Factorising quadratics through their zeros

The fac­tor the­o­rem says that every zero of a poly­no­mial hands us a fac­tor: if $p(\alpha) = 0$ then $x - \alpha$ divides $p(x)$. Put the other way round, once we know all the zeros we can write the poly­no­mial as a prod­uct. This les­son does that for qua­drat­ics. We first sharpen the split­ting method, includ­ing qua­drat­ics with surd coef­fi­cients, and then fac­torise any qua­dratic at all by find­ing its zeros with the for­mula, even when those zeros are irra­tional.

Split­ting the mid­dle term: the prod­uct rule#

To fac­torise $ax^2 + bx + c$ by split­ting, find two num­bers whose prod­uct is $ac$ and whose sum is $b$. Then split $bx$ into two terms and take out com­mon fac­tors in pairs.

Exam­ple 1#

Fac­torise $x^2 - 7x + 6$.

Prod­uct $1 \times 6 = 6$, sum $-7$. The pair is $-1$ and $-6$.

$$x^2 - x - 6x + 6 = x(x - 1) - 6(x - 1) = (x - 6)(x - 1)$$

Exam­ple 2#

Fac­torise $2x^2 - 9x - 5$.

Prod­uct $2 \times (-5) = -10$, sum $-9$. Since $1 \times (-10) = -10$ and $1 + (-10) = -9$, the pair is $1$ and $-10$.

$$2x^2 + x - 10x - 5 = x(2x + 1) - 5(2x + 1) = (x - 5)(2x + 1)$$

Exam­ple 3#

Fac­torise $7\sqrt{2}x^2 - 10x - 4\sqrt{2}$.

Prod­uct $7\sqrt{2} \times (-4\sqrt{2}) = -28 \times 2 = -56$, sum $-10$. Since $4 \times (-14) = -56$ and $4 - 14 = -10$, split $-10x$ as $4x - 14x$.

$$7\sqrt{2}x^2 + 4x - 14x - 4\sqrt{2}$$

To group the first pair, notice that $4 = 2 \times 2 = 2 \times \sqrt{2} \times \sqrt{2}$. So $7\sqrt{2}x^2 + 4x = \sqrt{2}x(7x + 2\sqrt{2})$. The sec­ond pair is $-14x - 4\sqrt{2} = -2(7x + 2\sqrt{2})$.

$$= \sqrt{2}x(7x + 2\sqrt{2}) - 2(7x + 2\sqrt{2}) = (\sqrt{2}x - 2)(7x + 2\sqrt{2})$$

Mul­ti­ply out to check: $7\sqrt{2}x^2 + 4x - 14x - 4\sqrt{2}$, which is the orig­i­nal.

Fac­toris­ing from the zeros#

The fac­tor the­o­rem gives these results.

Result 1. If $\alpha$ and $\beta$ are the zeros of $ax^2 + bx + c$ (that is, the roots of $ax^2 + bx + c = 0$), then

$$ax^2 + bx + c = a(x - \alpha)(x - \beta)$$

Result 2. If $\alpha$, $\beta$, $\gamma$ are the zeros of $ax^3 + bx^2 + cx + d$, then

$$ax^3 + bx^2 + cx + d = a(x - \alpha)(x - \beta)(x - \gamma)$$

Result 3. If $\alpha$, $\beta$, $\gamma$, $\delta$ are the zeros of $ax^4 + bx^3 + cx^2 + dx + e$, then

$$ax^4 + bx^3 + cx^2 + dx + e = a(x - \alpha)(x - \beta)(x - \gamma)(x - \delta)$$

The lead­ing coef­fi­cient $a$ stands in front, because each bracket $(x - \alpha)$ starts with $x$, and the prod­uct must start with $ax^2$ (or $ax^3$, or $ax^4$).

For a qua­dratic, the zeros come straight from the for­mula:

$$\displaystyle \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

So every qua­dratic with real zeros can be fac­torised this way, whether or not it splits nicely.

Exam­ple 4#

Fac­torise $2x^2 + x - 45$ in two ways.

By split­ting. Prod­uct $2 \times (-45) = -90$, sum $1$. Since $10 \times (-9) = -90$ and $10 - 9 = 1$:

$$2x^2 + 10x - 9x - 45 = 2x(x + 5) - 9(x + 5) = (x + 5)(2x - 9)$$

From the zeros. Here $a = 2$, $b = 1$, $c = -45$.

$$\displaystyle \begin{aligned} x &= \frac{-1 \pm \sqrt{1 - 4(2)(-45)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 360}}{4} \\ &= \frac{-1 \pm \sqrt{361}}{4} = \frac{-1 \pm 19}{4} \end{aligned}$$

So the zeros are $\displaystyle \frac{18}{4} = \frac{9}{2}$ and $\displaystyle \frac{-20}{4} = -5$. Then

$$\displaystyle 2x^2 + x - 45 = 2\left(x - \frac{9}{2}\right)(x + 5) = (2x - 9)(x + 5)$$

where the $2$ was mul­ti­plied into the first bracket. Both meth­ods give the same fac­tors.

Exam­ple 5#

Fac­torise $2x^2 - 3x - 4$.

The prod­uct $ac = -8$ and the sum $-3$: no pair of inte­gers works ($-8 \times 1$, $-4 \times 2$, $-2 \times 4$, $-1 \times 8$ give sums $-7$, $-2$, $2$, $7$). So we find the zeros. Here $a = 2$, $b = -3$, $c = -4$.

$$\displaystyle x = \frac{3 \pm \sqrt{9 - 4(2)(-4)}}{2(2)} = \frac{3 \pm \sqrt{9 + 32}}{4} = \frac{3 \pm \sqrt{41}}{4}$$

$$\displaystyle 2x^2 - 3x - 4 = 2\left(x - \frac{3 + \sqrt{41}}{4}\right)\left(x - \frac{3 - \sqrt{41}}{4}\right)$$

Exam­ple 6#

Fac­torise $x^2 + 8x + 5$.

Here $a = 1$, $b = 8$, $c = 5$.

$$\displaystyle \begin{aligned} x &= \frac{-8 \pm \sqrt{64 - 4(1)(5)}}{2(1)} = \frac{-8 \pm \sqrt{44}}{2} \\ &= \frac{-8 \pm \sqrt{4 \times 11}}{2} = \frac{-8 \pm 2\sqrt{11}}{2} = -4 \pm \sqrt{11} \end{aligned}$$

The zeros are $-4 + \sqrt{11}$ and $-4 - \sqrt{11}$, so

$$x^2 + 8x + 5 = \left(x - (-4 + \sqrt{11})\right)\left(x - (-4 - \sqrt{11})\right) = (x + 4 - \sqrt{11})(x + 4 + \sqrt{11})$$

Remem­ber. Sim­plify the surd before divid­ing: $\sqrt{44} = 2\sqrt{11}$, and only then can­cel the $2$ with the denom­i­na­tor.

Prac­tice#

Fac­torise each qua­dratic.

1. $x^2 + 10x + 24$
2. $3x^2 + 11x + 10$
3. $8x^2 + 2x - 3$
4. $14x^2 - 23x + 8$
5. $12x^2 - x - 35$
6. $15x^2 + 2x - 8$
7. $15 - 8x - 16x^2$
8. $\sqrt{3}x^2 + 10x + 7\sqrt{3}$
9. $x^2 + 6x + 4$ (use the zeros)
10. $3x^2 - 2x - 6$ (use the zeros)

Answers#