Factorisation of Algebraic Expressions and Polynomials
Factors of Algebraic Expressions
Understanding Factors in Algebra
In arithmetic, the concept of a factor is related to multiplication. A factor of a number is a number that divides it evenly, leaving no remainder. Equivalently, factors are the numbers that are multiplied together to produce another number. For example, the factors of $12$ are $1, 2, 3, 4, 6,$ and $12$ because $1 \times 12 = 12$, $2 \times 6 = 12$, and $3 \times 4 = 12$. We can also express $12$ as a product of its prime factors: $12 = 2 \times 2 \times 3 = 2^2 \times 3$.
The idea of factors extends to algebraic expressions. Factorisation (or factoring) of an algebraic expression is the process of writing the expression as a product of two or more other algebraic expressions. These expressions that are multiplied together are called the factors of the original expression.
For example, consider the expression $2x + 4$. We can see that both terms ($2x$ and $4$) have a common factor of $2$. Using the distributive property in reverse, we can write $2x + 4$ as $2(x + 2)$. In this product, $2$ and $(x + 2)$ are the factors of the algebraic expression $2x + 4$. When these factors are multiplied together, they reproduce the original expression: $2 \times (x + 2) = 2 \times x + 2 \times 2 = 2x + 4$.
Factorisation is essentially the reverse process of expanding or multiplying algebraic expressions. Just as multiplication combines factors to get a product, factorisation breaks down an expression into its constituent factors.
Factors of Terms vs. Factors of Expressions
It is important to distinguish between the factors of individual terms within an expression and the factors of the entire algebraic expression. While related, they refer to different levels of structure.
Factors of a Term:
These are the individual constants (numbers) and variables that are multiplied together to form a single term.Example: In the expression $5xy^2 + 7x$, the first term is $5xy^2$. The factors of the term $5xy^2$ are $5, x, y,$ and $y$. The second term is $7x$. The factors of the term $7x$ are $7$ and $x$.
Factors of an Expression:
These are the expressions (which can be monomials, binomials, trinomials, etc.) that, when multiplied together as a whole, result in the entire original expression.Example: Consider the expression $5xy^2 + 7x$. We can observe that the variable $x$ is present in both terms. We can factor out this common factor $x$ using the distributive property in reverse: $5xy^2 + 7x = x(5y^2 + 7)$. Here, $x$ and $(5y^2 + 7)$ are the factors of the algebraic expression $5xy^2 + 7x$. Multiplying these factors $x$ and $(5y^2 + 7)$ gives $x \times 5y^2 + x \times 7 = 5xy^2 + 7x$, the original expression.
When we talk about "factorising an algebraic expression," we are referring to finding the factors of the entire expression, which often involves identifying common factors among its terms or recognizing patterns that match algebraic identities.
Why is Factorisation Important?
Factorisation is one of the most fundamental and widely used skills in algebra. It unlocks numerous possibilities for simplifying expressions and solving problems.
Here are some key reasons for the importance of factorisation:
Simplifying Algebraic Expressions and Fractions:
Factorisation is essential for simplifying rational expressions (algebraic fractions). If there are common factors in the numerator and the denominator of a fraction, they can be cancelled out only after the numerator and denominator are expressed as products of their factors.Example: Simplify $\frac{x^2 - 4}{x - 2}$. By factorising the numerator using the difference of squares identity ($x^2 - 4 = (x-2)(x+2)$), we get $\frac{(x-2)(x+2)}{x - 2}$. For $x \neq 2$, we can cancel the common factor $(x-2)$, resulting in the simplified expression $x+2$.
Solving Polynomial Equations:
Finding the roots (or zeros) of polynomial equations is heavily dependent on factorisation. If a polynomial equation can be factored into linear factors, finding the roots becomes straightforward using the zero product property (if $A \times B = 0$, then $A=0$ or $B=0$).Example: Solve the quadratic equation $x^2 - 5x + 6 = 0$. By factorising the quadratic expression, we get $(x - 2)(x - 3) = 0$. Setting each factor to zero gives $x - 2 = 0 \implies x=2$ and $x - 3 = 0 \implies x=3$. The roots of the equation are $2$ and $3$. Without factoring, solving quadratic equations often requires using the quadratic formula or completing the square.
Analyzing Polynomials and Functions:
Factorisation helps reveal the zeros of a polynomial, which are the x-intercepts of its graph. The factors also provide information about the behavior of the polynomial near its zeros.Working with Inequalities:
Solving polynomial and rational inequalities often involves factoring the expressions to find the critical points (zeros and undefined points) that divide the number line into intervals.Advanced Mathematical Topics:
Factorisation is a prerequisite skill for topics such as partial fraction decomposition, working with polynomial functions, and concepts in abstract algebra.
In summary, factorisation is the process of breaking down an algebraic expression into a product of simpler expressions (its factors). It is a reverse operation to multiplication and is a fundamental technique required for simplifying expressions, solving equations, and analysing polynomials.
Factorising by Taking Out Common Factors
The Reverse of the Distributive Property
The simplest and most basic method of factorisation is by identifying and extracting common factors from the terms that make up an algebraic expression. This technique is a direct application of the distributive property of multiplication over addition or subtraction, used in reverse.
The distributive property states that $a(b + c) = ab + ac$. When we factor by taking out a common factor, we are starting with an expression in the form $ab + ac$ and rewriting it as the product of the common factor $a$ and the remaining expression $(b+c)$, i.e., $ab + ac = a(b + c)$. The factor $a$ is a common factor to both terms $ab$ and $ac$.
In practice, the common factor 'a' can be a number, a variable, or a product of numbers and variables.
Procedure for Factorising by Taking Out Common Factors
To factorise an algebraic expression by taking out the greatest common factor (GCF) from all its terms, follow these steps:
Find the Greatest Common Factor (GCF):
Identify the greatest common factor of all the individual terms in the expression.- Find the GCF of the numerical coefficients of all the terms.
- For each variable, find the lowest power that is common to ALL terms. If a variable is not present in every term, it is not part of the common factor.
- The GCF of the terms is the product of the GCF of the coefficients and the lowest common powers of all variables.
Write the GCF Outside:
Write the identified GCF outside a pair of parentheses.Divide Each Term by the GCF:
Inside the parentheses, write the result obtained by dividing each term of the original expression by the GCF. The operation signs (+ or -) connecting the terms inside the parentheses will be the same as those in the original expression.
The resulting expression will be the product of the GCF and the polynomial inside the parentheses. This is the factored form of the original expression.
Examples
Example 1. Factorise $6x + 12$.
Answer:
The given expression is $6x + 12$. The terms are $6x$ and $12$.
Step 1: Find the GCF of $6x$ and $12$.
- GCF of numerical coefficients $6$ and $12$: The factors of 6 are 1, 2, 3, 6. The factors of 12 are 1, 2, 3, 4, 6, 12. The greatest common factor is $6$.
