=3x$.

$(3x+4)(3x-5) = (3x)^2 + (4-5)(3x) + (4)(-5) = 9x^2 - 1(3x) - 20 \ $$ = \bf{9x^2 - 3x - 20}$.

---

(iv) (y² + $\frac{3}{2}$) (y² - $\frac{3}{2}$)

Using the identity $(a+b)(a-b) = a^2 - b^2$:

$(y^2 + \frac{3}{2})(y^2 - \frac{3}{2}) = (y^2)^2 - (\frac{3}{2})^2 = \bf{y^4 - \frac{9}{4}}$.

---

(v) (3 – 2x) (3 + 2x)

Using the identity $(a-b)(a+b) = a^2 - b^2$:

$(3-2x)(3+2x) = (3)^2 - (2x)^2 = \bf{9 - 4x^2}$.

(i) 103 × 107

We can write this as $(100+3)(100+7)$.

Using the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$:

$(100+3)(100+7) = (100)^2 + (3+7)(100) + (3)(7) \ $$ = 10000 + 10(100) + 21 = 10000 + 1000 + 21 = \bf{11021}$.

---

(ii) 95 × 96

We can write this as $(100-5)(100-4)$.

Using the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$:

$(100-5)(100-4) = (100)^2 + (-5-4)(100) + (-5)(-4) \ $$ = 10000 - 9(100) + 20 = 10000 - 900 + 20 = \bf{9120}$.

---

(iii) 104 × 96

We can write this as $(100+4)(100-4)$.

Using the identity $(a+b)(a-b) = a^2 - b^2$:

$(100+4)(100-4) = (100)^2 - (4)^2 = 10000 - 16 = \bf{9984}$.

(i) 9x² + 6xy + y²

This expression is in the form $a^2 + 2ab + b^2$, which is the expansion of $(a+b)^2$.

$9x^2 = (3x)^2$

$y^2 = (y)^2$

$6xy = 2(3x)(y)$

So, the factorization is $\bf{(3x + y)^2}$.

---

(ii) 4y² – 4y + 1

This expression is in the form $a^2 - 2ab + b^2$, which is the expansion of $(a-b)^2$.

$4y^2 = (2y)^2$

$1 = (1)^2$

$-4y = -2(2y)(1)$

So, the factorization is $\bf{(2y - 1)^2}$.

---

(iii) x² – $\frac{y^2}{100}$

This is a difference of two squares, $a^2 - b^2$, which factors to $(a-b)(a+b)$.

$x^2 = (x)^2$

$\frac{y^2}{100} = \left(\frac{y}{10}\right)^2$

So, the factorization is $\bf{\left(x - \frac{y}{10}\right)\left(x + \frac{y}{10}\right)}$.

We use the identity $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$.

---

(i) (x + 2y + 4z)²

$= (x)^2 + (2y)^2 + (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)$

$= \bf{x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx}$

---

(ii) (2x – y + z)²

$= (2x)^2 + (-y)^2 + (z)^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)$

$= \bf{4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx}$

---

(iii) (–2x + 3y + 2z)²

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

$= \bf{4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx}$

---

(iv) (3a – 7b – c)²

$= (3a)^2 + (-7b)^2 + (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a)$

$= \bf{9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ca}$

---

(v) (–2x + 5y – 3z)²

$= (-2x)^2 + (5y)^2 + (-3z)^2 + 2(-2x)(5y) + 2(5y)(-3z) \ $$ + 2(-3z)(-2x)$

$= \bf{4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12zx}$

---

(vi) $\left[ \frac{1}{4}a - \frac{1}{2}b + 1 \right]^2$

$= \left(\frac{1}{4}a\right)^2 + \left(-\frac{1}{2}b\right)^2 + (1)^2 + 2\left(\frac{1}{4}a\right)\left(-\frac{1}{2}b\right) + 2\left(-\frac{1}{2}b\right)(1) \ $$ + 2(1)\left(\frac{1}{4}a\right)$

$= \frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab - b + \frac{1}{2}a$

Rearranging, we get: $\bf{\frac{1}{16}a^2 + \frac{1}{4}b^2 + 1 - \frac{1}{4}ab + \frac{1}{2}a - b}$

The relevant identity is: a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²

(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz

We look for squares: 4x² = (2x)², 9y² = (3y)², 16z² = (4z)².

