Lesson 13: Factorising ax2+bx+c to double brackets

Introduction

You certainly need to be confident with Lesson 12 before attempting this. It will also help if you've completed Lesson 11 as the method I use for expanding double brackets links in nicely with this lesson.

So, the general form of a quadratic expression is \(~ax^2+bx+c~\), where \(~a~\), \(~b~\) and \(~c~\) can be any numbers (positive or negative). For the whole of Lesson 12, the value of \(~a~\) was 1. That's about to change!

SECTION A

OK, let's say we want to factorise \(~3x^2+14x-5~\).

In this case, \(~a=3~\), \(~b=14~\) and \(~c=-5~\)

Step 1

Find two magical numbers that multiply together to make \(~a\times c~\) and add to make \(~b~\).

Now \(~a\times c=3\times -5=-15~\) and \(~b=14~\). So we want two numbers that multiply together to make \(-\)15 and add together to make 14. After a bit of thinking and possibly some scribbling, you should come up with \(-\)1 and 15.

Step 2

Those two magical numbers tell you how to split the \(~x~\) term. That's all you need them for. In this case, they tell you to split \(~14x~\) into \(~-1x+15x~\) and it has to be split just so.

Now we can rewrite the question: \(~3x^2-1x+15x-5~\)

Important!

Those two new \(~x~\) term bits can be written either way round. So I could have written \(~3x^2+15x-1x-5~\) without it affecting my final answer. BUT if the two \(~x~\) term bits have different signs, write the negative one first. It's not essential but you will be less likely to make a mistake in step 3. I'm just sayin'!

Step 3

Factorise the first pair of terms:  \(~3x^2-1x~=~x(3x-1)\)

Then the second pair of terms:  \(~+15x-5~=~+5(3x-1)\)

And put back together: \(x(3x-1)+5(3x-1)\)

Important!

Both brackets must be identical. If they're not, you've either factorised wrongly or just not factorised fully. You must correct this before you continue.

Also, always expand the brackets back out (in your head is fine) to double check that your factorisation is correct. This is particularly useful for ensuring that signs are correct.

Step 4

Answer: \((x+5)(3x-1)\)

The first bracket is made from the 'outside' bits and the second bracket is a copy of one of the brackets in step 3. This step might seem like some kind of mysterious trick but it's actually not. If you want to understand it, revisit Lesson 11 and read under 'Coming up with a method'. This is that backwards!

One more example!

Factorise \(~15x^2-11x+2~\)

Step 1

Work out those magical numbers

\(a\times c=15\times 2=30~\) and \(~b=-11\)

Two magical numbers \(\times\) to make 30 and \(+\) to make \(-\)11.

\(-5\times -6=30~\) and \(~-5+(-6)=-11\)

Our magical numbers are \(-\)5 and \(-\)6

Step 2

Split the \(~x~\) term

\(15x^2-5x-6x+2\)

Step 3

Factorise ... carefully!

\(5x(3x-1)-2(3x-1)\)

Step 4

The easy bit!

\((5x-2)(3x-1)\)

In step 3, might you mistakenly write \(~-2(3x+1)~\) instead of \(~-2(3x-1)~\)?  I guess if you double check your factorisation, you'll spot any wobbles and correct them, right?

Well, what can I say? Get stuck in, be patient and persevere. One day, you'll wonder what all the fuss was about.

Practise to master

SECTION A

Factorise the following expressions.

The first six questions are negative-free!

01) \(~2x^2+9x+4~\)

\(a\times c=2\times 4=8~\)  and  \(~b=9\)
Since\(~1\times 8=8~\) and \(~1+8=9\)
Those magical numbers are 1 and 8
Split: \(2x^2+1x+8x+4\)
Factorise: \(x(2x+1)+4(2x+1)\)
Answer: \((2x+1)(x+4)\)

02) \(~4x^2+16x+15~\)

\(a\times c=4\times 15=60~\)  and  \(~b=16\)
Since\(~10\times 6=60~\) and \(~10+6=16\)
Those magical numbers are 10 and 6
Split: \(4x^2+10x+6x+15\)
Factorise: \(2x(2x+5)+3(2x+5)\)
Answer: \((2x+5)(2x+3)\)

03) \(~9x^2+18x+5~\)

