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Class 8th Chapters
1. Rational Numbers	2. Linear Equations in One Variable	3. Understanding Quadrilaterals
4. Practical Geometry	5. Data Handling	6. Squares and Square Roots
7. Cubes and Cube Roots	8. Comparing Quantities	9. Algebraic Expressions and Identities
10. Visualising Solid Shapes	11. Mensuration	12. Exponents and Powers
13. Direct and Inverse Proportions	14. Factorisation	15. Introduction to Graphs
16. Playing with Numbers

Content On This Page
Factors of Algebraic Expressions	Factorising by Taking Out Common Factors	Factorising by Grouping of Terms
Factorising using Algebraic Identities	Factorising of Trinomials	Division of Polynomial by Polynomial

Chapter 14 Factorisation (Concepts)

Welcome to the pivotal chapter on Factorisation! Think of this as the art of 'un-multiplying' algebraic expressions. In previous chapters, particularly the one on Algebraic Expressions and Identities, we spent considerable time learning how to multiply expressions together – for example, using the distributive property to expand $2x(x+2)$ into $2x^2 + 4x$, or multiplying binomials like $(x+y)(a+b)$ to get $xa+xb+ya+yb$. Factorisation is precisely the reverse process. Given an algebraic expression, our goal is to break it down into a product of simpler expressions (its factors), which, when multiplied back together, yield the original expression. These factors can be numbers, variables, or simpler algebraic expressions themselves. Mastering factorisation is a cornerstone skill in algebra, essential for simplifying complex expressions, solving polynomial equations, and working efficiently with fractions involving algebraic terms (rational expressions) in more advanced mathematics.

We begin with the most fundamental technique: finding the Common Factors. This method involves scrutinizing each term within a given expression to identify the Greatest Common Factor (GCF) – the largest monomial that divides exactly into every single term. Once the GCF is identified, we use the distributive property ($a(b+c) = ab+ac$) in reverse. We 'pull out' the GCF from each term, leaving the remaining factors inside parentheses. For instance, consider the expression $6a^2b + 9ab^2$. The GCF of the terms $6a^2b$ and $9ab^2$ is $3ab$. Factoring this out gives us $3ab(2a) + 3ab(3b)$, which simplifies to $3ab(2a + 3b)$. Verifying by multiplication: $3ab \times 2a = 6a^2b$ and $3ab \times 3b = 9ab^2$. Success!

Sometimes, an expression might not have a factor common to all its terms. In such cases, we might employ Factorisation by Regrouping Terms. This technique is particularly useful for expressions with four terms. We group the terms strategically, usually in pairs, such that each pair shares a common factor. After factoring out the common factor from each pair, we aim to reveal a common binomial factor across the groups, which can then be factored out. Consider $ax + bx + ay + by$. Grouping the first two and last two terms gives $(ax + bx) + (ay + by)$. Factoring out the common factor from each pair yields $x(a+b) + y(a+b)$. Now, we observe that $(a+b)$ is a common binomial factor. Factoring this out leaves us with the final factorised form: $(a+b)(x+y)$.

A major focus lies in leveraging the Algebraic Identities we learned previously, but applying them in reverse to factorise expressions efficiently. Recognizing patterns corresponding to these identities is key:

If an expression is in the form $a^2 + 2ab + b^2$, it factorises directly into $(a + b)^2$.
If an expression matches the form $a^2 - 2ab + b^2$, it factorises into $(a - b)^2$.
An expression in the form of a difference of two squares, $a^2 - b^2$, immediately factorises into $(a + b)(a - b)$.

We also extend this to trinomials of the form $x^2 + px + q$. This often involves finding two numbers, say $a$ and $b$, such that their sum is the coefficient of the middle term ($a+b = p$) and their product is the constant term ($ab = q$). If such numbers are found, the expression factorises according to the identity $x^2 + (a+b)x + ab = (x+a)(x+b)$. This technique is often referred to as 'splitting the middle term'.

Finally, the chapter establishes the intrinsic link between factorisation and the division of algebraic expressions. When dividing polynomials, factorisation often provides the simplest route. We cover:

Division of a monomial by another monomial (dividing coefficients, subtracting exponents of like variables).
Division of a polynomial by a monomial (dividing each term of the polynomial by the monomial).
Division of a polynomial by another polynomial. In many cases, this is effectively handled by factorising both the numerator (dividend) and the denominator (divisor) completely, and then cancelling out any common factors present in both. For example, to divide $(x^2 + 5x + 6)$ by $(x+2)$, we factorise the numerator as $(x+2)(x+3)$. The division becomes $\frac{(x+2)(x+3)}{(x+2)}$. Cancelling the common factor $(x+2)$ leaves the result $(x+3)$.