- Variable $x$: It appears in $6x$ but not in $12$. So, $x$ is not a common factor.
The GCF of $6x$ and $12$ is $6$.
Step 2: Write the GCF ($6$) outside the parentheses:
$$ 6(\phantom{\text{something inside}}) $$Step 3: Divide each term of the original expression by the GCF ($6$) and write the results inside the parentheses:
$$ \frac{6x}{6} = x $$ $$ \frac{12}{6} = 2 $$Write these results inside with the original plus sign:
$$ 6x + 12 = 6(x + 2) $$The factored form is $\textbf{6(x + 2)}$. The factors are $6$ and $(x+2)$.
Check: Multiply the factors: $6 \times (x+2) = 6 \times x + 6 \times 2 = 6x + 12$. This matches the original expression.
Example 2. Factorise $10a^2 b - 15ab^2$.
Answer:
The given expression is $10a^2 b - 15ab^2$. The terms are $10a^2 b$ and $-15ab^2$.
Step 1: Find the GCF of $10a^2 b$ and $-15ab^2$.
- GCF of numerical coefficients $10$ and $-15$: The GCF of $10$ and $15$ is $5$. We can choose to factor out $+5$ or $-5$. Usually, we factor out a positive GCF unless the leading term is negative and we want the factor inside the parentheses to start with a positive term. Let's use $+5$.
- Variable $a$: Appears as $a^2$ and $a^1$. The lowest power common to both is $a^1$.
- Variable $b$: Appears as $b^1$ and $b^2$. The lowest power common to both is $b^1$.
The GCF of $10a^2 b$ and $-15ab^2$ is $5a^1 b^1 = 5ab$.
Step 2: Write the GCF ($5ab$) outside the parentheses:
$$ 5ab(\phantom{\text{something inside}}) $$Step 3: Divide each term of the original expression by the GCF ($5ab$) and write the results inside:
$$ \frac{10a^2 b}{5ab} = \frac{10}{5} \times \frac{a^2}{a} \times \frac{b}{b} = 2 \times a^{2-1} \times b^{1-1} = 2a^1 b^0 = 2a $$ $$ \frac{-15ab^2}{5ab} = \frac{-15}{5} \times \frac{a}{a} \times \frac{b^2}{b} = -3 \times a^{1-1} \times b^{2-1} = -3a^0 b^1 = -3b $$Write these results inside with the original subtraction sign between the terms:
$$ 10a^2 b - 15ab^2 = 5ab(2a - 3b) $$The factored form is $\textbf{5ab(2a - 3b)}$. The factors are $5ab$ and $(2a-3b)$.
Check: Multiply the factors: $5ab \times (2a - 3b) = (5ab \times 2a) - (5ab \times 3b) = 10a^{1+1} b - 15a b^{1+1} = 10a^2 b - 15ab^2$. This matches the original expression.
Example 3. Factorise $7x^2 y + 14xy^2 - 21xy$.
Answer:
The given expression is $7x^2 y + 14xy^2 - 21xy$. The terms are $7x^2 y$, $14xy^2$, and $-21xy$.
Step 1: Find the GCF of $7x^2 y$, $14xy^2$, and $-21xy$.
- GCF of numerical coefficients $7, 14,$ and $-21$: The GCF of $7, 14,$ and $21$ is $7$. Let's use $+7$.
- Variable $x$: Appears as $x^2, x^1,$ and $x^1$. The lowest power common to all is $x^1$.
- Variable $y$: Appears as $y^1, y^2,$ and $y^1$. The lowest power common to all is $y^1$.
The GCF of $7x^2 y$, $14xy^2$, and $-21xy$ is $7x^1 y^1 = 7xy$.
Step 2: Write the GCF ($7xy$) outside the parentheses:
$$ 7xy(\phantom{\text{something inside}}) $$Step 3: Divide each term of the original expression by the GCF ($7xy$) and write the results inside:
$$ \frac{7x^2 y}{7xy} = \frac{7}{7} \times \frac{x^2}{x} \times \frac{y}{y} = 1 \times x^{2-1} \times y^{1-1} = 1x^1 y^0 = x $$ $$ \frac{14xy^2}{7xy} = \frac{14}{7} \times \frac{x}{x} \times \frac{y^2}{y} = 2 \times x^{1-1} \times y^{2-1} = 2x^0 y^1 = 2y $$ $$ \frac{-21xy}{7xy} = \frac{-21}{7} \times \frac{x}{x} \times \frac{y}{y} = -3 \times x^{1-1} \times y^{1-1} = -3x^0 y^0 = -3 $$Write these results inside with the original operation signs:
$$ 7x^2 y + 14xy^2 - 21xy = 7xy(x + 2y - 3) $$The factored form is $\textbf{7xy(x + 2y - 3)}$. The factors are $7xy$ and $(x+2y-3)$.
Check: Multiply the factors: $7xy(x + 2y - 3) = (7xy \times x) + (7xy \times 2y) - (7xy \times 3) = 7x^2 y + 14xy^2 - 21xy$. This matches the original expression.
Factorising by taking out the common factor is the essential first step in factorising many algebraic expressions. Always look for a common factor before attempting more advanced factorisation techniques.
Factorising by Grouping of Terms
Applying Common Factors to Groups
Factorisation by grouping is a technique used for expressions that typically have four or more terms. This method is applicable when there isn't a single common factor shared by all the terms in the expression, but we can form groups of terms where each group has its own common factor. The goal is that after factoring out the common factor from each group, a common binomial (or polynomial) factor appears across the different groups. This common binomial can then be factored out, leading to the complete factorisation of the original expression.
Procedure for Factorising by Grouping
To factorise an algebraic expression using the grouping method, follow these steps:
Group the Terms:
Arrange the terms in the expression (if necessary) and group them into pairs or sets that share common factors. The way you group might matter; the key is to form groups such that, after factoring within each group, you are left with an identical binomial or polynomial factor from each group. Using parentheses can help visually define the groups. Be careful with signs when grouping terms preceded by a minus sign.Factor within Each Group:
Factor out the greatest common factor (GCF) from each individual group formed in Step 1.Identify and Factor Out the Common Binomial:
After factoring within each group, examine the resulting expression. If the grouping was successful, you should now see a common binomial or polynomial factor present in every term of this new expression. Factor out this common binomial/polynomial factor.Write the Remaining Factors:
The terms that are left inside the parentheses after factoring out the common binomial form the other factor.
The final factored form will be the product of the common binomial/polynomial factor and the expression formed by the terms that were factored out from the groups.
Examples
Example 1. Factorise $ax + ay + bx + by$.
Answer:
The expression is $ax + ay + bx + by$. It has four terms. There is no single common factor for all four terms.