\(a\times c=9\times 5=45~\)  and  \(~b=18\)
Since\(~3\times 15=45~\) and \(~3+15=18\)
Those magical numbers are 3 and 15
Split: \(9x^2+3x+15x+5\)
Factorise: \(3x(3x+1)+5(3x+1)\)
Answer: \((3x+5)(3x+1)\)

04) \(~3x^2+11x+6~\)

\(a\times c=3\times 6=18~\)  and  \(~b=11\)
Since\(~2\times 9=18~\) and \(~2+9=11\)
Those magical numbers are 2 and 9
Split: \(3x^2+2x+9x+6\)
Factorise: \(x(3x+2)+3(3x+2)\)
Answer: \((3x+2)(x+3)\)

05) \(~6x^2+19x+10~\)

\(a\times c=6\times 10=60~\)  and  \(~b=19\)
Since\(~4\times 15=60~\) and \(~4+15=19\)
Those magical numbers are 4 and 15
Split: \(6x^2+4x+15x+10\)
Factorise: \(2x(3x+2)+5(3x+2)\)
Answer: \((3x+2)(2x+5)\)

06) \(~12x^2+31x+7~\)

\(a\times c=12\times 7=84~\)  and  \(~b=31\)
Since\(~3\times 28=84~\) and \(~3+28=31\)
Those magical numbers are 3 and 28
Split: \(12x^2+3x+28x+7\)
Factorise: \(3x(4x+1)+7(4x+1)\)
Answer: \((4x+1)(3x+7)\)

07) \(~4x^2-15x-4~\)

\(a\times c=4\times -4=-16~\)  and  \(~b=-15\)
Since\(~1\times -16=-16~\) and \(~1+(-16)=-15\)
Those magical numbers are 1 and \(-\)16
Split: \(4x^2-16x+1x-4\)
Factorise: \(4x(x-4)+1(x-4)\)
Answer: \((4x+1)(x-4)\)

08) \(~8x^2+6x-9~\)

\(a\times c=8\times -9=-72~\)  and  \(~b=6\)
Since\(~-6\times 12=-72~\) and \(~-6+12=6\)
Those magical numbers are \(-\)6 and 12
Split: \(8x^2-6x+12x-9\)
Factorise: \(2x(4x-3)+3(4x-3)\)
Answer: \((4x-3)(2x+3)\)

09) \(~15x^2-16x+4~\)

\(a\times c=15\times 4=60~\)  and  \(~b=-16\)
Since\(~-6\times -10=60~\) and \(~-6+(-10)=-16\)
Those magical numbers are \(-\)6 and \(-\)10
Split: \(15x^2-6x-10x+4\)
Factorise: \(3x(5x-2)-2(5x-2)\)
Answer: \((5x-2)(3x-2)\)

10) \(~5x^2-8x-4~\)

\(a\times c=5\times -4=-20~\)  and  \(~b=-8\)
Since\(~2\times -10=-20~\) and \(~2+(-10)=-8\)
Those magical numbers are 2 and \(-\)10
Split: \(5x^2-10x+2x-4\)
Factorise: \(5x(x-2)+2(x-2)\)
Answer: \((5x+2)(x-2)\)

11) \(~10x^2+21x-10~\)

\(a\times c=10\times -10=-100~\)  and  \(~b=21\)
Since\(~25\times -4=-100~\) and \(~25+(-4)=21\)
Those magical numbers are 25 and \(-\)4
Split: \(10x^2-4x+25x-10\)
Factorise: \(2x(5x-2)+5(5x-2)\)
Answer: \((5x-2)(2x+5)\)

12) \(~18x^2-15x+2~\)

\(a\times c=18\times 2=36~\)  and  \(~b=-15\)
Since\(~-3\times -12=36~\) and \(~-3+(-12)=-15\)
Those magical numbers are \(-\)3 and \(-\)12
Split: \(18x^2-3x-12x+2\)
Factorise: \(3x(6x-1)-2(6x-1)\)
Answer: \((6x-1)(3x-2)\)

13) \(~6x^2-19x+3~\)

\(a\times c=6\times 3=18~\)  and  \(~b=-19\)
Since\(~-1\times -18=18~\) and \(~-1+(-18)=-19\)
Those magical numbers are \(-\)1 and \(-\)18
Split: \(6x^2-1x-18x+3\)
Factorise: \(x(6x-1)-3(6x-1)\)
Answer: \((6x-1)(x-3)\)

14) \(~12x^2-28x-5~\)

