Factorising Non-Monic Quadratic Trinomials
Theory
A non-monic quadratic such as \(6x^2 + 11x - 10\) has an \(x^2\) coefficient other than \(1\). This Year 10 maths page shows the ac method: find two numbers with product \(ac\) and sum \(b\), split the middle term, then group in pairs. It also covers common factors, a negative \(x^2\) term, two pronumerals, missing coefficients and proofs.
A quadratic trinomial \(ax^2 + bx + c\) is non-monic when the coefficient \(a\) of \(x^2\) is not \(1\), as in \(5x^2 - 13x - 6\).
The product-and-sum shortcut for monic trinomials does not work directly, because the \(x\) terms in the brackets now have coefficients. The \(ac\) method fixes this: it rewrites the middle term as two terms so the expression can be factorised by grouping.
Grouping in pairs means taking a common factor out of the first two terms and out of the last two, so that the same bracket appears twice and can itself be taken out: \(3x(x + 4) - 2(x + 4)\) \(= (3x - 2)(x + 4)\).
Always check by expanding. A swapped sign gives the right \(x^2\) and constant terms but the wrong middle term.
For \(ax^2 + bx + c\), find \(m\) and \(n\) with
then
and factorise the four terms by grouping in pairs.
How to factorise \(ax^2 + bx + c\)
- Tidy: descending powers; take out any common factor; take out \(-1\) if \(a\) is negative.
- Multiply \(a \times c\) and find two numbers with that product and sum \(b\).
- Split the middle term using those two numbers.
- Group in pairs, take out the common bracket, and write the answer with any common factor in front.
- Check by expanding.
Variations
- Two pronumerals: \(ax^2 + bxy + cy^2\) works the same way; each number carries \(y\).
- Quartics: \(ax^4 + bx^2 + c\) is a quadratic in \(u = x^2\). Factorise, put \(x^2\) back, and look for differences of two squares.
- Repeated brackets: if expressions like \(x + 1\) and \(x - 2\) repeat, let \(u\) and \(v\) stand for them, factorise in \(u\) and \(v\), then substitute and simplify each bracket.
- Areas and perimeters: factorise an area to get the length and width, then use them in a perimeter equation.
- Finding an error: expand the suggested answer; if only the sign of the middle term is wrong, swap the signs in the brackets.
- Proofs: factorise an expression in \(n\) to show it is a product of two consecutive odd numbers, or of two numbers greater than \(1\) (Example 4).
(i) \(4x^2 - 4x - 15\)
(ii) \(9x^2 - 21x + 10\)
(i) \(ac = 4 \times (-15) = -60\). Two numbers with product \(-60\) and sum \(-4\): \(-10\) and \(6\).
\(4x^2 - 4x - 15\)
| \(\) | \(=\) | \(4x^2 - 10x + 6x - 15\) |
| \(\) | \(=\) | \(2x(2x - 5) + 3(2x - 5)\) |
| \(\) | \(=\) | \(\boldsymbol{(2x + 3)(2x - 5)}\) |
(ii) \(ac = 9 \times 10 = 90\). Product \(90\), sum \(-21\): both negative, \(-15\) and \(-6\).
\(9x^2 - 21x + 10\)
| \(\) | \(=\) | \(9x^2 - 15x - 6x + 10\) |
| \(\) | \(=\) | \(3x(3x - 5) - 2(3x - 5)\) |
| \(\) | \(=\) | \(\boldsymbol{(3x - 2)(3x - 5)}\) |
When you group, the second bracket must match the first. In (ii) take out \(-2\), not \(2\), so that \(-6x + 10\) becomes \(-2(3x - 5)\).
(i) \(12x^2 + 2x - 30\)
(ii) \(5 + 3x - 2x^2\)
(i) Take out the common factor \(2\), then use \(ac = 6 \times (-15) = -90\): numbers \(10\) and \(-9\).