Now, let's check the cross terms. We have +12xy, -24yz, -16xz.

Since the yz and xz terms are negative, the term involving 'z' must be negative in the expansion. So, we consider 16z² as (-4z)².

Let's test with a = 2x, b = 3y, c = -4z:

a² = (2x)² = 4x²

b² = (3y)² = 9y²

c² = (-4z)² = 16z²

2ab = 2(2x)(3y) = 12xy

2bc = 2(3y)(-4z) = -24yz

2ca = 2(-4z)(2x) = -16xz

These terms match the given expression.

Therefore, the expression is equivalent to (2x + 3y + (-4z))².

= (2x + 3y - 4z)²

---

(ii) 2x² + y² + 8z² – $2\sqrt{2}$xy + $4\sqrt{2}$yz – 8xz

We look for squares: 2x² = ($\sqrt{2}$x)², y² = (y)², 8z² = ($\sqrt{8}$z)² = ($2\sqrt{2}$z)².

Now, let's check the cross terms: -$2\sqrt{2}$xy, +$4\sqrt{2}$yz, -8xz.

Since the xy and xz terms are negative, the term involving 'x' must be negative in the expansion. So, we consider 2x² as ($-\sqrt{2}$x)².

Let's test with a = $-\sqrt{2}$x, b = y, c = $2\sqrt{2}$z:

a² = ($-\sqrt{2}$x)² = 2x²

b² = (y)² = y²

c² = ($2\sqrt{2}$z)² = 8z²

2ab = 2($-\sqrt{2}$x)(y) = -$2\sqrt{2}$xy

2bc = 2(y)($2\sqrt{2}$z) = $4\sqrt{2}$yz

2ca = 2($2\sqrt{2}$z)($-\sqrt{2}$x) = -2(2)( $\sqrt{2}$ $\sqrt{2}$)xz = -8xz

These terms match the given expression.

Therefore, the expression is equivalent to ($-\sqrt{2}$x + y + $2\sqrt{2}$z)².

= ($-\sqrt{2}$x + y + $2\sqrt{2}$z)²

We will use the binomial cube expansion identities:

For sums: (a + b)³ = a³ + 3a²b + 3ab² + b³

For differences: (a - b)³ = a³ - 3a²b + 3ab² - b³

---

(i) Expand (2x + 1)³

Here, we use the identity (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = 2x and b = 1.

= (2x)³ + 3(2x)²(1) + 3(2x)(1)² + (1)³

Now, calculate the powers and products:

= (8x³) + 3(4x²)(1) + 3(2x)(1) + 1

Simplify the terms:

= 8x³ + 12x² + 6x + 1

---

(ii) Expand (2a – 3b)³

Here, we use the identity (a - b)³ = a³ - 3a²b + 3ab² - b³ with a = 2a and b = 3b.

= (2a)³ - 3(2a)²(3b) + 3(2a)(3b)² - (3b)³

Now, calculate the powers and products:

= (8a³) - 3(4a²)(3b) + 3(2a)(9b²) - (27b³)

Simplify the terms:

= 8a³ - 36a²b + 54ab² - 27b³

---

(iii) Expand $\left[ \frac{3}{2} x \;+\; 1\right]^3$

Here, we use the identity (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = $\frac{3}{2}x$ and b = 1.

= ($\frac{3}{2}$x)³ + 3($\frac{3}{2}$x)²(1) + 3($\frac{3}{2}$x)(1)² + (1)³

Now, calculate the powers and products:

= ($\frac{27}{8}$x³) + 3($\frac{9}{4}$x²)(1) + 3($\frac{3}{2}$x)(1) + 1

Simplify the terms:

= $\frac{27}{8}$x³ + $\frac{27}{4}$x² + $\frac{9}{2}$x + 1

---

(iv) Expand $\left[ x \;-\; \frac{2}{3} \right]^3$

Using the identity for $(a-b)^3$ with $a=x$ and $b=\frac{2}{3}y$:

