Factorising Quadratic Trinomials
A quadratic trinomial is what a binomial product collapses to. Factorising it means finding the two numbers the expansion hid, and there is a reliable method whether or not the leading coefficient is one.
Find two numbers that add to $9$ and multiply to $20$. Now find two that add to $-9$ and multiply to $20$. Then two that add to $1$ and multiply to $-20$. Describe, in words, how the SIGNS of the sum and the product told you what signs the two numbers had, before you found them.
For a monic trinomial, one whose $x^2$ has coefficient $1$, factorising is a sum-and-product puzzle: find two numbers that add to the middle coefficient and multiply to the constant. For a non-monic one, the same two numbers exist but they are found from a different product, and they are used to split the middle term rather than written straight into brackets.
$$x^2 + bx + c = (x + p)(x + q) \quad \text{where } p + q = b \text{ and } pq = c$$
The signs tell you a great deal before you search. If the product $c$ is positive, both numbers share the sign of $b$. If $c$ is negative, the numbers have opposite signs and the larger one carries the sign of $b$. Reading the signs first cuts the search roughly in half.
Know
- That a monic trinomial factorises using two numbers with sum $b$ and product $c$
- That a non-monic trinomial uses two numbers with sum $b$ and product $ac$, used to split the middle term
- That a perfect-square trinomial is the case where the two numbers are equal
Understand
- Why the signs of $b$ and $c$ determine the signs of the two numbers before any searching
- Why splitting the middle term turns a non-monic trinomial into a four-term grouping problem
Can Do
- Factorise monic trinomials, including ones with negative coefficients
- Factorise non-monic trinomials by splitting the middle term and grouping
- Recognise a perfect-square trinomial and factorise it in one step
Because $(x+p)(x+q) = x^2 + (p+q)x + pq$, the two numbers you need are the ones that add to the middle coefficient and multiply to the constant.
For $x^2 + 7x + 12$: two numbers adding to $7$ and multiplying to $12$. The factor pairs of $12$ are $1 \times 12$, $2 \times 6$ and $3 \times 4$. Only $3$ and $4$ add to $7$:
$$x^2 + 7x + 12 = (x + 3)(x + 4)$$
For $x^2 - 5x + 6$: the product is positive and the sum is negative, so both numbers are negative. The pair is $-2$ and $-3$:
$$x^2 - 5x + 6 = (x - 2)(x - 3)$$
For $x^2 + 2x - 15$: the product is negative, so the signs differ, and the sum is positive, so the larger number is positive. The pair is $5$ and $-3$:
$$x^2 + 2x - 15 = (x + 5)(x - 3)$$
Before listing any factor pairs, the two signs tell you what you are looking for.
| Sign of $c$ | Sign of $b$ | The two numbers | Example |
|---|---|---|---|
| positive | positive | both positive | $x^2 + 8x + 15 = (x+3)(x+5)$ |
| positive | negative | both negative | $x^2 - 8x + 15 = (x-3)(x-5)$ |
| negative | positive | opposite, larger positive | $x^2 + 2x - 15 = (x+5)(x-3)$ |
| negative | negative | opposite, larger negative | $x^2 - 2x - 15 = (x-5)(x+3)$ |
This is not a table to memorise. Each row follows from the arithmetic: two negatives multiply to a positive and add to a negative, and so on. Deriving the row you need takes a moment and never goes stale.
When the leading coefficient is not $1$, the two numbers no longer go straight into the brackets. They still exist, but they multiply to $a \times c$ rather than to $c$, and their job is to split the middle term.
Factorise $6x^2 + 7x + 2$. Here $a = 6$, $b = 7$, $c = 2$, so $ac = 12$. Two numbers adding to $7$ and multiplying to $12$: those are $3$ and $4$.
Split the middle term into $3x + 4x$:
$$6x^2 + 3x + 4x + 2$$
Now it has four terms, which is a grouping problem from Lesson 3:
$$= 3x(2x + 1) + 2(2x + 1) = (2x + 1)(3x + 2)$$
The order of the split does not matter. Writing $4x + 3x$ instead gives $2x(3x+2) + 1(3x+2)$, the same answer.
A trinomial is a perfect square when the two numbers are equal. For $x^2 + 10x + 25$, two numbers adding to $10$ and multiplying to $25$ are $5$ and $5$:
$$x^2 + 10x + 25 = (x+5)(x+5) = (x+5)^2$$
You can spot these before searching: the first and last terms are perfect squares, and the middle term is twice the product of their roots. For $9x^2 - 24x + 16$, the roots are $3x$ and $4$, and $2(3x)(4) = 24x$, so it is $(3x - 4)^2$.