\(12x^2 + 2x - 30\)
| \(\) | \(=\) | \(2(6x^2 + x - 15)\) |
| \(\) | \(=\) | \(2(6x^2 + 10x - 9x - 15)\) |
| \(\) | \(=\) | \(2[2x(3x + 5) - 3(3x + 5)]\) |
| \(\) | \(=\) | \(\boldsymbol{2(2x - 3)(3x + 5)}\) |
(ii) Write in descending powers and take out \(-1\). Then \(ac = 2 \times (-5) = -10\): numbers \(-5\) and \(2\).
| \(5 + 3x - 2x^2\) | \(=\) | \(-(2x^2 - 3x - 5)\) |
| \(\) | \(=\) | \(-(2x^2 - 5x + 2x - 5)\) |
| \(\) | \(=\) | \(-(2x - 5)(x + 1)\) |
| \(\) | \(=\) | \(\boldsymbol{(5 - 2x)(x + 1)}\) |
(i) \(6p^2 + pq - 12q^2\)
(ii) \(9x^4 - 10x^2 + 1\)
(i) \(ac = 6 \times (-12) = -72\). Product \(-72\), sum \(1\): \(9\) and \(-8\), each carrying \(pq\).
\(6p^2 + pq - 12q^2\)
| \(\) | \(=\) | \(6p^2 + 9pq - 8pq - 12q^2\) |
| \(\) | \(=\) | \(3p(2p + 3q) - 4q(2p + 3q)\) |
| \(\) | \(=\) | \(\boldsymbol{(3p - 4q)(2p + 3q)}\) |
(ii) Let \(u = x^2\): \(9u^2 - 10u + 1\), with \(ac = 9\), numbers \(-9\) and \(-1\).
\(9u^2 - 10u + 1 = (9u - 1)(u - 1)\)
Put \(u = x^2\) back:
\(9x^4 - 10x^2 + 1\)
| \(\) | \(=\) | \((9x^2 - 1)(x^2 - 1)\) |
| \(\) | \(=\) | \(\boldsymbol{(3x - 1)(3x + 1)(x - 1)(x + 1)}\) |
Both brackets in (ii) were differences of two squares, so keep factorising.
(i) Factorise \(6n^2 + 7n + 2\).
(ii) Hence explain why \(6n^2 + 7n + 2\) is never a prime number when \(n\) is a positive integer.
(i) \(ac = 12\). Product \(12\), sum \(7\): \(4\) and \(3\).
\(6n^2 + 7n + 2\)
| \(\) | \(=\) | \(6n^2 + 4n + 3n + 2\) |
| \(\) | \(=\) | \(2n(3n + 2) + 1(3n + 2)\) |
| \(\) | \(=\) | \(\boldsymbol{(2n + 1)(3n + 2)}\) |
(ii) For a positive integer \(n\), \(2n + 1 \ge 3\) and \(3n + 2 \ge 5\). So the number is a product of two whole numbers, each greater than \(1\). It has a factor other than \(1\) and itself, so it is not prime.
Check: \(n = 2\) gives \(24 + 14 + 2 = 40 = 5 \times 8\).
Common pitfalls
Frequently asked questions
How do you factorise a non-monic quadratic?
Multiply a and c. Find two numbers that multiply to ac and add to b. Use them to split the middle term into two terms, then factorise the four terms by grouping in pairs.
What is the ac method?
It is a way to factorise ax squared + bx + c when a is not 1. You find two numbers with product ac and sum b, rewrite bx as the sum of two terms using those numbers, and factorise by grouping.
What does non-monic mean?
A non-monic quadratic has a coefficient of \(x^2\) other than \(1\), such as \(3x^2 + 5x - 2\).
Does it matter which way round I split the middle term?
No. Writing 15x + 2x or 2x + 15x gives the same factors after grouping. Only the common factors you take out change.
How do I check my factorisation?
Expand the brackets. You must get back the original trinomial exactly, including the sign of the middle term.