\(a\times c=12\times -5=-60~\)  and  \(~b=-28\)
Since\(~2\times -30=-60~\) and \(~2+(-30)=-28\)
Those magical numbers are 2 and \(-\)30
Split: \(12x^2-30x+2x-5\)
Factorise: \(6x(2x-5)+1(2x-5)\)
Answer: \((6x+1)(2x-5)\)

15) \(~9x^2+3x-2~\)

\(a\times c=9\times -2=-18~\)  and  \(~b=3\)
Since\(~6\times -3=-18~\) and \(~6+(-3)=3\)
Those magical numbers are 6 and \(-\)3
Split: \(9x^2-3x+6x-2\)
Factorise: \(3x(3x-1)+2(3x-1)\)
Answer: \((3x+2)(3x-1)\)

16) \(~5x^2-32x+12~\)

\(a\times c=5\times 12=60~\)  and  \(~b=-32\)
Since\(~-2\times -30=60~\) and \(~-2+(-30)=-32\)
Those magical numbers are \(-\)2 and \(-\)30
Split: \(5x^2-2x-30x+12\)
Factorise: \(x(5x-2)-6(5x-2)\)
Answer: \((5x-2)(x-6)\)

17) \(~4x^2-8x-21~\)

\(a\times c=4\times -21=-84~\)  and  \(~b=-8\)
Since\(~6\times -14=-84~\) and \(~6+(-14)=-8\)
Those magical numbers are 6 and \(-\)14
Split: \(4x^2-14x+6x-21\)
Factorise: \(2x(2x-7)+3(2x-7)\)
Answer: \((2x+3)(2x-7)\)

18) \(~12x^2+x-1~\)

\(a\times c=12\times -1=-12~\)  and  \(~b=1\)
Since\(~-3\times 4=-12~\) and \(~-3+4=1\)
Those magical numbers are \(-\)3 and 4
Split: \(12x^2-3x+4x-1\)
Factorise: \(3x(4x-1)+1(4x-1)\)
Answer: \((4x-1)(3x+1)\)

19) \(~3x^2+20x-7~\)

\(a\times c=3\times -7=-21~\)  and  \(~b=20\)
Since\(~-1\times 21=-21~\) and \(~-1+21=20\)
Those magical numbers are \(-\)1 and 21
Split: \(3x^2-1x+21x-7\)
Factorise: \(x(3x-1)+7(3x-1)\)
Answer: \((3x-1)(x+7)\)

20) \(~6x^2-13x+6~\)

\(a\times c=6\times 6=36~\)  and  \(~b=-13\)
Since\(~-4\times -9=36~\) and \(~-4+(-9)=-13\)
Those magical numbers are \(-\)4 and \(-\)9
Split: \(6x^2-4x-9x+6\)
Factorise: \(2x(3x-2)-3(3x-2)\)
Answer: \((3x-2)(2x-3)\)

21) \(~15x^2-19x-10~\)

\(a\times c=15\times -10=-150~\)  and  \(~b=-19\)
Since\(~6\times -25=-150~\) and \(~6+(-25)=-19\)
Those magical numbers are 6 and \(-\)25
Split: \(15x^2-25x+6x-10\)
Factorise: \(5x(3x-5)+2(3x-5)\)
Answer: \((5x+2)(3x-5)\)

22) \(~4x^2+27x-7~\)

\(a\times c=4\times -7=-28~\)  and  \(~b=27\)
Since\(~-1\times 28=-28~\) and \(~-1+28=27\)
Those magical numbers are \(-\)1 and 28
Split: \(4x^2-1x+28x-7\)
Factorise: \(x(4x-1)+7(4x-1)\)
Answer: \((4x-1)(x+7)\)

23) \(~8x^2-18x+9~\)

\(a\times c=8\times 9=72~\)  and  \(~b=-18\)
Since\(~-6\times -12=72~\) and \(~-6+(-12)=-18\)
Those magical numbers are \(-\)6 and \(-\)12
Split: \(8x^2-6x-12x+9\)
Factorise: \(2x(4x-3)-3(4x-3)\)
Answer: \((4x-3)(2x-3)\)

24) \(~18x^2-6x-4~\)

\(a\times c=18\times -4=-72~\)  and  \(~b=-6\)
Since\(~6\times -12=-72~\) and \(~6+(-12)=-6\)
Those magical numbers are 6 and \(-\)12
Split: \(18x^2-12x+6x-4\)
Factorise: \(6x(3x-2)+2(3x-2)\)
Answer: \((6x+2)(3x-2)\)
But wait!
We can factorise further: \(2(3x+1)(3x-2)\)
What? Don't worry, the next section makes light work of this!

