Factorising quadratics — two brackets — Mathematics

Before this

Expanding double brackets
Taking out a common factor · coming soon
Expanding triple brackets

This leads to

Running the film backwards

Expanding is going forwards. You start with two brackets and multiply them out:

(x + 3) (x + 4) = x^{2} + 7 x + 12

Factorising is the same film run backwards. You are given $x^{2} + 7 x + 12$ , and you must find the two brackets it came from.

Going backwards is harder. I am not going to pretend otherwise. Multiplying 13 by 14 is easy: 182. But if someone hands you 182 and asks which two numbers made it, you have to search. Factorising has the same kind of difficulty. The good news is that the search is short, because the last two lectures already told you exactly what to look for.

Here is a small surprise to show the link. Put $x = 10$ into $x^{2} + 7 x + 12$ . You get $100 + 70 + 12 = 182$ . And $182 = 13 \times 14$ , which is $(10 + 3) (10 + 4)$ . Factorising a quadratic is like factorising a number, done for every value of $x$ at once.

The clue the forward direction left behind

The lecture on expanding double brackets ended with a pattern. For any two brackets of the form $(x + p) (x + q)$ :

(x + p) (x + q) = x^{2} + (p + q) x + pq

The middle number is the sum of $p$ and $q$ . The last number is their product. That lecture promised this pattern would be the key to going backwards. Here it is. To factorise $x^{2} + 7 x + 12$ , you need two numbers whose sum is 7 and whose product is 12.

List the pairs of whole numbers that multiply to 12: 1 and 12, 2 and 6, 3 and 4. Their sums are 13, 8 and 7. Only 3 and 4 add to 7. So:

x^{2} + 7 x + 12 = (x + 3) (x + 4)

Why list the products first and not the sums? Because there are only a few ways to multiply to 12, but endless ways to add to 7 (6 and 1, 10 and −3, 7.5 and −0.5, and so on). Searching the short list is faster. That is it. That is the whole idea for these quadratics: find two numbers with the right product and the right sum.

What the signs tell you

Negative numbers make the search look harder, but the signs actually shorten it. The product and the sum each give a clue.

The number term is positive. Two numbers multiply to a positive number only when they have the same sign. If the middle term is also positive, both are positive. If the middle term is negative, both are negative. Example: $x^{2} - 8 x + 15$ needs product 15 and sum −8, so −3 and −5. The answer is $(x - 3) (x - 5)$ .
The number term is negative. The two numbers must have opposite signs, because only then is the product negative. Their sum is then the difference of their sizes, and the larger one carries the sign of the middle term. Example: $x^{2} - 5 x - 14$ needs product −14 and sum −5. The pair 2 and −7 works: $2 \times (- 7) = - 14$ and $2 + (- 7) = - 5$ . So the answer is $(x + 2) (x - 7)$ .

These are not rules to memorise. Each one is the sign rule for multiplying and adding, used in reverse. Try the search yourself below.

Explore — hunting for the pair

Choose the middle number and the last number of x² + bx + c. Every pair of whole numbers with the right product is listed, with its sum. Find the pair whose sum is b. Then look for quadratics where no pair works.

b: 7

c: 12

The search for three quadratics. For each, every pair of whole numbers with the right product is listed with its sum.

x² + 7x + 12 (product 12, sum 7): 1 and 12 (sum 13); 2 and 6 (sum 8); 3 and 4 (sum 7 ✓); and the negative pairs (sums −13, −8, −7). Answer: (x + 3)(x + 4).
x² − 5x − 14 (product −14, sum −5): 1 and −14 (sum −13); −1 and 14 (sum 13); 2 and −7 (sum −5 ✓); −2 and 7 (sum 5). Answer: (x + 2)(x − 7).
x² + x + 1 (product 1, sum 1): 1 and 1 (sum 2); −1 and −1 (sum −2). No pair works, so it does not factorise with whole numbers.

