Splitting the middle term

Most qua­dratic expres­sions, such as $3x^{2} + 10x + 3$, are not per­fect squares and not dif­fer­ences of squares, so the iden­ti­ties of the last les­son do not fit. The method that works for all of them that can be fac­torised is called split­ting the mid­dle term. It turns the three terms into four, and then group­ing from the sec­ond les­son fin­ishes the job.

The idea#

A qua­dratic expres­sion in $x$ has the form $ax^{2} + bx + c$, where $a$, $b$, $c$ are num­bers and $a \neq 0$. Here $a$ is the coef­fi­cient of $x^{2}$, $b$ the coef­fi­cient of $x$ (the mid­dle term), and $c$ the con­stant term.

To fac­torise $ax^{2} + bx + c$:

1. Mul­ti­ply the first and last coef­fi­cients: work out $a \times c$.
2. Find two num­bers whose prod­uct is $a \times c$ and whose sum is $b$.
3. Split the mid­dle term $bx$ into two terms using those num­bers.
4. Group the four terms in pairs and take out com­mon fac­tors. A com­mon bracket appears.
5. Take out the com­mon bracket.

Why it works#

Look at a sim­ple case, $x^{2} + 7x + 12$, where $a = 1$. We need two num­bers with prod­uct $1 \times 12 = 12$ and sum $7$. They are $3$ and $4$.

$$\begin{aligned} x^{2} + 7x + 12 &= x^{2} + 3x + 4x + 12 \\ &= x(x + 3) + 4(x + 3) \\ &= (x + 3)(x + 4) \end{aligned}$$

Expand­ing $(x + 3)(x + 4)$ gives $x^{2} + 3x + 4x + 12$: the two mid­dle num­bers add to give the mid­dle term and mul­ti­ply to give the last term. Split­ting the mid­dle term sim­ply runs the expan­sion back­wards.

Worked exam­ples#

Exam­ple 1#

Fac­torise $3x^{2} + 10x + 3$.

Here $a \times c = 3 \times 3 = 9$. List the pairs of fac­tors of $9$: $1 \times 9$ and $3 \times 3$. The pair that adds to $10$ is $1$ and $9$, since $1 + 9 = 10$.

$$\begin{aligned} 3x^{2} + 10x + 3 &= 3x^{2} + x + 9x + 3 \\ &= x(3x + 1) + 3(3x + 1) \\ &= (3x + 1)(x + 3) \end{aligned}$$

Exam­ple 2: a neg­a­tive mid­dle term#

Fac­torise $4y^{2} - 8y + 3$.

Here $a \times c = 4 \times 3 = 12$. The prod­uct is pos­i­tive but the sum, $-8$, is neg­a­tive, so both num­bers must be neg­a­tive. Fac­tor pairs of $12$: $1 \times 12$, $2 \times 6$, $3 \times 4$. The pair $-2$ and $-6$ works, since $(-2) \times (-6) = 12$ and $-2 - 6 = -8$.

$$\begin{aligned} 4y^{2} - 8y + 3 &= 4y^{2} - 2y - 6y + 3 \\ &= 2y(2y - 1) - 3(2y - 1) \\ &= (2y - 1)(2y - 3) \end{aligned}$$

In the sec­ond line, $-3$ was taken out of $-6y + 3$, which leaves $(2y - 1)$. Tak­ing out $+3$ would have left $(-2y + 1)$, which does not match.

Exam­ple 3: a neg­a­tive last term#

Fac­torise $2x^{2} - 3x - 2$.

Here $a \times c = 2 \times (-2) = -4$. A neg­a­tive prod­uct means the two num­bers have oppo­site signs. Pairs with prod­uct $-4$: $-1$ and $4$, $1$ and $-4$, $-2$ and $2$. The pair with sum $-3$ is $1$ and $-4$.

$$\begin{aligned} 2x^{2} - 3x - 2 &= 2x^{2} + x - 4x - 2 \\ &= x(2x + 1) - 2(2x + 1) \\ &= (2x + 1)(x - 2) \end{aligned}$$

Exam­ple 4#

Fac­torise $6x^{2} + 5x - 6$.

Here $a \times c = 6 \times (-6) = -36$. Pairs of fac­tors of $36$: $1 \times 36$, $2 \times 18$, $3 \times 12$, $4 \times 9$, $6 \times 6$. We need oppo­site signs with sum $+5$, and $-4 + 9 = 5$.

$$\begin{aligned} 6x^{2} + 5x - 6 &= 6x^{2} - 4x + 9x - 6 \\ &= 2x(3x - 2) + 3(3x - 2) \\ &= (3x - 2)(2x + 3) \end{aligned}$$

A sign guide#

The signs of $a \times c$ and $b$ tell you the signs of the two num­bers before you start search­ing:

When the prod­uct is large#

If $a \times c$ is a big num­ber, do not list every pair. Break it into primes, then build pairs from the primes.

Exam­ple 5#

Fac­torise $11x^{2} - 54x + 63$.

Here $a \times c = 11 \times 63 = 693$. In primes, $693 = 11 \times 7 \times 9 = 11 \times 3 \times 3 \times 7$. We need two neg­a­tive num­bers adding to $-54$. Try group­ing the primes: $77 \times 9$ gives sum $86$, too big. $33 \times 21$ gives sum $54$. So the num­bers are $-33$ and $-21$.

$$\begin{aligned} 11x^{2} - 54x + 63 &= 11x^{2} - 33x - 21x + 63 \\ &= 11x(x - 3) - 21(x - 3) \\ &= (x - 3)(11x - 21) \end{aligned}$$

Exam­ple 6: terms in the wrong order, neg­a­tive lead­ing term#

Fac­torise $28 - 31x - 5x^{2}$.

First write the terms in descend­ing pow­ers of $x$: $-5x^{2} - 31x + 28$. Here $a \times c = (-5) \times 28 = -140$. Since $140 = 5 \times 4 \times 7 = 35 \times 4$, and we need sum $-31$, the num­bers are $-35$ and $4$.

$$\begin{aligned} -5x^{2} - 31x + 28 &= -5x^{2} - 35x + 4x + 28 \\ &= -5x(x + 7) + 4(x + 7) \\ &= (x + 7)(-5x + 4) \end{aligned}$$

It is neater to write the answer as $(x + 7)(4 - 5x)$. Check: $(x + 7)(4 - 5x) = 4x - 5x^{2} + 28 - 35x = 28 - 31x - 5x^{2}$.

Be care­ful when you rewrite a fac­tori­sa­tion with the brack­ets turned round. $(x + 7)(4 - 5x)$ is cor­rect, but $(7 - x)(4 - 5x)$ is not: it expands to $28 - 39x + 5x^{2}$. Chang­ing the signs inside one bracket changes the prod­uct, unless you also change the signs inside the other bracket.

Prac­tice#

Fac­torise.

1. $x^{2} + 9x + 20$
2. $x^{2} - 2x - 15$
3. $5x^{2} + 13x + 6$
4. $3x^{2} - 14x + 8$
5. $10x^{2} + x - 3$
6. $6x^{2} - 13x + 6$
7. $-2x^{2} - 3x + 2$
8. $7x - 6 - 2x^{2}$
9. $7x^{2} - 19x - 6$
10. $12 + x - 6x^{2}$

Answers#