SECTION B

Occasionally, a quadratic expression can be factorised to a single bracket at the start, making the factorisation as a whole a bit easier.

Example 1

Factorise \(~2x^2+26x+44~\)

What's different here is that all three numbers have a common factor, which is 2. If you spot this, start by factorising to a single bracket.

\(2(x^2+13x+22)\)

Inside the brackets, the coefficient of the \(~x^2~\) term is 1 so we can factorise this part as we did in Lesson 12 and we get \((x+2)(x+11)\).

But you must then put the common factor (taken out at the start) back in front of this expression.

Final answer: \(2(x+2)(x+11)\)

The coefficient of \(~x^2~\) doesn't always reduce to 1 when you take out the common factor at the start. It's nice when it does, though!

What happens if you don't spot this at the start?

Factorise \(~2x^2+26x+44~\)

\(a\times c=2\times 44=88~\)  and  \(~b=26\)
Since\(~4\times 22=88~\) and \(~4+22=26\)
Those magical numbers are 4 and 22
Split: \(2x^2+4x+22x+44\)
Factorise: \(2x(x+2)+22(x+2)\)
Answer: \((2x+22)(x+2)\)

If you didn't notice the common factor of 2 at the start, you'll hopefully notice it now in the first bracket. And it needs to come out one way or another!

Final answer: \(2(x+11)(x+2)\)

In the practice questions that follow, there is always a common factor that can be taken out at the start. Sometimes this will reduce the coefficient of \(~x^2~\) to 1 and the rest of the factorisation will be a doddle; sometimes it will just make it slightly easier by making the numbers smaller. Take out the common factor at the end only if you realise that you forgot to do it at the start!

Practise to master

SECTION B

Factorise the following expressions.

01) \(~12x^2+38x+6~\)

Factorise: \(2(6x^2+19x+3)\)
\(a\times c=6\times 3=18~\)  and  \(~b=19\)
Since\(~1\times 18=18~\) and \(~1+18=19\)
Those magical numbers are 1 and 18
Split: \(6x^2+1x+18x+3\)
Factorise: \(x(6x+1)+3(6x+1)\)
Answer: \(2(6x+1)(x+3)\)

02) \(~2x^2+6x-8~\)

Factorise: \(2(x^2+3x-4)\)
\(a\times c=1\times -4=-4~\)  and  \(~b=3\)
Since\(~-1\times 4=-4~\) and \(~-1+4=3\)
Those beautiful(!) numbers are \(-\)1 and 4
Answer: \(2(x-1)(x+4)\)
This time it was really worth factorising right at the start!

03) \(~6x^2+9x-6~\)

Factorise: \(3(2x^2+3x-2)\)
\(a\times c=2\times -2=-4~\)  and  \(~b=3\)
Since\(~-1\times 4=-4~\) and \(~-1+4=3\)
Those magical numbers are \(-\)1 and 4
Split: \(2x^2-1x+4x-2\)
Factorise: \(x(2x-1)+2(2x-1)\)
Answer: \(3(2x-1)(x+2)\)

04) \(~3x^2-3x-18~\)

Factorise: \(3(x^2-x-6)\)
\(a\times c=1\times -6=-6~\)  and  \(~b=-1\)
Since\(~2\times -3=-6~\) and \(~2+(-3)=-1\)
Those beautiful numbers are 2 and \(-\)3
Answer: \(3(x-3)(x+2)\)

05) \(~6x^2-12x+6~\)

Factorise: \(6(x^2-2x+1)\)
\(a\times c=1\times 1=1~\)  and  \(~b=-2\)
Since\(~-1\times -1=1~\) and \(~-1+(-1)=-2\)
Those beautiful numbers are \(-\)1 and \(-\)1
Answer: \(6(x-1)^2\)

06) \(~40x^2-36x-36~\)

Factorise: \(4(10x^2-9x-9)\)
\(a\times c=10\times -9=-90~\)  and  \(~b=-9\)
Since\(~-15\times 6=-90~\) and \(~-15+6=-9\)
Those magical numbers are \(-\)15 and 6
Split: \(10x^2-15x+6x-9\)
Factorise: \(5x(2x-3)+3(2x-3)\)
Answer: \(4(5x+3)(2x-3)\)

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