Technique

Pair four linear factors so one quadratic repeats

Algebra • Expressions • Factorisation

Statement

(x + a) (x + b) (x + c) (x + d) + k

, if the constants can be paired so that

a + d = b + c

, then multiplying those pairs gives

(x^{2} + (a + d) x + a d)

and

(x^{2} + (a + d) x + b c)

, which differ only in their constant term; putting

t = x^{2} + (a + d) x

turns the whole expression into a quadratic in

t

How to use it

1Add the four constants in pairs and pick the pairing whose two sums agree; that is the pairing to multiply out.
2Substitute $t$ for the repeated quadratic, factorise in $t$ , then replace $t$ and factorise each quadratic in $x$ further where possible.
3A quadratic in $x$ that is left with a negative discriminant is irreducible and is a legitimate final factor.

Factorise

(x + 1) (x + 3) (x + 5) (x + 7) + 15

Step-by-step

1
$1 + 7 = 8 = 3 + 5$ , so pair $(x + 1) (x + 7)$ with $(x + 3) (x + 5)$
Why: Equal constant sums are what make the two products share their $x$ terms.
2
$(x^{2} + 8 x + 7) (x^{2} + 8 x + 15) + 15$
3
$t = x^{2} + 8 x$ : $(t + 7) (t + 15) + 15 = t^{2} + 22 t + 120 = (t + 10) (t + 12)$
Why: The substitution reduces a quartic to a quadratic.
4
$(x^{2} + 8 x + 10) (x^{2} + 8 x + 12) = (x^{2} + 8 x + 10) (x + 2) (x + 6)$
Why: Put $t$ back; the second quadratic factorises, the first has discriminant $24 < 0$ over the rationals and stops there.

Answer

(x^{2} + 8 x + 10) (x + 2) (x + 6)

Watch out

Only the pairing with equal constant sums works; pairing the brackets in the order they are printed usually leaves two quadratics with different

x

coefficients and no substitution is possible.

Builds on

Also known as

quadratic in disguise, product of four linear factors, pair the outer and inner factors, substitute t for x^2+bx, reduce a quartic to a quadratic