Step 1: Group the terms. A natural grouping here is the first two and the last two terms:
$$ (ax + ay) + (bx + by) $$Step 2: Factor out the GCF from each group:
- In the first group $(ax + ay)$, the common factor is $a$. Factoring gives $a(x + y)$.
- In the second group $(bx + by)$, the common factor is $b$. Factoring gives $b(x + y)$.
Substitute the factored groups back into the expression:
$$ a(x + y) + b(x + y) $$Step 3: Observe the resulting expression. We see that the binomial $(x + y)$ is a common factor in both terms ($a(x+y)$ and $b(x+y)$).
Factor out the common binomial factor $(x + y)$:
$$ (x + y)(\phantom{\text{something inside}}) $$Step 4: Write the terms that were factored out from each group ($a$ and $+b$) inside the second set of parentheses:
$$ (x + y)(a + b) $$The factored form is $\textbf{(x + y)(a + b)}$. The factors are the binomials $(x+y)$ and $(a+b)$.
Check: Expand the factors: $(x+y)(a+b) = x(a+b) + y(a+b) = xa + xb + ya + yb = ax + bx + ay + by$. Rearranging terms gives $ax + ay + bx + by$. Correct.
Example 2. Factorise $x^2 + 2x + xy + 2y$.
Answer:
The expression is $x^2 + 2x + xy + 2y$. Four terms, no overall common factor.
Step 1: Group the terms. Let's try the first two and the last two:
$$ (x^2 + 2x) + (xy + 2y) $$Step 2: Factor out the GCF from each group:
- In $(x^2 + 2x)$, the GCF is $x$. Factoring gives $x(x + 2)$.
- In $(xy + 2y)$, the GCF is $y$. Factoring gives $y(x + 2)$.
Substitute back:
$$ x(x + 2) + y(x + 2) $$Step 3: The binomial $(x + 2)$ is common to both terms.
Step 4: Factor out the common binomial $(x + 2)$:
$$ (x + 2)(x + y) $$The factored form is $\textbf{(x + 2)(x + y)}$.
Check: $(x+2)(x+y) = x(x+y) + 2(x+y) = x^2 + xy + 2x + 2y = x^2 + 2x + xy + 2y$. Correct.
Example 3. Factorise $6xy - 4y + 3x - 2$.
Answer:
The expression is $6xy - 4y + 3x - 2$. Four terms, no overall common factor.
Step 1: Group the terms. Let's try the first two and the last two:
$$ (6xy - 4y) + (3x - 2) $$Step 2: Factor out the GCF from each group:
- In $(6xy - 4y)$, the GCF is $2y$. Factoring gives $2y(3x - 2)$.
- In $(3x - 2)$, there's no apparent common factor other than $1$. Factoring gives $1(3x - 2)$.
Substitute back:
$$ 2y(3x - 2) + 1(3x - 2) $$Step 3: The binomial $(3x - 2)$ is common to both terms.
Step 4: Factor out the common binomial $(3x - 2)$:
$$ (3x - 2)(2y + 1) $$The factored form is $\textbf{(3x - 2)(2y + 1)}$.
Check: $(3x-2)(2y+1) = 3x(2y+1) - 2(2y+1) = 6xy + 3x - 4y - 2 = 6xy - 4y + 3x - 2$. Correct.
When Grouping Doesn't Work Initially
Sometimes, the terms of the expression might be arranged in such a way that an initial grouping does not yield a common binomial factor. In such cases, rearranging the terms before grouping is necessary to find a successful grouping.
Example 4. Factorise $ab - cd + ad - bc$.
Answer:
The expression is $ab - cd + ad - bc$. Four terms, no overall common factor.
Step 1: Try an initial grouping, say, the first two and the last two:
$$ (ab - cd) + (ad - bc) $$Factor out GCF from each group:
- In $(ab - cd)$, the GCF is $1$. Factoring gives $1(ab - cd)$.
- In $(ad - bc)$, the GCF is $1$. Factoring gives $1(ad - bc)$.
Substitute back: $1(ab - cd) + 1(ad - bc)$. The binomials $(ab-cd)$ and $(ad-bc)$ are not the same. This grouping does not work.
Step 1 (Attempt 2): Rearrange the terms and try a different grouping. Let's group terms that share a variable, for instance, group the terms with $a$ and the terms with $b$: $ab + ad - bc - cd$.
$$ (ab + ad) + (-bc - cd) $$Step 2 (Attempt 2): Factor out GCF from each new group:
- In $(ab + ad)$, the GCF is $a$. Factoring gives $a(b + d)$.
- In $(-bc - cd)$, we can factor out $c$ or $-c$. If we factor out $c$, we get $c(-b - d) = -c(b+d)$. If we factor out $-c$, we get $-c(b + d)$. Let's factor out $-c$ to make the binomial match $a(b+d)$. Factoring gives $-c(b + d)$.
Substitute back:
$$ a(b + d) - c(b + d) $$Step 3 (Attempt 2): The binomial $(b + d)$ is common to both terms.
Step 4 (Attempt 2): Factor out the common binomial $(b + d)$:
$$ (b + d)(a - c) $$The factored form is $\textbf{(b + d)(a - c)}$.
Check: Expand the factors: $(b+d)(a-c) = b(a-c) + d(a-c) = ba - bc + da - dc = ab - bc + ad - cd$. Rearranging gives $ab - cd + ad - bc$. Correct.
Factorisation by grouping is a valuable technique when direct common factoring is not possible. It requires careful grouping and factoring within groups to reveal a common polynomial factor.
Factorising using Algebraic Identities
Algebraic identities provide pre-established relationships between expanded and factored forms of specific types of expressions. When an algebraic expression matches the structure of the expanded side of an identity, we can use the identity in reverse to write the expression in its corresponding factored form. This method is particularly useful for factorising binomials that are differences of squares or sums/differences of cubes, and for factorising trinomials that are perfect squares.
Using Standard Identities for Factorisation
Let's revisit some standard algebraic identities and see how they are applied for factorisation. In these identities, $a$ and $b$ can represent any algebraic terms (constants, variables, or products of variables and constants).
Identity 1 (Reverse): Difference of Squares
The identity $(a + b)(a - b) = a^2 - b^2$ can be used in reverse. If we encounter an expression that is the difference of two perfect squares, we can factor it into the product of the sum and the difference of the square roots of those terms.
$a^2 - b^2 = (a + b)(a - b) $
[Difference of Squares Identity for Factorisation]
To use this identity for factorisation, the expression must be a binomial (two terms), the operation between the terms must be subtraction, and both terms must be perfect squares (meaning they can be written as the square of some other term).
Example 1. Factorise $x^2 - 49$.
Answer:
The expression $x^2 - 49$ is a binomial with subtraction. We check if each term is a perfect square:
- The first term is $x^2$. This is the square of $x$. So, $a = x$.