Did you find quadratics with no pair at all? Try $x^{2} + x + 1$ or $x^{2} + 2 x - 1$ . Many quadratics cannot be written as two brackets with whole numbers. That is not a failure of your search. Those quadratics simply do not factorise that way. A later lecture, on the quadratic formula, handles them.

When the x² has a number in front

Now try $2 x^{2} + 7 x + 3$ . The sum-and-product trick needs a change, because the brackets now look like $(2 x + ?) (x + ?)$ , and the pattern $x^{2} + (p + q) x + pq$ no longer fits.

Go back to the picture behind every expansion: the rectangle. Here is the rectangle for $(2 x + 1) (x + 3)$ .

The rectangle for (2x + 1)(x + 3) has four pieces: 2x², 6x, x and 3. The two corner pieces on one diagonal multiply to 2x² × 3 = 6x². The two on the other diagonal multiply to 6x × x = 6x². They are always equal.

Look at the two diagonals. One diagonal holds $2 x^{2}$ and $3$ . Their product is $6 x^{2}$ . The other diagonal holds the two middle pieces, $6 x$ and $x$ . Their product is also $6 x^{2}$ . This always happens. Each diagonal uses each of the four bracket terms exactly once, so both diagonals multiply to the same thing.

This is the hard part of the lecture. Spend time here. It tells you how to go backwards. In $2 x^{2} + 7 x + 3$ , the middle term $7 x$ is the two middle pieces added together. Those two pieces must:

add to $7 x$ , because together they make the middle term, and
multiply to $2 x^{2} \times 3 = 6 x^{2}$ , because of the diagonals.

So you need two numbers with sum 7 and product $2 \times 3 = 6$ . That product is $a \times c$ : the number in front of $x^{2}$ times the number at the end. This is why the method is called the ac method. The pair is 1 and 6. Split the middle term into those two pieces, $x$ and $6 x$ , and you have rebuilt the four pieces of the rectangle:

2 x^{2} + 7 x + 3 = 2 x^{2} + x + 6 x + 3

The last step is grouping: take out a common factor from each half.

2 x^{2} + x + 6 x + 3 = x (2 x + 1) + 3 (2 x + 1) = (2 x + 1) (x + 3)

In the first line, both groups contain the bracket $(2 x + 1)$ . That bracket is now a common factor of the whole expression, so it comes out, just as 5 comes out of $5 a + 5 b$ . Check with $x = 10$ : $21 \times 13 = 273$ , and $200 + 70 + 3 = 273$ .

Does it matter which way round you split the middle term? Try it. Writing $2 x^{2} + 6 x + x + 3$ instead gives $2 x (x + 3) + 1 (x + 3)$ , and then $(x + 3) (2 x + 1)$ . That is the same answer with the brackets in the other order. The order of the split only changes which piece of the rectangle you read first. So choose whichever split makes the common factors easiest to see.

Factorising ax² + bx + c (the ac method)

0. Take out any common factor first.

1. Work out $a \times c$ .

2. Find two numbers with product $a c$ and sum $b$ .

3. Split $b x$ into two terms using those numbers.

4. Group in pairs, take out a common factor from each pair, then take out the common bracket.

When $a = 1$ , steps 3 and 4 can be skipped: the two numbers go straight into the brackets.

Here is the method once more, one move at a time, on a harder example.

Step through — factorising 6x² + x − 2

Press Next to add one step. Read the reason before you go on.

Each step of factorising 6x² + x − 2 gives one line and the reason it is allowed.

a × c = 6 × (−2) = −12. The two middle pieces must multiply to −12x².
Pairs with product −12: 1 and −12, −1 and 12, 2 and −6, −2 and 6, 3 and −4, −3 and 4. The product is negative, so the signs are opposite.
4 and −3: product −12, sum 1. This matches b = 1.
6x² + 4x − 3x − 2. The middle term x is split into 4x and −3x; the value has not changed.
2x(3x + 2) − 1(3x + 2). Common factor from each pair. Taking −1 out of −3x − 2 leaves +2 inside the bracket.
(3x + 2)(2x − 1). The bracket (3x + 2) is common to both groups.
Check with x = 10: 32 × 19 = 608, and 600 + 10 − 2 = 608.