Factorisation is thus presented not just as a standalone technique but as a fundamental precursor to simplifying rational expressions and solving equations.

Factors of Algebraic Expressions

In previous chapters, you have learned about algebraic expressions, their terms, and operations like addition, subtraction, and multiplication. Now, we will explore the concept of factorisation for algebraic expressions. Factorisation is an inverse process of multiplication.

Factors of Natural Numbers (Revision)

Let's recall the concept of factors from arithmetic. For a natural number, its factors are the numbers that divide it exactly, leaving no remainder. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12.

We can express a number as a product of its factors. For instance, $12 = 3 \times 4$. Here, 3 and 4 are factors of 12. We can also write $12 = 2 \times 6$, so 2 and 6 are factors. We can continue breaking down composite factors until we are left only with prime factors. This is called prime factorisation. For 12, the prime factorisation is $12 = 2 \times 2 \times 3$. Here, 2, 2, and 3 are prime factors of 12.

Factors of Algebraic Expressions

Similar to natural numbers, algebraic expressions can also be written as a product of expressions. When an algebraic expression is written as a product of two or more expressions, each of these expressions in the product is called a factor of the given algebraic expression.

Example: Consider the algebraic expression $5x$. It can be written as a product of 5 and $x$. So, 5 and $x$ are factors of $5x$. We can say $5x = 5 \times x$.

Example: Consider the expression $3xy$. It can be written as $3 \times x \times y$. The factors are 3, $x$, and $y$.

Example: Consider the expression $x(x+2)$. This expression is already written as a product of two expressions, $x$ and $(x+2)$. So, $x$ and $(x+2)$ are factors of $x(x+2)$.

Example: Consider the expression $x^2 + 2x$. We can rewrite this expression by taking $x$ common from both terms: $x^2 + 2x = x(x+2)$. Now the expression is written as a product of $x$ and $(x+2)$. So, $x$ and $(x+2)$ are factors of $x^2 + 2x$.

The process of writing an algebraic expression as a product of its factors is called factorisation.

When we factorise an algebraic expression, we look for factors that are typically numbers (constants) or other algebraic expressions (with integer coefficients). We usually aim to break down the expression as much as possible into simpler factors.

Irreducible Factors

When factorising an algebraic expression, we continue the process until the factors cannot be factored further into simpler algebraic expressions (except for numerical factors like 2, 3, -1, etc.). These factors are called irreducible factors.

Think of irreducible factors as the algebraic equivalent of prime numbers in arithmetic. Prime numbers (like 2, 3, 5, 7) cannot be factored into smaller integers (other than 1 and themselves). Similarly, irreducible factors cannot be factored into simpler algebraic expressions.

Example: Consider the expression $12x^2y$. We can factorise it as a product of its prime and variable factors:

$12x^2y = (2 \times 2 \times 3) \times (x \times x) \times y$

The irreducible factors of $12x^2y$ are 2, 2, 3, $x$, $x$, and $y$. None of these can be broken down further into simpler algebraic factors.

Example: Consider the expression $x(x+2)$. This is already in factored form.

$x(x+2) = x \times (x+2)$

The factors are $x$ and $(x+2)$. Can $x$ be factored further? Not into algebraic expressions of lower degree. Can $(x+2)$ be factored further? No, not into simpler expressions with integer coefficients. So, $x$ and $(x+2)$ are irreducible factors of $x(x+2)$.

Example: Consider the expression $x^2 - 4$. We know the identity $a^2 - b^2 = (a+b)(a-b)$.

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

The factors are $(x+2)$ and $(x-2)$. These are irreducible factors because they cannot be factored further into simpler algebraic expressions with integer coefficients.

The goal of factorisation is usually to express an algebraic expression as a product of its irreducible factors.

Factorising by Taking Out Common Factors

Factorisation is the process of writing an algebraic expression as a product of its factors. The simplest method of factorisation involves identifying and taking out common factors from the terms of the expression. This method is based on the distributive property of multiplication in reverse.

Using the Distributive Property in Reverse

We know the distributive property: $a(b+c) = ab + ac$. If we look at this equality from right to left, $ab + ac = a(b+c)$, we see that if a factor 'a' is common to both terms 'ab' and 'ac', we can "take out" the common factor 'a' and write the expression as the product of 'a' and the remaining expression $(b+c)$. Similarly, $ab - ac = a(b-c)$.

To factorise an expression by taking out common factors, we need to find the Greatest Common Factor (GCF) or Highest Common Factor (HCF) of all the terms in the expression. The GCF is the product of the GCF of the numerical coefficients and the lowest power of each common variable.