- The second term is $49$. This is the square of $7$ (since $7^2 = 49$). So, $b = 7$.
The expression is in the form $a^2 - b^2$, where $a=x$ and $b=7$. Apply the difference of squares identity:
$$ x^2 - 49 = x^2 - 7^2 = (x + 7)(x - 7) $$The factored form is $\textbf{(x + 7)(x - 7)}$.
Example 2. Factorise $25a^2 - 16b^2$.
Answer:
The expression $25a^2 - 16b^2$ is a binomial with subtraction. Check for perfect squares:
- The first term is $25a^2$. We need to find a term whose square is $25a^2$. Since $5^2 = 25$ and $a^2 = (a)^2$, $(5a)^2 = 5^2 a^2 = 25a^2$. So, $a = 5a$.
- The second term is $16b^2$. Similarly, $16b^2 = (4b)^2$ (since $4^2 = 16$ and $b^2 = (b)^2$). So, $b = 4b$.
The expression is in the form $a^2 - b^2$, where $a=5a$ and $b=4b$. Apply the difference of squares identity:
$$ 25a^2 - 16b^2 = (5a)^2 - (4b)^2 = (5a + 4b)(5a - 4b) $$The factored form is $\textbf{(5a + 4b)(5a - 4b)}$.
Identity 2 (Reverse): Perfect Square Trinomials
The identities for the square of a sum and the square of a difference, $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$, can be used in reverse. If a trinomial (three terms) matches the pattern of a perfect square expansion, it can be factored into the square of a binomial.
$a^2 + 2ab + b^2 = (a + b)^2 = (a+b)(a+b) $
[Perfect Square Trinomial Identity 1]
$a^2 - 2ab + b^2 = (a - b)^2 = (a-b)(a-b) $
[Perfect Square Trinomial Identity 2]
To use these identities, look for a trinomial where:
- Two of the terms are perfect squares and have positive signs (these will be $a^2$ and $b^2$).
- The third term (the middle term) is equal to either $+2$ times the product of the square roots of the other two terms ($+2ab$) or $-2$ times the product of the square roots of the other two terms ($-2ab$).
Example 3. Factorise $x^2 + 6x + 9$.
Answer:
This is a trinomial. Let's check for perfect square terms:
- $x^2$ is the square of $x$. So, $a=x$.
- $9$ is the square of $3$. So, $b=3$. Both $x^2$ and $9$ are positive.
Now, check the middle term ($6x$). Is it equal to $2ab$ or $-2ab$ using $a=x$ and $b=3$? $2ab = 2(x)(3) = 6x$. The middle term matches $+2ab$.
Since the expression is in the form $a^2 + 2ab + b^2$, where $a=x$ and $b=3$, we can apply the identity $a^2 + 2ab + b^2 = (a + b)^2$:
$$ x^2 + 6x + 9 = x^2 + 2(x)(3) + 3^2 = (x + 3)^2 $$The factored form is $\textbf{(x + 3)\textsuperscript{2}}$ or $\textbf{(x + 3)(x + 3)}$.
Example 4. Factorise $4y^2 - 12y + 9$.
Answer:
This is a trinomial. Check for perfect square terms with positive signs:
- $4y^2$ is the square of $2y$. So, $a=2y$. (It has a positive sign).
- $9$ is the square of $3$. So, $b=3$. (It has a positive sign).
- The remaining term is $-12y$.
Check if the middle term ($-12y$) is equal to $2ab$ or $-2ab$ using $a=2y$ and $b=3$. $2ab = 2(2y)(3) = 12y$. The middle term is $-12y$, which matches $-2ab$.
Since the expression is in the form $a^2 - 2ab + b^2$, where $a=2y$ and $b=3$, we can apply the identity $a^2 - 2ab + b^2 = (a - b)^2$:
$$ 4y^2 - 12y + 9 = (2y)^2 - 2(2y)(3) + 3^2 = (2y - 3)^2 $$The factored form is $\textbf{(2y - 3)\textsuperscript{2}}$ or $\textbf{(2y - 3)(2y - 3)}$.
Identity 3 (Reverse): Sum/Difference of Cubes
The identities for the sum and difference of cubes, $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ and $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, can be used to factorise binomials that are sums or differences of perfect cubes.
$a^3 + b^3 = (a + b)(a^2 - ab + b^2) $
[Sum of Cubes Identity for Factorisation]
$a^3 - b^3 = (a - b)(a^2 + ab + b^2) $
[Difference of Cubes Identity for Factorisation]
To use these identities, check if the binomial expression is a sum or difference of terms that are perfect cubes (i.e., they can be written as the cube of some other term).
Example 5. Factorise $x^3 - 64$.
Answer:
This is a binomial with subtraction. Check for perfect cubes:
- The first term is $x^3$. This is the cube of $x$. So, $a=x$.
- The second term is $64$. This is the cube of $4$ (since $4 \times 4 \times 4 = 64$, $4^3 = 64$). So, $b=4$.
The expression is in the form $a^3 - b^3$, where $a=x$ and $b=4$. Apply the difference of cubes identity:
$$ x^3 - 64 = x^3 - 4^3 = (x - 4)(x^2 + x(4) + 4^2) $$Simplify the trinomial factor:
$$ = (x - 4)(x^2 + 4x + 16) $$The factored form is $\textbf{(x - 4)(x\textsuperscript{2} + 4x + 16)}$.
Example 6. Factorise $8p^3 + 125q^3$.
Answer:
This is a binomial with addition. Check for perfect cubes:
- The first term is $8p^3$. We need a term whose cube is $8p^3$. Since $2^3 = 8$ and $p^3 = (p)^3$, $(2p)^3 = 2^3 p^3 = 8p^3$. So, $a=2p$.
- The second term is $125q^3$. Similarly, $(5q)^3 = 5^3 q^3 = 125q^3$ (since $5^3 = 125$ and $q^3 = (q)^3$). So, $b=5q$.
The expression is in the form $a^3 + b^3$, where $a=2p$ and $b=5q$. Apply the sum of cubes identity:
$$ 8p^3 + 125q^3 = (2p)^3 + (5q)^3 = ((2p) + (5q))((2p)^2 - (2p)(5q) + (5q)^2) $$Simplify the terms inside the second parentheses:
$$ = (2p + 5q)(4p^2 - 10pq + 25q^2) $$The factored form is $\textbf{(2p + 5q)(4p\textsuperscript{2} - 10pq + 25q\textsuperscript{2})}$.
Using algebraic identities for factorisation is a powerful and efficient technique for specific forms of polynomials. It complements other factorisation methods like taking out common factors and grouping.
