Surds and fractions as coefficients

So far every coef­fi­cient has been a whole num­ber. In higher classes you meet qua­drat­ics like $7\sqrt{2}x^{2} - 10x - 4\sqrt{2}$, with square roots in the coef­fi­cients, and $\displaystyle 2x^{2} - \frac{5}{6}x + \frac{1}{12}$, with frac­tions. The method of split­ting the mid­dle term does not change at all. What needs care is the arith­metic of find­ing the two num­bers. This les­son shows how.

Square roots in the coef­fi­cients#

A square root that does not work out to a whole num­ber, such as $\sqrt{2}$ or $\sqrt{5}$, is called a surd. Two facts do most of the work:

$$\sqrt{2} \times \sqrt{2} = 2 \qquad \sqrt{5} \times \sqrt{5} = 5$$

So when you mul­ti­ply $a \times c$, the surds often dis­ap­pear, and you are left look­ing for two num­bers just as before. When you split the mid­dle term, you may need to write a whole num­ber in terms of surds, for exam­ple $10 = \sqrt{5} \times \sqrt{5} \times 2 = 2\sqrt{5} \times \sqrt{5}$, so that the com­mon bracket appears.

Exam­ple 1#

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

Here $a \times c = 7\sqrt{2} \times (-4\sqrt{2}) = -28 \times 2 = -56$. We need two num­bers with prod­uct $-56$ and sum $-10$: they are $-14$ and $4$.

$$\begin{aligned} 7\sqrt{2}x^{2} - 10x - 4\sqrt{2} &= 7\sqrt{2}x^{2} - 14x + 4x - 4\sqrt{2} \\ &= 7\sqrt{2}x(x - \sqrt{2}) + 4(x - \sqrt{2}) \\ &= (x - \sqrt{2})(7\sqrt{2}x + 4) \end{aligned}$$

In the sec­ond line, tak­ing $7\sqrt{2}x$ out of $-14x$ leaves $-\sqrt{2}$, because $7\sqrt{2} \times \sqrt{2} = 14$.

Exam­ple 2#

Fac­torise $5\sqrt{5}x^{2} + 30x + 8\sqrt{5}$.

Here $a \times c = 5\sqrt{5} \times 8\sqrt{5} = 5 \times 8 \times 5 = 200$. We need prod­uct $200$ and sum $30$: since $200 = 5 \times 4 \times 2 \times 5 = 10 \times 20$, the num­bers are $10$ and $20$.

$$\begin{aligned} 5\sqrt{5}x^{2} + 30x + 8\sqrt{5} &= 5\sqrt{5}x^{2} + 10x + 20x + 8\sqrt{5} \\ &= 5x(\sqrt{5}x + 2) + 4\sqrt{5}(\sqrt{5}x + 2) \\ &= (\sqrt{5}x + 2)(5x + 4\sqrt{5}) \end{aligned}$$

To see the sec­ond line, write $10 = 5 \times 2$ and $20 = 4 \times 5 = 4 \times \sqrt{5} \times \sqrt{5}$. Then $5\sqrt{5}x^{2} + 10x = 5x(\sqrt{5}x + 2)$ and $20x + 8\sqrt{5} = 4\sqrt{5}(\sqrt{5}x + 2)$.

Exam­ple 3: surds in the two num­bers#

Fac­torise $x^{2} + 3\sqrt{3}x - 30$.

Here $a \times c = -30$ and the sum must be $3\sqrt{3}$. The two num­bers must both be mul­ti­ples of $\sqrt{3}$, since their sum is. Try $5\sqrt{3}$ and $-2\sqrt{3}$: their prod­uct is $-10 \times 3 = -30$ and their sum is $3\sqrt{3}$.

$$\begin{aligned} x^{2} + 3\sqrt{3}x - 30 &= x^{2} + 5\sqrt{3}x - 2\sqrt{3}x - 30 \\ &= x(x + 5\sqrt{3}) - 2\sqrt{3}(x + 5\sqrt{3}) \\ &= (x + 5\sqrt{3})(x - 2\sqrt{3}) \end{aligned}$$

Here $-2\sqrt{3} \times 5\sqrt{3} = -10 \times 3 = -30$, which is why the last term comes out right.

Frac­tions in the coef­fi­cients#

There are two good ways to han­dle frac­tions.

Method A: take the frac­tions out. Write the whole expres­sion over the L.C.M. of the denom­i­na­tors, take that frac­tion out­side as a fac­tor, fac­torise what is left, then share the frac­tion back into the brack­ets if you wish.