Steps to factorise by taking out common factors:

Identify the terms in the expression.
Find the GCF of the numerical coefficients of all the terms.
Find the common variables among all the terms. For each common variable, find the lowest power it has in any of the terms. The product of these lowest powers is the common variable factor.
The GCF of the terms is the product of the GCF of the numerical coefficients and the common variable factor (if any).
Rewrite each term of the expression as a product of the GCF and the remaining factors.
Use the distributive property in reverse to write the expression as the product of the GCF and the sum/difference of the remaining factors.

Example 1. Factorise $5x + 10y$.

Answer:

The terms are $5x$ and $10y$.

Step 1 & 2: Find the GCF of the numerical coefficients 5 and 10. The GCF of 5 and 10 is 5.

Step 3: Find the common variable factor. The variable $x$ is only in the first term. The variable $y$ is only in the second term. There are no variables common to both terms. The common variable factor is 1.

Step 4: The GCF of the terms is the GCF of numerical coefficients times the common variable factor $= 5 \times 1 = 5$.

Step 5: Rewrite each term as a product of the GCF and the remaining factor:

$5x = 5 \times x$

$10y = 5 \times 2y$

Step 6: Use the distributive property in reverse $ab+ac = a(b+c)$ where $a=5$, $b=x$, $c=2y$:

$5x + 10y = 5(x + 2y)$

The factorised form is $5(x + 2y)$. The factors are 5 and $(x+2y)$.

Example 2. Factorise $6a^2 - 9ab$.

Answer:

The given expression is $6a^2 - 9ab$.

To factorise this expression, we will use the method of common factors. This involves finding the highest common factor (HCF) of the terms in the expression and then using the distributive law.

Step 1: Write each term as a product of its irreducible factors.

The first term is $6a^2$. Its factors are:

$6a^2 = 2 \times 3 \times a \times a$

The second term is $9ab$. Its factors are:

$9ab = 3 \times 3 \times a \times b$

Step 2: Identify the common factors in both terms.

Comparing the factorizations:

$6a^2 = 2 \times \underline{3} \times \underline{a} \times a$

$9ab = 3 \times \underline{3} \times \underline{a} \times b$

The common factors are $3$ and $a$.

Step 3: Determine the Highest Common Factor (HCF).

The HCF is the product of the common factors.

HCF = $3 \times a = 3a$

Step 4: Rewrite the expression by taking out the HCF.

We can rewrite each term of the original expression as a product of the HCF and the remaining factors.

$6a^2 = 3a \times (2a)$

$9ab = 3a \times (3b)$

Now, substitute these back into the expression and use the distributive law ($ac - ad = a(c-d)$):

$6a^2 - 9ab = (3a \times 2a) - (3a \times 3b)$

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

Thus, the factors of $6a^2 - 9ab$ are $3a$ and $(2a - 3b)$.

The factorised form is $3a(2a - 3b)$.

Example 3. Factorise $12x^2y + 18xy^2 - 6xy$.

Answer:

The given expression is $12x^2y + 18xy^2 - 6xy$.

We will factorise this trinomial by finding the highest common factor (HCF) of all three terms.

Step 1: Write the irreducible factors of each term.

The first term is $12x^2y$.

$12x^2y = 2 \times 2 \times 3 \times x \times x \times y$

The second term is $18xy^2$.

$18xy^2 = 2 \times 3 \times 3 \times x \times y \times y$

The third term is $6xy$.

$6xy = 2 \times 3 \times x \times y$

Step 2: Identify the common factors across all three terms.

Let's underline the common factors in each term's factorization:

$12x^2y = \underline{2} \times 2 \times \underline{3} \times \underline{x} \times x \times \underline{y}$

$18xy^2 = \underline{2} \times \underline{3} \times 3 \times \underline{x} \times \underline{y} \times y$

$6xy = \underline{2} \times \underline{3} \times \underline{x} \times \underline{y}$

The factors common to all three terms are $2$, $3$, $x$, and $y$.

Step 3: Determine the Highest Common Factor (HCF).

The HCF is the product of all the common factors.

HCF = $2 \times 3 \times x \times y = 6xy$

Step 4: Rewrite the expression using the distributive law.

We take the HCF, $6xy$, out as a common factor. To find the terms inside the bracket, we divide each original term by the HCF.

$12x^2y + 18xy^2 - 6xy = 6xy(\frac{12x^2y}{6xy} + \frac{18xy^2}{6xy} - \frac{6xy}{6xy})$

Performing the division for each term:

$\frac{12x^2y}{6xy} = 2x$
$\frac{18xy^2}{6xy} = 3y$
$\frac{6xy}{6xy} = 1$

Substituting these back, we get:

$ = 6xy(2x + 3y - 1)$

Thus, the factors of $12x^2y + 18xy^2 - 6xy$ are $6xy$ and $(2x + 3y - 1)$.