Factorising of Trinomials (Quadratic Form)
Factorising Trinomials of the Form $ax^2 + bx + c$
A trinomial is a polynomial with three terms. Trinomials of the form $ax^2 + bx + c$, where $a, b,$ and $c$ are constants and $x$ is the variable, are particularly common. These are often referred to as trinomials of quadratic form because the highest power of the variable is $2$. Factorising such a trinomial means expressing it as a product of simpler factors, typically two linear binomials (e.g., $(px+q)(rx+s)$) or, in special cases (perfect square trinomials), the square of a single binomial (e.g., $(px+q)^2$). This skill is essential for solving quadratic equations, simplifying algebraic fractions, and working with parabolas.
The approach to factorising $ax^2 + bx + c$ depends on whether the leading coefficient $a$ is $1$ or not.
Factorising Trinomials of the Form $x^2 + bx + c$ (where $a=1$)
This is the simplest type of quadratic trinomial to factorise. We are looking for two linear factors of the form $(x+p)$ and $(x+q)$ such that their product is the given trinomial $x^2 + bx + c$.
Let's expand the product $(x+p)(x+q)$:
$$ (x+p)(x+q) = x(x+q) + p(x+q) $$ $$ = x^2 + xq + px + pq $$ $$ = x^2 + (q+p)x + pq $$Using the commutative property of addition, $q+p = p+q$. So,
$$ (x+p)(x+q) = x^2 + (p+q)x + pq $$Now, compare this expanded form to the general form $x^2 + bx + c$. For the two polynomials to be identical, the coefficients of corresponding powers of $x$ must be equal:
- Coefficient of $x^2$: $1 = 1$ (Consistent).
- Coefficient of $x$: $b = p+q$.
- Constant term: $c = pq$.
So, to factorise $x^2 + bx + c$, we need to find two numbers $p$ and $q$ whose sum is equal to the coefficient of the $x$ term ($b$) and whose product is equal to the constant term ($c$).
To factorise $x^2 + bx + c$, find $p, q$ such that $p + q = b$ and $p \times q = c$.
[Rule for $a=1$ trinomials]
The factored form will then be $(x + p)(x + q)$.
Steps for factorising $x^2 + bx + c$:
- Identify the values of $b$ (coefficient of $x$) and $c$ (constant term).
- Find pairs of numbers whose product is $c$.
- From these pairs, find the pair whose sum is $b$. These are your numbers $p$ and $q$.
- Write the factored form as $(x+p)(x+q)$.
Example 1. Factorise $x^2 + 5x + 6$.
Answer:
The trinomial is $x^2 + 5x + 6$. It is in the form $x^2 + bx + c$, with $b=5$ and $c=6$.
We need to find two numbers $p$ and $q$ such that $p + q = 5$ and $p \times q = 6$.
Let's list pairs of integer factors of $c = 6$:
- $1 \times 6 = 6$. Sum $= 1 + 6 = 7$. (Not 5)
- $2 \times 3 = 6$. Sum $= 2 + 3 = 5$. (This works!)
- $(-1) \times (-6) = 6$. Sum $= -1 + (-6) = -7$. (Not 5)
- $(-2) \times (-3) = 6$. Sum $= -2 + (-3) = -5$. (Not 5)
The two numbers are $p=2$ and $q=3$ (or vice versa). The factored form is $(x+p)(x+q)$.
$$ x^2 + 5x + 6 = (x + 2)(x + 3) $$The factors are $\textbf{(x + 2)}$ and $\textbf{(x + 3)}$.
Check: Expand $(x+2)(x+3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6$. Correct.
Example 2. Factorise $m^2 - 7m + 12$.
Answer:
The trinomial is $m^2 - 7m + 12$. It is in the form $m^2 + bm + c$, with $b=-7$ and $c=12$. The variable is $m$ instead of $x$, but the process is the same.
We need to find two numbers $p$ and $q$ such that $p + q = -7$ and $p \times q = 12$.
Let's list pairs of integer factors of $c = 12$: $(1, 12), (2, 6), (3, 4), (-1, -12), (-2, -6), (-3, -4)$.
Check which pair adds up to $b = -7$:
- $1 + 12 = 13$
- $2 + 6 = 8$
- $3 + 4 = 7$
- $-1 + (-12) = -13$
- $-2 + (-6) = -8$
- $-3 + (-4) = -7$. (This works!)
The two numbers are $p=-3$ and $q=-4$. The factored form is $(m+p)(m+q)$.
$$ m^2 - 7m + 12 = (m + (-3))(m + (-4)) = (m - 3)(m - 4) $$The factors are $\textbf{(m - 3)}$ and $\textbf{(m - 4)}$.
Check: $(m-3)(m-4) = m^2 - 4m - 3m + 12 = m^2 - 7m + 12$. Correct.
Example 3. Factorise $y^2 + y - 20$.
Answer:
The trinomial is $y^2 + y - 20$. It is in the form $y^2 + by + c$, with $b=1$ (since $y = 1y$) and $c=-20$.
We need to find two numbers $p$ and $q$ such that $p + q = 1$ and $p \times q = -20$.
Since the product $pq$ is negative ($-20$), one number must be positive and the other negative. The sum $p+q$ is positive ($1$), which means the positive number must have a larger absolute value than the negative number.
Let's list pairs of integer factors of $c = -20$ (where the positive factor has a larger absolute value): $(5, -4), (10, -2), (20, -1)$.
Check which pair adds up to $b = 1$:
- $5 + (-4) = 1$. (This works!)
- $10 + (-2) = 8$.
- $20 + (-1) = 19$.
The two numbers are $p=5$ and $q=-4$. The factored form is $(y+p)(y+q)$.
$$ y^2 + y - 20 = (y + 5)(y + (-4)) = (y + 5)(y - 4) $$The factors are $\textbf{(y + 5)}$ and $\textbf{(y - 4)}$.
Check: $(y+5)(y-4) = y^2 - 4y + 5y - 20 = y^2 + y - 20$. Correct.
Factorising Trinomials of the Form $ax^2 + bx + c$ (where $a \neq 1$)
When the leading coefficient $a$ is not $1$, the method is slightly different. We cannot directly use the property $p+q=b, pq=c$. Instead, we modify the trinomial by splitting the middle term ($bx$) into two terms, and then factor by grouping.
The key idea is to find two numbers, say $p$ and $q$, such that their sum is still the coefficient of the $x$ term ($b$), but their product is equal to the product of the leading coefficient ($a$) and the constant term ($c$).
To factorise $ax^2 + bx + c$, find $p, q$ such that $p + q = b$ and $p \times q = a \times c$.
[Rule for $a \neq 1$ trinomials]
Once you find such numbers $p$ and $q$, you rewrite the middle term $bx$ as the sum $px + qx$. The trinomial $ax^2 + bx + c$ is transformed into a four-term expression $ax^2 + px + qx + c$. You then factorise this four-term expression by grouping.
Steps for factorising $ax^2 + bx + c$ ($a \neq 1$):
- Identify the coefficients $a, b,$ and $c$.
- Calculate the product $ac$.