And the common factor still comes first. $2x^2 + 10x + 12$ has an HCF of $2$:
$$= 2(x^2 + 5x + 6) = 2(x+2)(x+3)$$
Removing it turned a non-monic problem into a monic one, which is much quicker. Always check for it before deciding you are in the harder case.
Watch Me Solve It · 3 examples
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1Read the signs$c = -28 < 0, \quad b = -3 < 0$A negative product means opposite signs; a negative sum means the larger number is negative.
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2List factor pairs of 28$1 \times 28, \quad 2 \times 14, \quad 4 \times 7$
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3Find the pair with a difference of 3$7 - 4 = 3 \Rightarrow \text{the numbers are } -7 \text{ and } 4$Opposite signs, larger negative, as the signs predicted.
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4Write the brackets and check$(x - 7)(x + 4)$$\text{middle term: } 4x - 7x = -3x \;\checkmark$
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1Check for a common factor$10, 11, 6 \text{ share no factor}$So the non-monic method is genuinely needed.
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2Compute ac and find the pair$ac = 10 \times (-6) = -60$$\text{two numbers with sum } -11 \text{ and product } -60: \; 4 \text{ and } -15$Opposite signs since the product is negative; larger negative since the sum is negative.
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3Split the middle term$10x^2 + 4x - 15x - 6$Either order works.
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4Group in pairs$2x(5x + 2) - 3(5x + 2) = (5x + 2)(2x - 3)$Taking $-3$ out of the second pair makes the brackets match.
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1Take out the HCF$4x^3 - 4x^2 - 24x = 4x(x^2 - x - 6)$Numbers share $4$, letters share $x$.
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2Note what it became$x^2 - x - 6 \text{ is MONIC}$Removing the common factor turned a hard problem into an easy one.
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3Find sum and product$\text{sum } -1, \quad \text{product } -6 \Rightarrow -3 \text{ and } 2$
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4Write it fully factorised$4x(x - 3)(x + 2)$Check by expanding: $(x-3)(x+2) = x^2 - x - 6$, then times $4x$ returns the original.
Brain Trainer · 4 problems
Four quick problems. Work each one, then reveal the answer.
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1 Factorise $x^2 + 9x + 20$.
Two numbers adding to $9$, multiplying to $20$: $4$ and $5$.$(x+4)(x+5)$ -
2 Factorise $x^2 - 7x + 10$.
Product positive, sum negative, so both negative: $-2$ and $-5$.$(x-2)(x-5)$ -
3 Factorise $3x^2 + 10x + 8$.
$ac = 24$; two numbers adding to $10$: $4$ and $6$. Split and group.$(3x+4)(x+2)$ -
4 Factorise $2x^2 + 12x + 18$ fully.
HCF of $2$ first, leaving a monic perfect square.$2(x+3)^2$
Multiple Choice · 5 questions
To factorise $x^2 + 11x + 24$, you need two numbers that:
In $x^2 - 4x - 21$, the two numbers are:
To factorise $4x^2 + 13x + 3$ by splitting, the two numbers must multiply to:
$x^2 - 14x + 49$ factorises to:
The quickest first step for $5x^2 - 20x + 15$ is to:
Short Answer · 3 questions
(a) $x^2 + 4x - 45$
(b) $x^2 - 12x + 36$
(c) $6x^2 - 5x - 4$
(d) $3x^3 + 3x^2 - 60x$
(b) Explain why the same shortcut fails for $2x^2 + 7x + 3$, and show what must be used instead.
(c) Show that your factorisation of $2x^2 + 7x + 3$ is correct.
(a) $x^2 - 9x + 18 = (x - 3)(x - 6)$
(b) $x^2 + 5x - 24 = (x + 3)(x - 8)$
(c) $4x^2 + 8x + 3 = (4x + 3)(x + 1)$
(b) The number $b^2 - 4ac$ is called the discriminant. Compute it for $x^2 + 3x + 5$ and for $x^2 + 6x + 5$, and comment on the difference.
(c) Predict, using the discriminant alone, whether $2x^2 + 5x - 3$ factorises over the integers, then confirm by factorising or by showing it cannot be done.
Monic
Sum $b$, product $c$, straight into the brackets
Non-monic
Sum $b$, product $ac$, split the middle term
Signs
Product's sign decides same or opposite
First
Common factor, which may make it monic
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