The factorised form is $6xy(2x + 3y - 1)$.

Factorising by Grouping of Terms

The method of taking out a common factor directly works only when a common factor exists in every term of the expression. For expressions that do not have such a common factor, but typically have four terms, we can often use the method of factorising by grouping of terms.

The core idea is to break the expression into smaller groups, factorise each group individually, and then find a common factor that emerges from these groups.

Steps for Factorising by Grouping

Group Terms: Arrange the terms of the expression into convenient groups, usually pairs. The key is to form groups that have their own common factor. Sometimes, you may need to rearrange the original terms to create suitable groups.
Factorise Each Group: For each group, find the Highest Common Factor (HCF) and take it out.
Identify the Common Binomial Factor: After factorising each group, you should see a common factor that is a binomial (an expression in parentheses). This is the crucial step; if you don't get a common binomial factor, you may need to try a different grouping.
Factor Out the Common Binomial: Take out the common binomial factor using the distributive property in reverse. The remaining factors from each group will form the second binomial factor.

Example 1. Factorise $ax + bx + ay + by$.

Answer:

Given expression: $ax + bx + ay + by$.

Notice there is no single factor common to all four terms.

Solution:

Step 1: Group the terms.

Let's group the first two terms together and the last two terms together.

$(ax + bx) + (ay + by)$

Step 2: Factorise each group.

In the first group, $(ax + bx)$, the common factor is $x$.

$ax + bx = x(a + b)$

In the second group, $(ay + by)$, the common factor is $y$.

$ay + by = y(a + b)$

Step 3: Identify the common binomial factor.

After factorising the groups, the expression becomes:

$x(a + b) + y(a + b)$

We can clearly see that $(a + b)$ is a common factor to both parts of the expression.

Step 4: Factor out the common binomial.

We treat $(a+b)$ as a single entity and factor it out.

$(a+b)(x + y)$

Thus, the factorised form is $(a+b)(x + y)$.

Alternate Grouping

Let's see what happens if we group the terms differently, for example, $(ax + ay) + (bx + by)$.

Factorising the new groups:

From $(ax+ay)$, the common factor is $a$, giving $a(x+y)$.

From $(bx+by)$, the common factor is $b$, giving $b(x+y)$.

The expression becomes $a(x+y) + b(x+y)$.

The common binomial is $(x+y)$, and factoring it out gives $(x+y)(a+b)$, which is the same result.

Example 2. Factorise $15xy - 6x + 5y - 2$.

Answer:

Given expression: $15xy - 6x + 5y - 2$.

Step 1 & 2: Group and Factorise.

Group the first two and last two terms: $(15xy - 6x) + (5y - 2)$.

Factor the first group. The HCF of $15xy$ and $6x$ is $3x$.

$15xy - 6x = 3x(5y - 2)$

Factor the second group. The terms $5y$ and $-2$ have no common factor other than 1. We can write this as:

$5y - 2 = 1(5y - 2)$

Step 3 & 4: Identify and Factor Out the Common Binomial.

The expression becomes:

$3x(5y - 2) + 1(5y - 2)$

The common binomial factor is $(5y - 2)$. Factoring it out, we get:

$(5y - 2)(3x + 1)$

The factorised form is $(5y - 2)(3x + 1)$.

Example 3. Factorise $p^2q - pr^2 - pq + r^2$.

Answer:

Given expression: $p^2q - pr^2 - pq + r^2$.

In this case, the initial grouping might not seem to work, so rearranging the terms is a good strategy.

Step 1: Rearrange and Group.

Let's rearrange the terms to group those with common factors together: $p^2q - pq - pr^2 + r^2$.

Now, group them into pairs:

$(p^2q - pq) + (-pr^2 + r^2)$

Step 2: Factorise Each Group.

In the first group, $(p^2q - pq)$, the common factor is $pq$.

$p^2q - pq = pq(p - 1)$

In the second group, $(-pr^2 + r^2)$, the common factor is $r^2$. To make the resulting binomial match $(p-1)$, we should factor out $-r^2$.

$-pr^2 + r^2 = -r^2(p - 1)$

(Check: $-r^2 \times p = -pr^2$ and $-r^2 \times -1 = +r^2$. It's correct.)

Step 3 & 4: Identify and Factor Out the Common Binomial.