- Find two numbers $p$ and $q$ such that their sum is $b$ ($p+q=b$) and their product is $ac$ ($pq=ac$).
- Rewrite the original trinomial by splitting the middle term $bx$ into $px + qx$. The expression becomes $ax^2 + px + qx + c$. (The order of $px$ and $qx$ usually doesn't matter for the final result).
- Factorise the resulting four-term expression by grouping. Group the first two terms and the last two terms, factor out the GCF from each group, and then factor out the common binomial factor.
Example 4. Factorise $2x^2 + 7x + 3$.
Answer:
The trinomial is $2x^2 + 7x + 3$. It is in the form $ax^2 + bx + c$, with $a=2, b=7,$ and $c=3$. Here $a \neq 1$.
Step 1 & 2: Identify $a, b, c$ and calculate $ac$. $a=2, b=7, c=3$. $ac = 2 \times 3 = 6$.
Step 3: Find two numbers $p$ and $q$ such that $p + q = 7$ and $p \times q = 6$.
Let's list pairs of integer factors of $ac = 6$: $(1, 6), (2, 3), (-1, -6), (-2, -3)$.
Check which pair adds up to $b = 7$: $1+6=7$. The pair $(1, 6)$ works.
So, $p=1$ and $q=6$.
Step 4: Rewrite the middle term $7x$ as $1x + 6x$ (or $6x + 1x$). The expression becomes:
$$ 2x^2 + 1x + 6x + 3 $$Step 5: Factorise this four-term expression by grouping. Group the first two and the last two terms:
$$ (2x^2 + 1x) + (6x + 3) $$Factor out the GCF from each group:
- In $(2x^2 + 1x)$, the GCF is $x$. Factoring gives $x(2x + 1)$.
- In $(6x + 3)$, the GCF is $3$. Factoring gives $3(2x + 1)$.
Substitute the factored groups back:
$$ x(2x + 1) + 3(2x + 1) $$Now, factor out the common binomial $(2x + 1)$:
$$ (2x + 1)(x + 3) $$The factored form is $\textbf{(2x + 1)(x + 3)}$.
Check: Expand $(2x+1)(x+3) = 2x(x+3) + 1(x+3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3$. Correct.
Example 5. Factorise $6y^2 - 5y - 6$.
Answer:
The trinomial is $6y^2 - 5y - 6$. It is in the form $ay^2 + by + c$, with $a=6, b=-5,$ and $c=-6$. Here $a \neq 1$.
Step 1 & 2: Identify $a, b, c$ and calculate $ac$. $a=6, b=-5, c=-6$. $ac = 6 \times (-6) = -36$.
Step 3: Find two numbers $p$ and $q$ such that $p + q = -5$ and $p \times q = -36$.
Since the product $pq = -36$ is negative, one number must be positive and the other negative. The sum $p+q = -5$ is negative, which suggests the negative number has a larger absolute value.
Let's list pairs of integer factors of $ac = -36$ where the negative factor has a larger absolute value: $(1, -36), (2, -18), (3, -12), (4, -9), (6, -6)$.
Check which pair adds up to $b = -5$:
- $1 + (-36) = -35$
- $2 + (-18) = -16$
- $3 + (-12) = -9$
- $4 + (-9) = -5$. (This works!)
- $6 + (-6) = 0$
The two numbers are $p=4$ and $q=-9$.
Step 4: Rewrite the middle term $-5y$ as $4y - 9y$. The expression becomes:
$$ 6y^2 + 4y - 9y - 6 $$Step 5: Factorise this four-term expression by grouping. Group the first two and the last two terms:
$$ (6y^2 + 4y) + (-9y - 6) $$Factor out the GCF from each group:
- In $(6y^2 + 4y)$, the GCF is $2y$. Factoring gives $2y(3y + 2)$.
- In $(-9y - 6)$, we factor out a negative GCF ($-3$) to make the binomial factor $(3y+2)$ match the first group. Factoring gives $-3(3y + 2)$.
Substitute the factored groups back:
$$ 2y(3y + 2) - 3(3y + 2) $$Now, factor out the common binomial $(3y + 2)$:
$$ (3y + 2)(2y - 3) $$The factored form is $\textbf{(3y + 2)(2y - 3)}$.
Check: Expand $(3y+2)(2y-3) = 3y(2y-3) + 2(2y-3) = 6y^2 - 9y + 4y - 6 = 6y^2 - 5y - 6$. Correct.
Factorising trinomials of quadratic form, especially $ax^2+bx+c$, is a fundamental skill. The method of splitting the middle term and then grouping is a systematic way to approach cases where the leading coefficient is not 1.
Factorisation of Quadratic Polynomials
A quadratic polynomial in one variable is a polynomial of degree $2$. It is generally expressed in the form $ax^2 + bx + c$, where $a, b,$ and $c$ are constants (with $a \neq 0$) and $x$ is the variable. Factorising a quadratic polynomial means writing it as a product of two or more simpler algebraic expressions. For quadratic polynomials with real coefficients, the factors are usually linear binomials with real coefficients (or the square of a linear binomial), provided the polynomial has real roots.
The methods used for factorising quadratic polynomials are the same as those discussed for factorising trinomials of the form $ax^2 + bx + c$ in the previous section (I5), as these two terms refer to the same type of expression.
Factorising Quadratic Polynomials ($P(x) = ax^2 + bx + c$)
The standard technique for factorising a quadratic polynomial $ax^2 + bx + c$ when direct identities are not immediately applicable (like difference of squares or perfect squares) is by splitting the middle term ($bx$).
The method works as follows:
Identify Coefficients:
Identify the values of the coefficients $a, b,$ and $c$ in the quadratic polynomial $ax^2 + bx + c$.Calculate Product $ac$:
Calculate the product of the leading coefficient $a$ and the constant term $c$.Find Two Numbers:
Find two numbers, let's call them $p$ and $q$, such that their sum is equal to the coefficient of the middle term ($b$), and their product is equal to the product $ac$ calculated in the previous step.We need $p + q = b$
And $p \times q = a \times c $
Split the Middle Term:
Rewrite the original trinomial $ax^2 + bx + c$ by replacing the middle term $bx$ with the sum of the two terms $px + qx$. The expression becomes a four-term polynomial: $ax^2 + px + qx + c$. (The order of $px$ and $qx$ does not affect the final result).Factor by Grouping:
Factorise the resulting four-term polynomial by grouping. Group the first two terms and the last two terms: $(ax^2 + px) + (qx + c)$. Factor out the greatest common factor (GCF) from each pair. If done correctly, you should obtain a common binomial factor in both resulting terms.Factor Out the Common Binomial:
Factor out the common binomial factor from the grouped expression. The remaining terms form the second factor.
The resulting expression is the factored form of the quadratic polynomial.
Example 1. Factorise the quadratic polynomial $x^2 - 10x + 25$.