$\left(x-\frac{2}{3}y\right)^3 = (x)^3 - 3(x)^2\left(\frac{2}{3}y\right) + 3(x)\left(\frac{2}{3}y\right)^2 - \left(\frac{2}{3}y\right)^3$

$= x^3 - \cancel{3}(x^2)\left(\frac{2}{\cancel{3}}y\right) + 3(x)\left(\frac{4}{9}y^2\right) - \frac{8}{27}y^3$

$= x^3 - 2x^2y + \frac{\cancel{3}(4)}{\cancel{9}_3}xy^2 - \frac{8}{27}y^3$

$= \bf{x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3}$

(i) (99)³ = (100 - 1)³

Using identity: (a - b)³ = a³ - 3a²b + 3ab² - b³

= (100)³ - 3(100)²(1) + 3(100)(1)² - (1)³

= 1000000 - 3(10000)(1) + 3(100)(1) - 1

= 1000000 - 30000 + 300 - 1

= 970000 + 300 - 1

= 970300 - 1

= 970299

---

(ii) (102)³ = (100 + 2)³

Using identity: (a + b)³ = a³ + 3a²b + 3ab² + b³

= (100)³ + 3(100)²(2) + 3(100)(2)² + (2)³

= 1000000 + 3(10000)(2) + 3(100)(4) + 8

= 1000000 + 60000 + 1200 + 8

= 1061208

---

(iii) (998)³ = (1000 - 2)³

Using identity: (a - b)³ = a³ - 3a²b + 3ab² - b³

= (1000)³ - 3(1000)²(2) + 3(1000)(2)² - (2)³

= 1000000000 - 3(1000000)(2) + 3(1000)(4) - 8

= 1000000000 - 6000000 + 12000 - 8

= 994000000 + 12000 - 8 = 994012000 - 8

= 994011992

We will use the cube expansion identities in reverse to factorise:

a³ + 3a²b + 3ab² + b³ = (a + b)³

a³ - 3a²b + 3ab² - b³ = (a - b)³

(i) 8a³ + b³ + 12a²b + 6ab²

We observe the expression has all positive terms, suggesting the form (a + b)³.

Let's identify the terms: a³ = 8a³ implies a = 2a. b³ = b³ implies b = b.

Rewrite the expression in the standard form a³ + 3a²b + 3ab² + b³:

= (2a)³ + 3(2a)²(b) + 3(2a)(b)² + (b)³

Check the middle terms:

3(2a)²(b) = 3(4a²)b = 12a²b (Matches).

3(2a)(b)² = 6ab² (Matches).

Since it matches the form (a + b)³ with a=2a and b=b:

= (2a + b)³

---

(ii) 8a³ – b³ – 12a²b + 6ab²

We observe mixed signs, suggesting the form (a - b)³ = a³ - 3a²b + 3ab² - b³.

Let's identify the terms: a³ = 8a³ implies a = 2a. -b³ = -b³ implies b = b.

Rewrite the expression in the standard form a³ - 3a²b + 3ab² - b³:

= (2a)³ - 3(2a)²(b) + 3(2a)(b)² - (b)³

Check the middle terms:

-3(2a)²(b) = -3(4a²)b = -12a²b (Matches).

+3(2a)(b)² = +6ab² (Matches).

Since it matches the form (a - b)³ with a=2a and b=b:

= (2a - b)³

---

(iii) 27 – 125a³ – 135a + 225a²

Rearrange to match the standard form a³ - 3a²b + 3ab² - b³: 27 + 225a² - 135a - 125a³

Identify the terms: a³ = 27 implies a = 3. -b³ = -125a³ implies b = 5a.

Rewrite the expression in the standard form (a - b)³:

= (3)³ - 3(3)²(5a) + 3(3)(5a)² - (5a)³

Check the middle terms:

-3(3)²(5a) = -3(9)(5a) = -27(5a) = -135a (Matches).

+3(3)(5a)² = +9(25a²) = +225a² (Matches).

Since it matches the form (a - b)³ with a=3 and b=5a:

= (3 - 5a)³

---

(iv) 64a³ – 27b³ – 144a²b + 108ab²

We observe mixed signs, suggesting the form (a - b)³ = a³ - 3a²b + 3ab² - b³.