The expression is now:

$pq(p - 1) - r^2(p - 1)$

The common binomial factor is $(p - 1)$. Factoring it out gives:

$(p - 1)(pq - r^2)$

The factorised form is $(p - 1)(pq - r^2)$.

Factorising using Algebraic Identities

In Chapter 9, we learned about standard algebraic identities which are equalities true for all values of the variables. These identities are very useful tools not only for multiplying expressions quickly but also for factorising certain types of expressions. We use the identities in reverse to write an expression as a product of factors.

Recall Standard Identities:

The standard identities we will use for factorisation are:

Identity I: Square of a sum: $(a+b)^2 = a^2 + 2ab + b^2$
Identity II: Square of a difference: $(a-b)^2 = a^2 - 2ab + b^2$
Identity III: Difference of squares: $(a+b)(a-b) = a^2 - b^2$

For factorisation, we read these identities from right to left:

Factorisation based on Identity I: $a^2 + 2ab + b^2 = (a+b)^2 = (a+b)(a+b)$
Factorisation based on Identity II: $a^2 - 2ab + b^2 = (a-b)^2 = (a-b)(a-b)$
Factorisation based on Identity III: $a^2 - b^2 = (a+b)(a-b)$

Factorising Trinomials of the form $a^2 \pm 2ab + b^2$

If a given algebraic expression is a trinomial (has three terms) and looks like the expansion of a perfect square binomial (Identities I or II), we can factorise it using these identities. We look for two terms that are perfect squares and check if the third term is twice the product of their square roots.

Steps to factorise $a^2 + 2ab + b^2$ or $a^2 - 2ab + b^2$:

Identify if two of the terms are perfect squares. These will be your $a^2$ and $b^2$ terms.
Find the square roots of these two terms. Let these be $a$ and $b$.
Check if the remaining term is equal to $2ab$ or $-2ab$.
If it is $+2ab$, the factorisation is $(a+b)^2$.
If it is $-2ab$, the factorisation is $(a-b)^2$.

Example 1. Factorise $x^2 + 6x + 9$.

Answer:

The expression is $x^2 + 6x + 9$. It is a trinomial.

Step 1: Check for perfect square terms. The first term $x^2$ is a perfect square $(x)^2$. The last term 9 is a perfect square $(3)^2$.

Step 2: Find the square roots. The square root of $x^2$ is $x$. The square root of 9 is 3. Let $a=x$ and $b=3$.

Step 3: Check the middle term. The middle term is $+6x$. Calculate $2ab = 2(x)(3) = 6x$. The middle term matches $+2ab$.

Step 4: Since the expression is of the form $a^2 + 2ab + b^2$, we use Identity I: $a^2 + 2ab + b^2 = (a+b)^2$.

Substitute $a=x$ and $b=3$:

$x^2 + 6x + 9 = (x+3)^2$

This means the factors are $(x+3)$ multiplied by itself.

$= (x+3)(x+3)$

The factorised form is $(x+3)^2$ or $(x+3)(x+3)$.

Example 2. Factorise $4p^2 - 20pq + 25q^2$.

Answer:

The expression is $4p^2 - 20pq + 25q^2$. It is a trinomial.

Step 1: Check for perfect square terms. The first term $4p^2$ is a perfect square $(2p)^2$. The last term $25q^2$ is a perfect square $(5q)^2$.

Step 2: Find the square roots. The square root of $4p^2$ is $2p$. The square root of $25q^2$ is $5q$. Let $a=2p$ and $b=5q$.

Step 3: Check the middle term. The middle term is $-20pq$. Calculate $2ab = 2(2p)(5q) = 20pq$. The middle term matches $-2ab$.

Step 5: Since the expression is of the form $a^2 - 2ab + b^2$, we use Identity II: $a^2 - 2ab + b^2 = (a-b)^2$.

Substitute $a=2p$ and $b=5q$:

$4p^2 - 20pq + 25q^2 = (2p - 5q)^2$

This means the factors are $(2p-5q)$ multiplied by itself.

$= (2p - 5q)(2p - 5q)$

The factorised form is $(2p-5q)^2$ or $(2p-5q)(2p-5q)$.

Factorising Expressions of the form $a^2 - b^2$

If a given algebraic expression is a binomial (has two terms) and can be written as the difference of two perfect squares, we can factorise it using Identity III.

Steps to factorise $a^2 - b^2$:

Identify if the expression is a binomial with a subtraction sign between the two terms.
Check if the first term is a perfect square. Find its square root ($a$).
Check if the second term is a perfect square. Find its square root ($b$).
If both terms are perfect squares and they are separated by a subtraction sign, the factorisation is $(a+b)(a-b)$.

Example 3. Factorise $m^2 - 49$.