Answer:
The polynomial is $x^2 - 10x + 25$. It is in the form $ax^2 + bx + c$, with $a=1, b=-10,$ and $c=25$.
Step 1 & 2: Identify $a, b, c$ and calculate $ac$. $a=1, b=-10, c=25$. $ac = 1 \times 25 = 25$.
Step 3: Find two numbers $p$ and $q$ such that $p + q = -10$ and $p \times q = 25$.
Consider pairs of integer factors of $ac = 25$: $(1, 25), (5, 5), (-1, -25), (-5, -5)$.
Check which pair adds up to $b = -10$: $1+25=26$, $5+5=10$, $-1+(-25)=-26$, $-5+(-5)=-10$. The pair $\textbf{(-5, -5)}$ works.
Step 4: Rewrite the middle term $-10x$ as $-5x - 5x$. The expression becomes:
$$ x^2 - 5x - 5x + 25 $$Step 5: Factorise by grouping:
$$ (x^2 - 5x) + (-5x + 25) $$Factor out the GCF from each group:
- In $(x^2 - 5x)$, the GCF is $x$. Factoring gives $x(x - 5)$.
- In $(-5x + 25)$, the GCF is $-5$ (factor out $-5$ to make the binomial match). Factoring gives $-5(x - 5)$.
Substitute the factored groups back:
$$ x(x - 5) - 5(x - 5) $$Step 6: Factor out the common binomial $(x - 5)$:
$$ (x - 5)(x - 5) $$This can be written as $(x - 5)^2$.
The factored form is $\textbf{(x - 5)\textsuperscript{2}}$.
Alternate Approach (Using Identity):
We could also recognise $x^2 - 10x + 25$ as a perfect square trinomial $a^2 - 2ab + b^2$, where $a=x$ and $b=5$. The identity is $a^2 - 2ab + b^2 = (a-b)^2$. Applying this directly gives $(x-5)^2$. This highlights how recognizing identity forms can simplify factorisation.
Example 2. Factorise the quadratic polynomial $4x^2 - 8x + 3$.
Answer:
The polynomial is $4x^2 - 8x + 3$. It is in the form $ax^2 + bx + c$, with $a=4, b=-8,$ and $c=3$. Here $a \neq 1$.
Step 1 & 2: Identify $a, b, c$ and calculate $ac$. $a=4, b=-8, c=3$. $ac = 4 \times 3 = 12$.
Step 3: Find two numbers $p$ and $q$ such that $p + q = -8$ and $p \times q = 12$.
Since the product $pq = 12$ is positive, $p$ and $q$ must have the same sign. Since the sum $p+q = -8$ is negative, both numbers must be negative.
Let's list pairs of negative integer factors of $ac = 12$: $(-1, -12), (-2, -6), (-3, -4)$.
Check which pair adds up to $b = -8$: $-1 + (-12) = -13$, $-2 + (-6) = -8$. The pair $\textbf{(-2, -6)}$ works.
So, $p=-2$ and $q=-6$.
Step 4: Rewrite the middle term $-8x$ as $-2x - 6x$. The expression becomes:
$$ 4x^2 - 2x - 6x + 3 $$Step 5: Factorise this four-term expression by grouping. Group the first two and the last two terms:
$$ (4x^2 - 2x) + (-6x + 3) $$Factor out the GCF from each group:
- In $(4x^2 - 2x)$, the GCF is $2x$. Factoring gives $2x(2x - 1)$.
- In $(-6x + 3)$, the GCF is $-3$ (factor out $-3$ to make the binomial factor $(2x-1)$ match the first group). Factoring gives $-3(2x - 1)$.
Substitute the factored groups back:
$$ 2x(2x - 1) - 3(2x - 1) $$Step 6: Factor out the common binomial $(2x - 1)$:
$$ (2x - 1)(2x - 3) $$The factored form is $\textbf{(2x - 1)(2x - 3)}$.
Check: Expand $(2x-1)(2x-3) = 2x(2x-3) - 1(2x-3) = 4x^2 - 6x - 2x + 3 = 4x^2 - 8x + 3$. Correct.
Factorising quadratic polynomials is a fundamental skill in algebra. It is often a precursor to finding the roots of quadratic equations, simplifying rational expressions, and analysing the properties of quadratic functions.
Factorisation of Cubic Polynomials
Approaches to Factorising Cubic Polynomials ($P(x) = ax^3 + bx^2 + cx + d$)
A cubic polynomial is a polynomial of degree $3$, generally written in the form $ax^3 + bx^2 + cx + d$, where $a, b, c,$ and $d$ are constants and $a \neq 0$. Factorising cubic polynomials can be more challenging than factorising quadratics because there is no single formula like the quadratic formula that directly gives the factors in a simple form (although there are formulas for the roots of cubic equations, they are very complex).
The most common and general approach to factorising cubic polynomials involves finding at least one linear factor, usually of the form $(x-c)$, and then using polynomial division to reduce the cubic to a linear factor and a quadratic factor. The quadratic factor can then be factorised using the methods discussed in the previous sections.
Methods for Factorising Cubic Polynomials
Here are the main methods used to factorise cubic polynomials:
Method 1: Using the Factor Theorem (Finding at least one zero)
This is the most general method. It relies on finding one root (zero) of the polynomial, which then gives a corresponding linear factor by the Factor Theorem. Once one linear factor is found, we can divide the cubic polynomial by this factor to obtain a quadratic polynomial, which we can then factorise.
Steps:
Find a Value $c$ such that $P(c) = 0$:
Try to find a number $c$ such that when you substitute $x=c$ into the polynomial $P(x)$, the result is $0$. If such a number $c$ is found, then by the Factor Theorem, $(x-c)$ is a factor of $P(x)$.For polynomials with integer coefficients, the Rational Root Theorem can help identify potential rational zeros to test. It states that if a rational number $p/q$ (in lowest terms) is a root, then $p$ must be a divisor of the constant term ($d$) and $q$ must be a divisor of the leading coefficient ($a$). For simple cubics, especially when the leading coefficient is 1, testing small integer divisors of the constant term ($\pm 1, \pm 2, \pm 3, \ldots$) is often successful.
Use Polynomial Division:
Once a zero $c$ is found, divide the polynomial $P(x)$ by the linear factor $(x-c)$ using polynomial long division (or synthetic division, which is a shortcut for dividing by a linear factor). Since $(x-c)$ is a factor, the remainder of this division will be $0$. The quotient will be a quadratic polynomial. Let the quotient be $Q(x)$.Factorise the Quadratic Quotient:
The polynomial $P(x)$ can now be written as $P(x) = (x-c)Q(x)$. The quotient $Q(x)$ is a quadratic polynomial (degree 2). Factorise $Q(x)$ using the methods for quadratic factorisation (splitting the middle term, identities, etc.).Write the Complete Factorisation:
The complete factorisation of the cubic polynomial $P(x)$ is the product of the linear factor $(x-c)$ and the factors of the quadratic quotient $Q(x)$.
