SPI Mathematics · Polynomials

Factorising Polynomials

Factorising writes a polynomial as a product of simpler ones: by splitting the middle term, by grouping, or by treating a repeated bracket as one letter.

All Polynomials notes

Why this matters

Five questions in the bank, three EASY and two MODERATE. The options are four sets of brackets that differ only in signs, so the skill is to check both the product and the sum before choosing.

Concept 1 of 3: Factorising a quadratic polynomial by splitting the middle term

Multiply out

(x + p)(x + q)

and you get

x^2 + (p + q)x + pq

. So to factorise, you look for two numbers that multiply to the constant and add to the middle coefficient. When the

x^2

coefficient is not 1, the same idea works with the product ac.

Definition

For $x^2 + bx + c$ : find p and q with $pq = c$ and $p + q = b$ . Then $x^2 + bx + c = (x + p)(x + q)$ .
For $ax^2 + bx + c$ : find two numbers with product ac and sum b, split bx into them, and group.
The zeroes come from setting each bracket to 0: $(x + 4)$ gives the zero $-4$ .

Product of two brackets

(x + p)(x + q) = x^2 + (p + q)x + pq

Worked example

Factorise

6x^2 + 7x - 3

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Polynomials · Factorising Polynomials

(x^2 - 4x - 21) = ?

Checking only the product

(x + 6)(x - 4)

and

(x - 6)(x + 4)

both give the constant

-24

. The middle term decides: the first gives

+2x

, the second

-2x

Drill 1 more on factorising a quadratic polynomial by splitting the middle term

Concept 2 of 3: Finding a coefficient by expanding and comparing

If two polynomials are equal for every x, their matching coefficients are equal. So expand the side with brackets and set the coefficient of x equal to the coefficient of x, and the constant equal to the constant.

Definition

Expand fully, then line up the terms by power of x.
Equal polynomials have equal coefficients for every power of x.
Use the term that contains the unknown, and check with another term.

Worked example

x^2 + px + 10 = (x + 2)(x + 5)

, find p.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Polynomials · Factorising Polynomials

x^2 + kx - 3 = (x - 3)(x + 1)

, then the value of 'k' is

Comparing terms of different powers

Match the coefficient of x with the coefficient of x, and the constant with the constant. In

x^2 + kx + 8 = (x + 2)(x + 4)

, k is 6, the x coefficient, not 8.

Concept 3 of 3: Factorising by grouping and by substitution

A four-term polynomial often splits into two pairs that share the same bracket. And an expression built from a repeated bracket, such as

(a + b)

, is just a quadratic in disguise: call the bracket t and factorise as usual.

Definition

Grouping: pair the terms, take a common factor out of each pair, and the same bracket appears twice: $x^2(x - 2) + 5(x - 2) = (x - 2)(x^2 + 5)$ .
If the second pair starts with a minus, take out a negative factor so the brackets match.
Substitution: let t stand for the repeated bracket, factorise the quadratic in t, then put the bracket back in place of t.

Grouping pattern

ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)

Worked example

Factorise

3(x - 1)^2 - 5(x - 1) - 2

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 3 · Polynomials · Factorising Polynomials

3x^3 + 2x^2 + 3x + 2 = ?

Taking out a positive factor from a negative pair

x^3 - 2x^2 - 5x + 10

, take

-5

out of the second pair:

x^2(x - 2) - 5(x - 2) = (x - 2)(x^2 - 5)

. Taking out

+5

leaves

(-x + 2)

, which does not match.

Leaving the answer in t

After factorising in t, replace t by the bracket it stood for.

(t - 3)(3t + 2)

with

t = a + b

becomes

(a + b - 3)(3a + 3b + 2)

Drill 1 more on factorising by grouping and by substitution

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Factorising a quadratic polynomial by splitting the middle term
Product of two brackets
$(x + p)(x + q) = x^2 + (p + q)x + pq$
Factorising by grouping and by substitution
Grouping pattern
$ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)$

Watch out for (4)

Checking only the product
→ Factorising a quadratic polynomial by splitting the middle term
Comparing terms of different powers
→ Finding a coefficient by expanding and comparing
Taking out a positive factor from a negative pair
→ Factorising by grouping and by substitution
Leaving the answer in t
→ Factorising by grouping and by substitution

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