Answer:

The expression is $m^2 - 49$. It is a binomial with a subtraction sign.

Step 1 & 2: The first term $m^2$ is a perfect square $(m)^2$. Its square root is $m$. Let $a=m$.

Step 3: The second term 49 is a perfect square $(7)^2$. Its square root is 7. Let $b=7$.

The expression is the difference of two perfect squares: $m^2 - 7^2$.

Step 4: Using Identity III: $a^2 - b^2 = (a+b)(a-b)$.

Substitute $a=m$ and $b=7$:

$m^2 - 49 = (m+7)(m-7)$

The factors are $(m+7)$ and $(m-7)$.

Example 4. Factorise $16x^2 - 81y^2$.

Answer:

The expression is $16x^2 - 81y^2$. It is a binomial with a subtraction sign.

Step 1 & 2: The first term $16x^2$ is a perfect square $(4x)^2$ because $(4x)^2 = 4^2 \times x^2 = 16x^2$. Its square root is $4x$. Let $a=4x$.

Step 3: The second term $81y^2$ is a perfect square $(9y)^2$ because $(9y)^2 = 9^2 \times y^2 = 81y^2$. Its square root is $9y$. Let $b=9y$.

The expression is the difference of two perfect squares: $(4x)^2 - (9y)^2$.

Step 4: Using Identity III: $a^2 - b^2 = (a+b)(a-b)$.

Substitute $a=4x$ and $b=9y$:

$16x^2 - 81y^2 = (4x+9y)(4x-9y)$

The factors are $(4x+9y)$ and $(4x-9y)$.

Example 5. Factorise $(a+b)^2 - c^2$.

Answer:

The expression is $(a+b)^2 - c^2$. It is a binomial with a subtraction sign.

Step 1 & 2: The first term is $(a+b)^2$, which is a perfect square. Its square root is $(a+b)$. Let $X = (a+b)$.

Step 3: The second term is $c^2$, which is a perfect square. Its square root is $c$. Let $Y = c$.

The expression is in the form $X^2 - Y^2$, which is a difference of two perfect squares.

Step 4: Using Identity III: $X^2 - Y^2 = (X+Y)(X-Y)$.

Substitute $X = (a+b)$ and $Y = c$ back into the factorised form:

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

Simplify the expressions inside the outer parentheses:

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

The factors are $(a+b+c)$ and $(a+b-c)$.

Factorising of Trinomials

A trinomial is an algebraic expression with three terms. A common and very important type of trinomial is the quadratic trinomial of the form $ax^2 + bx + c$. In this section, we will focus on the simpler case where $a=1$, i.e., trinomials of the form $x^2 + Px + Q$. We can factorise these by reversing the process of multiplying two binomials, using a method called "splitting the middle term".

Factorising Trinomials of the form $x^2 + Px + Q$

Recall the multiplication of two binomials like $(x+a)$ and $(x+b)$:

$(x+a)(x+b) = x(x+b) + a(x+b) \ $$ = x^2 + bx + ax + ab = x^2 + (a+b)x + ab$

By comparing the result $x^2 + (a+b)x + ab$ with the general form $x^2 + Px + Q$, we can see a clear relationship:

The coefficient of the middle term, $P$, is the sum of $a$ and $b$. ($P = a+b$)
The constant term, $Q$, is the product of $a$ and $b$. ($Q = a \times b$)

Therefore, to factorise $x^2 + Px + Q$ into $(x+a)(x+b)$, our goal is to find two numbers, $a$ and $b$, that add up to $P$ and multiply to give $Q$.

Steps for Factorising by Splitting the Middle Term

Identify Coefficients: In the trinomial $x^2 + Px + Q$, identify the values of $P$ (the coefficient of $x$) and $Q$ (the constant term).
Find Two Numbers: Find two numbers, let's call them $a$ and $b$, that satisfy two conditions:
- Their product is $Q$ ($a \times b = Q$).
- Their sum is $P$ ($a + b = P$).
Tip for finding the numbers: Start by listing the pairs of factors of the constant term $Q$. Then, check which pair adds up to the middle term coefficient $P$. Pay close attention to the signs.
Split the Middle Term: Rewrite the middle term $Px$ as the sum of $ax$ and $bx$. The expression will become a four-term polynomial: $x^2 + ax + bx + Q$.
Factor by Grouping: Group the first two terms and the last two terms. Factor out the common factor from each group. You should be left with a common binomial factor.
Final Factorisation: Factor out the common binomial to get the final answer in the form $(x+a)(x+b)$.

Example 1. Factorise $x^2 + 5x + 6$.