Identify the terms: a³ = 64a³ implies a = 4a. -b³ = -27b³ implies b = 3b.

Rewrite the expression in the standard form a³ - 3a²b + 3ab² - b³:

= (4a)³ - 3(4a)²(3b) + 3(4a)(3b)² - (3b)³

Check the middle terms:

-3(4a)²(3b) = -3(16a²)(3b) = -9(16a²)b = -144a²b (Matches).

+3(4a)(3b)² = +3(4a)(9b²) = +12a(9b²) = +108ab² (Matches).

Since it matches the form (a - b)³ with a=4a and b=3b:

= (4a - 3b)³

---

(v) 27p³ – $\frac{1}{216}$ – $\frac{9}{2}$ p² + $\frac{1}{4}$ p

Rearrange to match the standard form a³ - 3a²b + 3ab² - b³: 27p³ - $\frac{9}{2}$p² + $\frac{1}{4}$p - $\frac{1}{216}$

Identify the terms: a³ = 27p³ implies a = 3p. -b³ = -$\frac{1}{216}$ implies b = $\frac{1}{6}$ (since 6³ = 216).

Rewrite the expression in the standard form (a - b)³:

= (3p)³ - 3(3p)²($\frac{1}{6}$) + 3(3p)($\frac{1}{6}$)² - ($\frac{1}{6}$)³

Check the middle terms:

-3a²b = -3(3p)²($\frac{1}{6}$) = -3(9p²)($\frac{1}{6}$) = -$\frac{27}{6}$p² = -$\frac{9}{2}$p² (Matches).

+3ab² = +3(3p)($\frac{1}{6}$)² = +3(3p)($\frac{1}{36}$) = +$\frac{9p}{36}$ = +$\frac{1}{4}$p (Matches).

The original expression is indeed equal to (3p)³ - ($\frac{1}{6}$)³ - $\frac{9}{2}$p² + $\frac{1}{4}$p.

Since it matches the form (a - b)³ with a=3p and b=1/6:

= (3p - $\frac{1}{6}$)³

(i) To verify x³ + y³ = (x + y)(x² – xy + y²), expand the Right Hand Side (RHS):

RHS = (x + y)(x² – xy + y²)

= x(x² – xy + y²) + y(x² – xy + y²)

= (x³ – x²y + xy²) + (yx² – xy² + y³)

= x³ – x²y + xy² + x²y – xy² + y³

= x³ + (-x²y + x²y) + (xy² - xy²) + y³

= x³ + 0 + 0 + y³ = x³ + y³ = LHS (Left Hand Side)

Hence, verified.

---

(ii) To verify x³ – y³ = (x – y)(x² + xy + y²), expand the RHS:

RHS = (x - y)(x² + xy + y²)

= x(x² + xy + y²) - y(x² + xy + y²)

= (x³ + x²y + xy²) - (yx² + xy² + y³)

= x³ + x²y + xy² - x²y - xy² - y³

= x³ + (x²y - x²y) + (xy² - xy²) - y³

= x³ + 0 + 0 - y³ = x³ - y³ = LHS

Hence, verified.

(i) 27y³ + 125z³ = (3y)³ + (5z)³

Using identity: a³ + b³ = (a + b)(a² – ab + b²)

Here, a = 3y, b = 5z.

= (3y + 5z)((3y)² – (3y)(5z) + (5z)²)

= (3y + 5z)(9y² – 15yz + 25z²)

---

(ii) 64m³ – 343n³ = (4m)³ – (7n)³

Using identity: a³ - b³ = (a - b)(a² + ab + b²)

Here, a = 4m, b = 7n.

= (4m - 7n)((4m)² + (4m)(7n) + (7n)²)

= (4m - 7n)(16m² + 28mn + 49n²)

Assuming the expression is 27x³ + y³ + z³ – 9xyz.

= (3x)³ + (y)³ + (z)³ - 3(3x)(y)(z)

Using identity: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

Here, a = 3x, b = y, c = z.

