Repeated brackets and two-letter quadratics

Split­ting the mid­dle term works on more than plain qua­drat­ics in $x$. If a whole bracket keeps appear­ing, you can call it by a sin­gle let­ter and fac­torise as usual. And if the expres­sion has two let­ters, like $14x^{2} + 11xy - 15y^{2}$, the same method works with the sec­ond let­ter car­ried along. This les­son shows both.

Replac­ing a repeated bracket by one let­ter#

Look at $(x - 2y)^{2} - 5(x - 2y) + 6$. The bracket $(x - 2y)$ appears twice: once squared and once on its own. If we write $p$ for it, the expres­sion becomes $p^{2} - 5p + 6$, an ordi­nary qua­dratic. Fac­torise that, then put the bracket back in place of $p$. This is called sub­sti­tu­tion.

Exam­ple 1#

Fac­torise $(x - 2y)^{2} - 5(x - 2y) + 6$.

Let $x - 2y = p$.

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

To split $-5p$ we needed two num­bers with prod­uct $6$ and sum $-5$; the pair $2 \times 3$ becomes $-2$ and $-3$. Do not for­get the last step: the ques­tion was about $x$ and $y$, so the answer must be too.

Qua­drat­ics in two let­ters#

An expres­sion such as $14x^{2} + 11xy - 15y^{2}$ has an $x^{2}$ term, an $xy$ term and a $y^{2}$ term. Treat it exactly like $ax^{2} + bx + c$, with $a = 14$, $b = 11$ and $c = -15$: find two num­bers with prod­uct $a \times c$ and sum $b$, then split the mid­dle term. The let­ter $y$ sim­ply rides along with the num­bers.

Exam­ple 2#

Fac­torise $14x^{2} + 11xy - 15y^{2}$.

Here $a \times c = 14 \times (-15) = -210$. In primes, $210 = 7 \times 2 \times 5 \times 3$, and $21 \times 10 = 210$ with $21 - 10 = 11$. So the num­bers are $21$ and $-10$.

$$\begin{aligned} 14x^{2} + 11xy - 15y^{2} &= 14x^{2} + 21xy - 10xy - 15y^{2} \\ &= 7x(2x + 3y) - 5y(2x + 3y) \\ &= (2x + 3y)(7x - 5y) \end{aligned}$$

Exam­ple 3: a prod­uct of let­ters in the mid­dle#

Fac­torise $15x^{2} - 16xyz - 15y^{2}z^{2}$.

Here the "sec­ond let­ter" is the prod­uct $yz$: the last term is $-15(yz)^{2}$. So $a \times c = 15 \times (-15) = -225 = -(5 \times 3 \times 5 \times 3)$, and we want sum $-16$: the num­bers are $-25$ and $9$, since $25 \times 9 = 225$.

$$\begin{aligned} 15x^{2} - 16xyz - 15y^{2}z^{2} &= 15x^{2} - 25xyz + 9xyz - 15y^{2}z^{2} \\ &= 5x(3x - 5yz) + 3yz(3x - 5yz) \\ &= (3x - 5yz)(5x + 3yz) \end{aligned}$$

Check by expand­ing: $(3x - 5yz)(5x + 3yz) = 15x^{2} + 9xyz - 25xyz - 15y^{2}z^{2}$, which is the expres­sion we started with. Expand­ing is the quick­est way to catch a mis­placed let­ter.

Both ideas together#

Exam­ple 4#

Fac­torise $8(a + 1)^{2} + 2(a + 1)(b + 2) - 15(b + 2)^{2}$.

Two brack­ets repeat, so give each a let­ter. Let $a + 1 = x$ and $b + 2 = y$. The expres­sion becomes a two-let­ter qua­dratic:

$$8x^{2} + 2xy - 15y^{2}$$

Here $a \times c = 8 \times (-15) = -120$, and $12 \times 10 = 120$ with $12 - 10 = 2$. So split $2xy$ as $12xy - 10xy$.

$$\begin{aligned} 8x^{2} + 2xy - 15y^{2} &= 8x^{2} + 12xy - 10xy - 15y^{2} \\ &= 4x(2x + 3y) - 5y(2x + 3y) \\ &= (2x + 3y)(4x - 5y) \end{aligned}$$

Now put the brack­ets back and sim­plify each fac­tor:

$$\begin{aligned} &= [2(a + 1) + 3(b + 2)][4(a + 1) - 5(b + 2)] \\ &= (2a + 2 + 3b + 6)(4a + 4 - 5b - 10) \\ &= (2a + 3b + 8)(4a - 5b - 6) \end{aligned}$$

Exam­ple 5: pow­ers of x as the repeated quan­tity#

Fac­torise $4x^{4} - 13x^{2} + 9$.

Here $x^{4} = (x^{2})^{2}$, so let $x^{2} = p$: the expres­sion is $4p^{2} - 13p + 9$. Prod­uct $36$, sum $-13$: the num­bers are $-4$ and $-9$.

$$\begin{aligned} 4p^{2} - 13p + 9 &= 4p^{2} - 4p - 9p + 9 \\ &= 4p(p - 1) - 9(p - 1) \\ &= (p - 1)(4p - 9) \\ &= (x^{2} - 1)(4x^{2} - 9) \\ &= (x + 1)(x - 1)(2x + 3)(2x - 3) \end{aligned}$$

Both brack­ets turned out to be dif­fer­ences of squares, so they fac­torise once more.

Prac­tice#

Fac­torise.

1. $x^{2} + 5xy + 6y^{2}$
2. $2p^{2} - 7pq + 3q^{2}$
3. $12x^{2} - 17xy + 6y^{2}$
4. $6 - x - 2x^{2}$
5. $36a^{2} + 12abc - 15b^{2}c^{2}$
6. $(x + 1)^{2} - 3(x + 1) - 10$
7. $(2a - b)^{2} + 2(2a - b) - 8$
8. $6(a - 1)^{2} + 5(a - 1)(b + 1) + (b + 1)^{2}$
9. $x^{4} - 5x^{2} + 4$

Answers#