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Factorising quadratics

Factorising quadratics

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Tutor: Bilal

Summary

Factorising quadratics

In a nutshell

A quadratic in the form ax2+bx+cax^2+bx+c can be factorised. You have already learnt how to factorise a quadratic when a=1a=1 by finding two numbers to multiply to give cc and add to give bb​. This lesson will also discuss how to factorise a quadratic expression where a>1a>1.



Factorise when a=1a=1​​

A quadratic is in the form

ax2+bx+cax^2 + bx +c​​

When a=1a=1 we can factorise a quadratic as follows:


PROCEDURE

​1.

Ensure the quadratic is in the form ax2+bx+cax^2 + bx + c and identify a,ba,b and cc​​

2.

Set up the answer with two brackets and an xx at the start of each bracket, as follows (x)(x)(x \qquad)(x \qquad)​​

3.

Find two numbers which multiply to give cc and add to give bb​​

4.

Fill these numbers in the double brackets 


Example 1

Factorise 

x2+10x+24x^2 + 10x + 24


The quadratic is in the correct form where b=10b=10 and c=24c=24.

6×4=24 and 6+4=106 \times 4 = 24 \space and \space 6 + 4 = 10, so the numbers are 6 and 46 \space and \space 4.

x2+10x+24=(x+6)(x+4)x^2 + 10x+24 = \underline{(x+6)(x+4)}



Factorise when a>1a>1

​When a>1a>1 you can factorise a quadratic as follows:


PROCEDURE

​1.

Ensure the quadratic is in the form ax2+bx+cax^2+bx+c, and identify a,ba, b and cc.

2.

Multiply aa and cc to give a new number dd.​

3.

Think of two numbers, mm and nn which multiply to give dd and add to give bb.​

​4.

Write the quadratic using mm and nn in the form ax2+mx+nx+cax^2+mx+nx+c.

​5.

Factorise ax2+mxax^2+mx into single brackets.

​6.

Factorise nx+cnx+c into single brackets.

​7.

Use both factorised single brackets expressions to find the double brackets expression. The first bracket consists of the terms outside the single brackets, and the second bracket consists of the expression inside the single brackets.


Example 2

Factorise

2x2+13x+152x^2+13x+15​​



The quadratic has a=2,b=13a=2, b=13 and c=15c=15. Multiply aa and cc to give 2×15=302 \times 15 = 30.

Think of two numbers which will multiply to give 3030 and add to give 1313. The two numbers are 1010 and 33, as 10×3=3010 \times 3 = 30 and 10+3=1310+3 = 13.

Write the quadratic expression using 1010 and 33, and factorise.

2x2+13x+152x2+10xfactorise+3x+15factorise2x(x+5)+3(x+5)\begin {aligned}&2x^2+13x+15 \\&\underbrace{2x^2 + 10x}_{\text{factorise}} + \underbrace{3x + 15}_{\text{factorise}} \\&2x(x+5) + 3(x+5)\end {aligned}​​


The two expressions inside both single brackets should match, in this case x+5x+5.

The final answer consists of two brackets, the first bracket contains the terms on the outside of each bracket, 2x2x and +3+3. And the second bracket is the repeated expression inside the single brackets factorised.

(2x+3)(x+5)\underline{(2x+3)(x+5)}​​


Example 3

Factorise and solve

3x2+11x4=03x^2+11x-4=0​​


The quadratic has a=3,b=11a=3, b=11 and c=4c=-4. Multiply aa and cc to give 12-12.

Think of two numbers which will multiply to give 12-12 and add to give +11+11. The two numbers are +12+12 and 1-1, as 12×1=1212\times -1 = -12 and 121=1112-1=11.

Write the quadratic expression using 1212 and 1-1, and factorise.

3x2+11x4=03x2+12xfactorise 1x4factorise=03x(x+4)1(x+4)=0(3x1)(x+4)=0\begin {aligned}3x^2+11x-4&=0 \\\underbrace{3x^2+12x}_{\text{factorise}}\space \underbrace{-1x-4}_{\text{factorise}}&=0 \\3x(x+4) -1(x+4)&=0 \\(3x-1)(x+4)&=0\end {aligned}​​


Now the quadratic is factorised, it can be solved. Find the solutions by making either bracket equal to 00.

x=13 or x=4\underline{x=\frac 1 3 \space or \space x = -4}​​

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FAQs - Frequently Asked Questions

How do you factorise a quadratic ax^2+bx+c when a>1?

How do you factorise into double brackets?

What do double brackets mean?

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