= (3x + y + z)((3x)² + (y)² + (z)² - (3x)(y) - (y)(z) - (z)(3x))

= (3x + y + z)(9x² + y² + z² - 3xy - yz - 3zx)

Start with the Right Hand Side (RHS):

RHS = $\frac{1}{2}$ (x + y + z)[(x – y)² + (y – z)² + (z – x)²]

Expand the squared terms inside the bracket:

= $\frac{1}{2}$ (x + y + z)[(x² - 2xy + y²) + (y² - 2yz + z²) + (z² - 2zx + x²)]

Combine like terms inside the bracket:

= $\frac{1}{2}$ (x + y + z)[x² + x² + y² + y² + z² + z² - 2xy - 2yz - 2zx]

= $\frac{1}{2}$ (x + y + z)[2x² + 2y² + 2z² - 2xy - 2yz - 2zx]

Factor out 2 from the bracket:

= $\frac{1}{2}$ (x + y + z) × 2 [x² + y² + z² - xy - yz - zx]

Cancel the $\frac{1}{2}$ and 2:

= (x + y + z)(x² + y² + z² - xy - yz - zx)

This is the known identity for x³ + y³ + z³ – 3xyz.

= x³ + y³ + z³ – 3xyz = LHS (Left Hand Side)

Hence, verified.

We know the identity: x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

Given that x + y + z = 0.

Substitute this value into the identity:

x³ + y³ + z³ – 3xyz = (0)(x² + y² + z² - xy - yz - zx)

x³ + y³ + z³ – 3xyz = 0

Add 3xyz to both sides:

x³ + y³ + z³ = 3xyz

Hence, shown.

We use the result from Question 13: If x + y + z = 0, then x³ + y³ + z³ = 3xyz.

(i) Let x = -12, y = 7, z = 5.

Check if x + y + z = 0:

x + y + z = (-12) + 7 + 5 = -12 + 12 = 0.

Since the sum is 0, we can apply the identity:

(–12)³ + (7)³ + (5)³ = 3(-12)(7)(5)

= 3(-12)(35)

= -36 × 35

= -1260

---

(ii) Let x = 28, y = -15, z = -13.

Check if x + y + z = 0:

x + y + z = 28 + (-15) + (-13) = 28 - 15 - 13 = 28 - 28 = 0.

Since the sum is 0, we can apply the identity:

(28)³ + (–15)³ + (–13)³ = 3(28)(-15)(-13)

= 3(28)(195)

= 84 × 195

= 16380

Area of a rectangle = Length × Breadth. We need to factorise the given quadratic expressions for the area.

(i) Area = 25a² – 35a + 12

We need to factorise by splitting the middle term. Product = 25 × 12 = 300. Sum = -35.

We look for two numbers whose product is 300 and sum is -35. These numbers are -15 and -20.

= 25a² - 15a - 20a + 12

= 5a(5a - 3) - 4(5a - 3)

= (5a - 3)(5a - 4)

Possible expressions:

Length = (5a - 3), Breadth = (5a - 4) (or vice versa)

---

(ii) Area = 35y² + 13y - 12

Product = 35 × (-12) = -420. Sum = 13.

We look for two numbers whose product is -420 and sum is 13. These numbers are 28 and -15.

= 35y² + 28y - 15y - 12

= 7y(5y + 4) - 3(5y + 4)

= (5y + 4)(7y - 3)

Possible expressions:

Length = (5y + 4), Breadth = (7y - 3) (or vice versa)

Volume of a cuboid = Length × Breadth × Height. We need to factorise the given expressions into three factors.

(i) Volume = 3x² – 12x

Factor out the common term 3x:

= 3x(x - 4)

= 3 × x × (x - 4)

Possible dimensions are: 3, x, (x - 4) (in any order)

---

(ii) Volume = 12ky² + 8ky – 20k

Factor out the common term 4k:

= 4k(3y² + 2y - 5)

Now, factorise the quadratic 3y² + 2y - 5.

Product = 3 × (-5) = -15. Sum = 2. Numbers are 5 and -3.

= 4k(3y² + 5y - 3y - 5)

= 4k[y(3y + 5) - 1(3y + 5)]

= 4k(3y + 5)(y - 1)

Possible dimensions are: 4k, (3y + 5), (y - 1) (in any order)