Answer:

The trinomial is $x^2 + 5x + 6$.

Step 1: Identify P and Q.

Comparing with $x^2 + Px + Q$, we have $P=5$ and $Q=6$.

Step 2: Find two numbers that multiply to 6 and add to 5.

Let's list the pairs of factors of 6 and find their sum:

Factors of 6	Sum of Factors
1 and 6	$1+6=7$
2 and 3	$2+3=5$ (Correct pair)
-1 and -6	$-1+(-6)=-7$
-2 and -3	$-2+(-3)=-5$

The two numbers are 2 and 3.

Step 3: Split the middle term.

We rewrite $5x$ as $2x + 3x$.

$x^2 + 5x + 6 = x^2 + 2x + 3x + 6$

Step 4: Factor by grouping.

Group the terms: $(x^2 + 2x) + (3x + 6)$.

Factor out the common factor from each group:

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

Step 5: Factor out the common binomial.

The common binomial is $(x+2)$. Factoring this out, we get:

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

The factorised form is $(x+2)(x+3)$.

Example 2. Factorise $y^2 - 7y + 12$.

Answer:

The trinomial is $y^2 - 7y + 12$. Here, $P=-7$ and $Q=12$.

Step 1 & 2: Find two numbers that multiply to 12 and add to -7.

Since the product is positive ($+12$) and the sum is negative ($-7$), both numbers must be negative.

Let's list the negative factor pairs of 12:

(-1) and (-12) $\implies$ Sum = -13
(-2) and (-6) $\implies$ Sum = -8
(-3) and (-4) $\implies$ Sum = -7 (Correct pair)

The two numbers are -3 and -4.

Step 3: Split the middle term.

We rewrite $-7y$ as $-3y - 4y$.

$y^2 - 7y + 12 = y^2 - 3y - 4y + 12$

Step 4: Factor by grouping.

Group the terms: $(y^2 - 3y) + (-4y + 12)$.

Factor out the common factor from each group. Be careful with the negative sign in the second group.

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

Step 5: Factor out the common binomial.

The common binomial is $(y-3)$.

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

The factorised form is $(y-3)(y-4)$.

Example 3. Factorise $m^2 - m - 20$.

Answer:

The trinomial is $m^2 - m - 20$. Here, $P=-1$ and $Q=-20$.

Step 1 & 2: Find two numbers that multiply to -20 and add to -1.

Since the product is negative ($-20$), one number must be positive and the other negative. Since the sum is negative ($-1$), the negative number must have a larger absolute value.

Let's list factor pairs of 20 and find their difference (we need a difference of 1):

1 and 20 $\implies$ Difference is 19
2 and 10 $\implies$ Difference is 8
4 and 5 $\implies$ Difference is 1 (This looks promising)

We need the sum to be -1. With the numbers 4 and 5, this is achieved if we use $+4$ and $-5$. Let's check: $4 + (-5) = -1$ and $4 \times (-5) = -20$. This is correct.

The two numbers are 4 and -5.

Step 3: Split the middle term.

We rewrite $-m$ as $+4m - 5m$.

$m^2 - m - 20 = m^2 + 4m - 5m - 20$

Step 4: Factor by grouping.

$= (m^2 + 4m) + (-5m - 20)$

$= m(m + 4) - 5(m + 4)$

Step 5: Factor out the common binomial.

The common binomial is $(m+4)$.

$= (m+4)(m - 5)$

The factorised form is $(m+4)(m-5)$.

Division of Polynomial by Polynomial

We have already learned how to divide algebraic expressions where the divisor is a monomial (in Chapter 9) and by factoring the dividend and cancelling common factors when the divisor is a polynomial (in Section I5 of this chapter). This section reiterates and demonstrates how factorisation can be used to perform the division of a polynomial by another polynomial, specifically when the divisor is a factor of the dividend.

Division using Factorisation

If we can factorise the dividend (the polynomial being divided) into a product of factors, and one of these factors is the same as the divisor (the polynomial we are dividing by), then the quotient is simply the other factor(s) of the dividend.

This is the inverse relationship between multiplication and division. If $P(x) = D(x) \times Q(x)$, then $\frac{P(x)}{D(x)} = Q(x)$ (provided $D(x) \neq 0$). Here, $P(x)$ is the dividend, $D(x)$ is the divisor, and $Q(x)$ is the quotient.

The key is to factorise the dividend completely or at least find the factor(s) that include the divisor.

Example 1. Divide $x^2 + 5x + 6$ by $x+2$.

Answer:

We need to calculate $\frac{x^2 + 5x + 6}{x + 2}$. Assume $x+2 \neq 0$, i.e., $x \neq -2$.