Method B: split directly. Find two frac­tions with the right prod­uct and sum and split the mid­dle term as usual.

Exam­ple 4: method A#

Fac­torise $\displaystyle 2x^{2} - \frac{5}{6}x + \frac{1}{12}$.

The L.C.M. of $6$ and $12$ is $12$.

$$\displaystyle \begin{aligned} 2x^{2} - \frac{5}{6}x + \frac{1}{12} &= \frac{24x^{2} - 10x + 1}{12} \\ &= \frac{1}{12}(24x^{2} - 10x + 1) \\ &= \frac{1}{12}[24x^{2} - 6x - 4x + 1] \\ &= \frac{1}{12}[6x(4x - 1) - 1(4x - 1)] \\ &= \frac{1}{12}(4x - 1)(6x - 1) \end{aligned}$$

For $24x^{2} - 10x + 1$ we needed prod­uct $24$ and sum $-10$: the num­bers $-6$ and $-4$. Finally, since $12 = 2 \times 6$, share the $\displaystyle \frac{1}{12}$ between the brack­ets:

$$\displaystyle \frac{1}{12}(4x - 1)(6x - 1) = \frac{4x - 1}{2} \times \frac{6x - 1}{6} = \left(2x - \frac{1}{2}\right)\left(x - \frac{1}{6}\right)$$

Exam­ple 5: method B on the same expres­sion#

Fac­torise $\displaystyle 2x^{2} - \frac{5}{6}x + \frac{1}{12}$ by split­ting directly.

Here $\displaystyle a \times c = 2 \times \frac{1}{12} = \frac{1}{6}$. We need two frac­tions with prod­uct $\displaystyle \frac{1}{6}$ and sum $\displaystyle -\frac{5}{6}$. Since $\displaystyle \frac{1}{6} = \frac{1}{2} \times \frac{1}{3}$ and $\displaystyle -\frac{1}{2} - \frac{1}{3} = \frac{-3 - 2}{6} = -\frac{5}{6}$, the frac­tions are $\displaystyle -\frac{1}{2}$ and $\displaystyle -\frac{1}{3}$.

$$\displaystyle \begin{aligned} 2x^{2} - \frac{5}{6}x + \frac{1}{12} &= 2x^{2} - \frac{1}{2}x - \frac{1}{3}x + \frac{1}{12} \\ &= x\left(2x - \frac{1}{2}\right) - \frac{1}{6}\left(2x - \frac{1}{2}\right) \\ &= \left(2x - \frac{1}{2}\right)\left(x - \frac{1}{6}\right) \end{aligned}$$

The same answer. In the sec­ond line, $\displaystyle -\frac{1}{6}$ was taken out because $\displaystyle -\frac{1}{6} \times 2x = -\frac{1}{3}x$ and $\displaystyle -\frac{1}{6} \times \left(-\frac{1}{2}\right) = \frac{1}{12}$.

Exam­ple 6: a per­fect square in dis­guise#

Fac­torise $\displaystyle 21x^{2} - 2x + \frac{1}{21}$.

$$\displaystyle \begin{aligned} 21x^{2} - 2x + \frac{1}{21} &= \frac{441x^{2} - 42x + 1}{21} \\ &= \frac{1}{21}(441x^{2} - 42x + 1) \\ &= \frac{1}{21}\left[(21x)^{2} + (1)^{2} - 2(21x)(1)\right] \\ &= \frac{1}{21}(21x - 1)^{2} \\ &= \frac{1}{21}(21x - 1)(21x - 1) = \left(x - \frac{1}{21}\right)(21x - 1) \end{aligned}$$

After clear­ing the frac­tion, $441x^{2} - 42x + 1$ has the shape $a^{2} + b^{2} - 2ab$ with $a = 21x$ and $b = 1$, because $441 = 21^{2}$ and $42 = 2 \times 21$. It always pays to check for an iden­tity before split­ting.

Prac­tice#

Fac­torise.

1. $\displaystyle x^{2} + \frac{12}{35}x + \frac{1}{35}$
2. $\sqrt{3}x^{2} + 10x + 7\sqrt{3}$
3. $2\sqrt{2}x^{2} + 5x + \sqrt{2}$
4. $x^{2} - 2\sqrt{2}x - 6$
5. $\displaystyle x^{2} - \frac{1}{6}x - \frac{1}{6}$
6. $\displaystyle 3x^{2} - 2x + \frac{1}{3}$
7. $\displaystyle x^{2} + \frac{1}{4}x - \frac{1}{8}$

Answers#