Stop and think

A student writes $x^{2} + 7 x + 12 = x (x + 3) + 4 (x + 3)$ and stops. Is the expression factorised?

Show the answer

Not yet. "Factorised" means written as a product: one thing times another. The student's line is still a sum of two parts, $x (x + 3)$ plus $4 (x + 3)$ . It becomes a product only after the common bracket comes out: $(x + 3) (x + 4)$ . The grouping line is a halfway point, not an answer.

Worked examples

Worked Example 1 — Both numbers negative

Factorise $x^{2} - 8 x + 15$ .

Product 15 and sum −8. The product is positive and the sum negative, so both numbers are negative. Pairs: −1 and −15 (sum −16), −3 and −5 (sum −8).

Answer: $(x - 3) (x - 5)$ . Check with $x = 10$ : $7 \times 5 = 35$ , and $100 - 80 + 15 = 35$ .

Worked Example 2 — The ac method with two negatives

Factorise $3 x^{2} - 10 x + 8$ .

$a c = 3 \times 8 = 24$ . Product 24 and sum −10, so both numbers are negative: −4 and −6.

Split and group: $3 x^{2} - 4 x - 6 x + 8 = x (3 x - 4) - 2 (3 x - 4)$ . Taking −2 out of $- 6 x + 8$ leaves $3 x - 4$ , because $- 2 \times (- 4) = + 8$ .

Answer: $(3 x - 4) (x - 2)$ . Check with $x = 10$ : $26 \times 8 = 208$ , and $300 - 100 + 8 = 208$ .

Worked Example 3 — A common factor first

Factorise $2 x^{2} + 10 x + 12$ completely.

Every term is even, so take out 2 first: $2 (x^{2} + 5 x + 6)$ . Now product 6 and sum 5: 2 and 3.

Answer: $2 (x + 2) (x + 3)$ . Check with $x = 10$ : $2 \times 12 \times 13 = 312$ , and $200 + 100 + 12 = 312$ .

The ac method would also work on $2 x^{2} + 10 x + 12$ directly, giving $(2 x + 4) (x + 3)$ . That is correct but not complete, because $2 x + 4$ still contains the factor 2. Taking out the common factor first keeps the numbers small and the answer complete.

Worked Example 4 — Many pairs to search

Factorise $4 x^{2} - 4 x - 3$ .

$a c = 4 \times (- 3) = - 12$ . The product is negative, so the signs are opposite. The sum is −4, so the larger number is negative. Go through the factor pairs of 12 with that rule: 1 and −12 (sum −11), 2 and −6 (sum −4 ✓). There is no need to try 3 and −4.

Split and group: $4 x^{2} + 2 x - 6 x - 3 = 2 x (2 x + 1) - 3 (2 x + 1)$ .

Answer: $(2 x + 1) (2 x - 3)$ . Check with $x = 10$ : $21 \times 17 = 357$ , and $400 - 40 - 3 = 357$ .

Notice that neither bracket starts with $4 x$ . When $a$ is 4, the brackets could start $4 x$ and $x$ , or $2 x$ and $2 x$ . Guessing would mean trying both, with every sign. The ac method decides for you.

Three classic mistakes

Trap 1 — Checking the product but not the sum

For $x^{2} + 5 x + 6$ , the pair 1 and 6 has the right product but sum 7. It gives $(x + 1) (x + 6) = x^{2} + 7 x + 6$ , the wrong quadratic. Always check both: product $a c$ and sum $b$ .