Recall or factorise the dividend $x^2 + 5x + 6$. From Example 11 in Section I5, we factorised this trinomial by splitting the middle term:

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

Now substitute this factorised form into the division expression:

$\frac{x^2 + 5x + 6}{x + 2} = \frac{(x+2)(x+3)}{x + 2}$

Since we assumed $x+2 \neq 0$, we can cancel out the common factor $(x+2)$ from the numerator and the denominator:

$= \frac{\cancel{(x+2)}^{\normalsize 1}(x+3)}{\cancel{(x+2)}_{\normalsize 1}}$

$= x+3$

The quotient is $x+3$ (provided $x \neq -2$).

Example 2. Divide $a^2 - 49$ by $a-7$.

Answer:

We need to calculate $\frac{a^2 - 49}{a-7}$. Assume $a-7 \neq 0$, i.e., $a \neq 7$.

Factorise the dividend $a^2 - 49$. This is a difference of squares ($a^2 - 7^2$). From Example 8 in Section I4, we used Identity III:

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

Now substitute this factorised form into the division expression:

$\frac{a^2 - 49}{a - 7} = \frac{(a+7)(a-7)}{a - 7}$

Since we assumed $a-7 \neq 0$, we can cancel out the common factor $(a-7)$ from the numerator and the denominator:

$= \frac{(a+7)\cancel{(a-7)}^{\normalsize 1}}{\cancel{(a-7)}_{\normalsize 1}}$

$= a+7$

The quotient is $a+7$ (provided $a \neq 7$).

Example 3. Divide $12xy(2x+3y-1)$ by $6xy$.

Answer:

We need to calculate $\frac{12xy(2x+3y-1)}{6xy}$. Assume $6xy \neq 0$, i.e., $x \neq 0$ and $y \neq 0$.

The dividend is already in factored form as a product of $12xy$ and $(2x+3y-1)$. The divisor is $6xy$.

We can rewrite the expression and cancel the common factors in the numerical and variable parts of the monomials:

$\frac{12xy(2x+3y-1)}{6xy} = (\frac{12xy}{6xy}) \times (2x+3y-1)$

Simplify the monomial division:

$\frac{12xy}{6xy} = \frac{\cancel{12}^{\normalsize 2} \cancel{x}\cancel{y}}{\cancel{6}_{\normalsize 1} \cancel{x}\cancel{y}} = 2$

(assuming $x \neq 0, y \neq 0$)

Substitute this back into the expression:

$= 2 \times (2x+3y-1)$

The quotient is $2(2x+3y-1)$ (provided $x \neq 0, y \neq 0$). You can also distribute the 2 if needed: $4x + 6y - 2$.

When the dividend cannot be easily factored such that the divisor is one of the factors, we use the method of polynomial long division, which is similar to the long division of numbers. This method is typically covered in more detail in higher classes, but here is a simple example using the format:

Polynomial Long Division Example (from Chapter 9):

Example 7 (from Chapter 9). Divide $(y^2 + 7y + 10)$ by $(y + 2)$ using long division.

Answer:

We want to divide $y^2 + 7y + 10$ by $y + 2$. Assume $y+2 \neq 0$, i.e., $y \neq -2$.

Arrange in long division format:

$\begin{array}{r} \phantom{y+2\overline{\smash{\big)}\,}} y+5\phantom{)} \\ y+2{\overline{\smash{\big)}\,y^2+7y+10}} \\ \underline{-~\phantom{(}(y^2+2y)} \\ \phantom{0+ }5y+10 \\ \underline{-~\phantom{()}(5y+10)} \\ \phantom{0+0+}0 \end{array}$

Steps followed in the long division:

Divide the first term of the dividend ($y^2$) by the first term of the divisor ($y$): $\frac{y^2}{y} = y$. This is the first term of the quotient.
Multiply the divisor $(y+2)$ by the first term of the quotient ($y$): $y(y+2) = y^2 + 2y$.
Subtract this result from the dividend: $(y^2 + 7y) - (y^2 + 2y) = 5y$. Bring down the next term (+10). The new dividend is $5y+10$.
Divide the first term of the new dividend ($5y$) by the first term of the divisor ($y$): $\frac{5y}{y} = 5$. This is the next term of the quotient (+5).
Multiply the divisor $(y+2)$ by this term of the quotient (5): $5(y+2) = 5y + 10$.
Subtract this result from the new dividend: $(5y + 10) - (5y + 10) = 0$. The remainder is 0.

The quotient is $y+5$, and the remainder is 0.

Thus, $\frac{y^2 + 7y + 10}{y + 2} = y+5$ (provided $y \neq -2$).