Last updated: September 28, 2021

## How do you factorise quadratics?

In algebra, a polynomial consisting of variables and coefficients having the highest degree value of $$2$$ is termed as a quadratic polynomial. The general form of a quadratic polynomial is $$ax^2+bx+c$$, where $$a$$, $$b$$ and $$c$$ are real numbers.

There are various methods to factorise quadratic polynomial or quadratics. Some of them are discussed later in the article.

## E2.2: Factorise where possible expressions of the form: $$a^2x^2-b^2y^2$$, $$a^2+2ab+b^2$$, $$ax^2+bx+c$$.

### Factorisation of the quadratic polynomial of the form $$a^2x^2-b^2y^2$$

Consider $$(ax+by)(ax-by)$$. When the given product is expanded, it gives $$a^2x^2+axby-axby+b^2y^2$$. On further solving, the resultant becomes $$a^2x^2-b^2y^2$$. So, it is clear that the factorisation of the quadratic polynomial of the form $$a^2x^2-b^2y^2$$ is $$(ax+by)(ax-by)$$. It is also termed as the sum of difference method.

### Factorisation of the quadratic polynomial of the form $$a^2+2ab+b^2$$

The factorisation of the quadratic polynomial of the form $$a^2+2ab+b^2$$ is $$(a+b)^2$$. It is termed the perfect square trinomials method. To apply this method, first, check whether the two terms are perfect squares of $$a$$ and $$b$$, and the third term is twice the product of $$a$$ and $$b$$.

### Factorisation of the quadratic polynomial of the form $$ax^2+bx+c$$

This form of a quadratic polynomial can be factorised by using the middle term split method. For a quadratic polynomial of the form $$ax^2+bx+c$$, find two numbers $$p$$ and $$q$$ such that the product of $$p$$ and $$q$$ is equal to the product of $$a$$ and $$c$$, and the sum of $$p$$ and $$q$$ is equal to $$b$$. Then, write the polynomial in the form of $$ax^2+px+qx+c$$. After this, group the terms and get the desired factors.

### Worked examples of factorising quadratic polynomials

Example 1: Factorise $$25x^2-9y^2$$.

Step 1: Recall the factorisation of quadratic polynomial $$a^2x^2-b^2y^2$$

$$a^2x^2-b^2y^2 =(ax+by)(ax-by)$$

Step 2: Change the polynomial in appropriate form

$$25x^2-9y^2=5^2x^2-3^2y^2$$

Step 3: Compare it with the general form to get $$a$$ and $$b$$

$$5^2x^2-3^2y^2= a^2x^2-b^2y^2$$

On comparing, it is clear that $$a=5$$, $$b=3$$.

Step 4: Put the value of $$a$$ and $$b$$ in the factorisation of the general form to get the factors of the given polynomial

Thus, $$25x^2-9y^2=(5x+3y)(5x-3y)$$

Example 2: Find the factor of the quadratic polynomial $$x^2+8x+16$$.

Step 1: Recall the factorisation of quadratic polynomial of the form $$a^2+2ab+b^2$$

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

Step 2: Change the given polynomial in the appropriate form

$$x^2+8x+16=x^2+2\times x \times4+4^2$$

Step 3: Compare it with the general form to get the value of $$a$$, $$b$$

$$x^2+2\times x \times 4+4^2= a^2+2ab+b^2$$

On comparing, it is clear that $$a=x$$ and $$b=4$$.

Step 4: Put the value of $$a$$, $$b$$ in the factorisation of the general form to get the factors of the given polynomial

Thus, $$x^2+2\times x \times 4+4^2=(x+4)^2$$.

Example 3: Factorise the quadratic expression: $$x^2+7x+12$$.

Step 1: Recall the factorisation of quadratic polynomial of the form $$ax^2+bx+c$$

$$ax^2+bx+c=(mx+p)(nx+q)$$

Step 2: Find 2 numbers such that their product is $$12$$ and sum is $$7$$

The multiples of $$12$$ are $$1$$, $$2$$, $$3$$, $$4$$, $$6$$, $$12$$. From these multiples, the combination of $$4$$ and $$3$$ is suitable for the condition. Step 3: Write the equation by splitting the middle term in the form of $$ax^2+px+qx+c$$

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

Step 4: Take $$x$$ from the first two terms and $$3$$ from the other two terms and solve further

$$x^2+4x+3x+12=x(x+4)+3(x+4)$$

$$(x+4)(x+3)$$

Thus, the factorisation of $$x^2+7x+12$$ is $$(x+4)(x+3)$$.

Example 4: Factorise the quadratic expression: $$6x^2+7x-20$$.

Step 1: Recall the factorisation of quadratic polynomial of the form $$ax^2+bx+c$$

$$ax^2+bx+c=(mx+p)(nx+q)$$

Step 2: Find $$2$$ numbers such that their product is $$120$$ and difference is $$7$$

The multiples of $$120$$ are $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$8$$, $$10$$, $$12$$, $$15$$, $$20$$, $$24$$, $$30$$, $$40$$, $$60$$, $$120$$. From these multiples, the combination of $$15$$ and $$8$$ is suitable for the condition.

Step 3: Write the equation by splitting the middle term in the form of $$ax^2+px+qx+c$$

$$6x^2+7x-20=6x^2+15x-8x+20$$

Step 4: Take $$3x$$ from the first two terms and $$-4$$ from the other two terms and solve further

$$6x^2+15x-8x+20=3x(2x+5)-4(2x+5)$$

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

Thus, the factorisation of $$6x^2+7x-20$$ is $$(2x+5)(3x-4)$$.

The foremost significance in factorising quadratics that need to be noted is to always check thoroughly that the factors used for the term are correctly used and placed. To check whether the answer you found is correct or not, you can again multiply the factors derived if you get the result you started with.