Trap 2 — A sign lost in grouping

When you take a negative number out of a pair, every sign inside changes. $- 7 x - 14 = - 7 (x + 2)$ , not $- 7 (x - 2)$ . If the two brackets in the grouping line do not match exactly, check the signs before anything else.

Trap 3 — Not factorising completely

$(2 x + 4) (x + 3)$ is not complete, because 2 is still a factor of $2 x + 4$ . The complete answer is $2 (x + 2) (x + 3)$ . Look for a common factor at the start, and look again at the end.

Try it yourself

Work each one on paper first. Check every answer by expanding, or with $x = 10$ .

Factorise $x^{2} + 9 x + 20$ .

Hint 1
Find two numbers with product 20 and sum 9.

Hint 2
The pairs for 20 are 1 and 20, 2 and 10, 4 and 5.

Full answer
4 and 5. So $x^{2} + 9 x + 20 = (x + 4) (x + 5)$ . Check with $x = 10$ : $14 \times 15 = 210$ , and $100 + 90 + 20 = 210$ .
Factorise $x^{2} - x - 12$ .

Hint 1
The product is −12, so the signs are opposite. The sum is −1, so the larger number is negative.

Hint 2
Which pair of factors of 12 differ by 1?

Full answer
3 and −4: product −12, sum −1. So $x^{2} - x - 12 = (x + 3) (x - 4)$ . Check with $x = 10$ : $13 \times 6 = 78$ , and $100 - 10 - 12 = 78$ .
Factorise $5 x^{2} - 13 x - 6$ .

Hint 1
$a c = 5 \times (- 6) = - 30$ . You need product −30 and sum −13.

Hint 2
The pair is 2 and −15. Split: $5 x^{2} + 2 x - 15 x - 6$ .

Full answer
$5 x^{2} + 2 x - 15 x - 6 = x (5 x + 2) - 3 (5 x + 2) = (5 x + 2) (x - 3)$ . Check with $x = 10$ : $52 \times 7 = 364$ , and $500 - 130 - 6 = 364$ .
Find every whole number $k$ for which $x^{2} + k x + 12$ can be factorised into two brackets with whole numbers. Explain why there are exactly that many.

Hint 1
Work backwards: $k$ must be the sum of a pair of whole numbers whose product is 12.

Hint 2
List all pairs with product 12, including negative pairs. Each pair gives one value of $k$ .

Full answer
The pairs with product 12 are 1 and 12, 2 and 6, 3 and 4, and the same pairs with both numbers negative. Their sums give $k = 13, 8, 7, - 7, - 8, - 13$ . So there are exactly six values. Every factorisation $(x + p) (x + q)$ needs $pq = 12$ , and 12 has only three positive factor pairs, each of which can also be made negative. For any other $k$ , such as $k = 5$ , no pair works, and the quadratic does not factorise with whole numbers.

The bigger idea

Factorising turns a sum into a product, and products are powerful for one reason: if a product is zero, one of its factors must be zero. So $(x + 3) (x + 4) = 0$ means $x + 3 = 0$ or $x + 4 = 0$ . The next unit uses exactly this to solve quadratic equations. A hard equation splits into two easy ones.

Notice also what the method really is. You did not guess brackets and hope. You used a fact about multiplication (the diagonals of the rectangle multiply to the same thing) to turn a guessing problem into a short, certain search. That is how mathematicians work: when a problem is hard backwards, look at the forward direction until you find a clue it always leaves behind. The next lecture finds one more clue, in quadratics with no middle term.

Key concepts in this lesson

Factorising is expanding in reverse: it writes a quadratic as a product of two brackets.
For x² + bx + c, find two numbers with product c and sum b, and they go straight into the brackets.
For ax² + bx + c, find two numbers with product ac and sum b, split the middle term, then group.
The ac method works because the two diagonals of the expansion rectangle always multiply to the same thing.
Take out common factors first, check every answer by expanding, and expect some quadratics not to factorise with whole numbers.