Example 1. Factorise the cubic polynomial $P(x) = x^3 - 6x^2 + 11x - 6$.
Answer:
The polynomial is $P(x) = x^3 - 6x^2 + 11x - 6$. Here $a=1$ and the constant term $d=-6$.
Step 1: Find a value $c$ such that $P(c) = 0$. Let's test small integer divisors of the constant term $-6$: $\pm 1, \pm 2, \pm 3, \pm 6$.
- Test $x=1$: $P(1) = (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6(1) + 11 - 6 = 1 - 6 + 11 - 6 = 12 - 12 = 0$.
Since $P(1) = 0$, by the Factor Theorem, $(x - 1)$ is a factor of $P(x)$. So, $c=1$ is a zero.
Step 2: Divide $P(x)$ by $(x - 1)$ using polynomial long division:
$$ \begin{array}{r} x^2 - 5x + 6 \\ x-1{\overline{\smash{\big)}\,x^3-6x^2+11x-6}} \\ \underline{-~\phantom{(}(x^3-x^2)\phantom{+11x-6)}} \\ -5x^2+11x\phantom{-6} \\ \underline{-~\phantom{()}(-5x^2+5x)\phantom{-6}} \\ 6x-6 \\ \underline{-~\phantom{()}(6x-6)} \\ 0 \end{array} $$The quotient polynomial is $Q(x) = x^2 - 5x + 6$, and the remainder is $R(x) = 0$.
So, $P(x) = (x - 1)(x^2 - 5x + 6)$.
Step 3: Factorise the quadratic quotient $Q(x) = x^2 - 5x + 6$. This is a trinomial of the form $x^2+bx+c$ where $b=-5$ and $c=6$. We need two numbers that multiply to $6$ and add to $-5$. These numbers are $-2$ and $-3$.
$$ x^2 - 5x + 6 = (x - 2)(x - 3) $$Step 4: Write the complete factorisation of $P(x)$. It is the product of the linear factor found in Step 1 and the factors of the quadratic quotient from Step 3.
$$ P(x) = (x - 1)(x - 2)(x - 3) $$The factors of the cubic polynomial are $\textbf{(x - 1), (x - 2)}$, and $\textbf{(x - 3)}$.
Method 2: Using Sum/Difference of Cubes Identities (for specific forms)
If a cubic polynomial is a binomial (two terms) that fits the structure of a sum of cubes or a difference of cubes, it can be factorised directly using the corresponding identity:
$a^3 + b^3 = (a + b)(a^2 - ab + b^2) $
[Sum of Cubes Identity]
$a^3 - b^3 = (a - b)(a^2 + ab + b^2) $
[Difference of Cubes Identity]
Here $a$ and $b$ can be variables or constants whose cubes form the terms of the binomial.
Example 2. Factorise the cubic polynomial $x^3 + 125$.
Answer:
The expression is $x^3 + 125$. This is a binomial (two terms). We check if the terms are perfect cubes. $x^3$ is the cube of $x$, and $125$ is the cube of $5$ ($5^3 = 125$). The terms are added, so it's a sum of cubes.
It matches the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ with $a=x$ and $b=5$.
Apply the identity:
$$ x^3 + 125 = x^3 + 5^3 = (x + 5)(x^2 - x(5) + 5^2) $$Simplify the terms in the trinomial factor:
$$ = (x + 5)(x^2 - 5x + 25) $$The quadratic factor $x^2 - 5x + 25$ cannot be factored further into linear factors with real coefficients because its discriminant ($b^2 - 4ac$) is $(-5)^2 - 4(1)(25) = 25 - 100 = -75$, which is negative. It has complex roots. It is irreducible over real numbers.
The factors over real numbers are $\textbf{(x + 5)}$ and $\textbf{(x\textsuperscript{2} - 5x + 25)}$.
Method 3: Factorising by Grouping (for specific cubics)
Some cubic polynomials with four terms can be factorised using the grouping method, similar to how we factorise certain four-term quadratic-form expressions.
Steps:
- Group the four terms into two pairs.
- Factor out the GCF from each pair.
- If a common binomial factor appears, factor it out.
- If one of the resulting factors is a quadratic, check if it can be factored further.
Example 3. Factorise the cubic polynomial $x^3 - 2x^2 - 9x + 18$.
Answer:
The expression is $x^3 - 2x^2 - 9x + 18$. Four terms, no overall common factor.
Step 1: Group the terms. Let's group the first two and the last two:
$$ (x^3 - 2x^2) + (-9x + 18) $$Step 2: Factor out the GCF from each group:
- In the first group $(x^3 - 2x^2)$, the GCF is $x^2$. Factoring gives $x^2(x - 2)$.
- In the second group $(-9x + 18)$, the GCF is $-9$. Factoring gives $-9(x - 2)$. (Factoring out $-9$ makes the binomial $(x-2)$ match the first group).
Substitute the factored groups back:
$$ x^2(x - 2) - 9(x - 2) $$Step 3: The binomial $(x - 2)$ is common to both terms. Factor out the common binomial:
$$ (x - 2)(x^2 - 9) $$Step 4: The second factor is $x^2 - 9$. This is a quadratic expression. We recognise it as a difference of squares: $x^2 - 3^2$. It can be factored further using the identity $a^2 - b^2 = (a+b)(a-b)$ with $a=x$ and $b=3$.
$$ x^2 - 9 = (x + 3)(x - 3) $$Substitute this factored form back into the overall expression:
$$ (x - 2)(x + 3)(x - 3) $$The factors of the cubic polynomial are $\textbf{(x - 2), (x + 3)}$, and $\textbf{(x - 3)}$.
Verification (Optional):
The zeros corresponding to these factors are $x=2, x=-3,$ and $x=3$. Let's check if $P(2)$, $P(-3),$ and $P(3)$ are zero.
$P(2) = (2)^3 - 2(2)^2 - 9(2) + 18 = 8 - 2(4) - 18 + 18 = 8 - 8 - 18 + 18 = 0$.
$P(-3) = (-3)^3 - 2(-3)^2 - 9(-3) + 18 = -27 - 2(9) + 27 + 18 = -27 - 18 + 27 + 18 = 0$.
$P(3) = (3)^3 - 2(3)^2 - 9(3) + 18 = 27 - 2(9) - 27 + 18 = 27 - 18 - 27 + 18 = 0$.
All three values are zeros, confirming the factors.
In summary, factorising cubic polynomials often involves finding one root (usually by testing likely candidates or using the Rational Root Theorem), using polynomial division to get a quadratic quotient, and then factorising the quadratic quotient. Specific forms might allow using sum/difference of cubes identities or factorisation by grouping directly.