Factorising Non-Monic Quadratic Trinomials
A non-monic quadratic such as \(3x^2 + 10x + 8\) has a leading coefficient other than \(1\). In NSW Year 9 Advanced maths these are factorised by splitting the middle term and grouping in pairs, or by testing pairs of factors (NESA outcome MA5‑ALG‑P‑02).
You will factorise trinomials with every sign pattern, take out a common factor first, and deal with a negative leading term. You will factorise expressions in \(x\) and \(y\), with a bracket as the variable or with a letter in a coefficient, find an unknown coefficient from a known factor, prove that an expression is never prime, and find the dimensions of a prism from its volume. It is part of Year 9 Advanced Quadratics.
Every question with a fully worked solution.
- Factorising Non-Monic Quadratic Trinomials - Video - Factorising Non-Monic Quadratics Watch
Theory
A non-monic quadratic such as \(3x^2 + 10x + 8\) has a leading coefficient other than \(1\). The reliable way to factorise it is the ac method: split the middle term, then group in pairs. This Year 9 Advanced maths page shows the method step by step and the variations you will meet. It leads on to Advanced and Extension maths.
A quadratic trinomial \(ax^2 + bx + c\) is non-monic when \(a \ne 1\), such as \(2x^2 - 5x - 12\).
When it factorises, it becomes \((px + q)(rx + s)\) with \(pr = a\) and \(qs = c\). Guessing these four numbers is slow, so use the ac method.
Splitting the middle term means writing \(bx\) as the sum of two terms, using the two numbers whose product is \(ac\) and whose sum is \(b\). The four terms then group in pairs with a common bracket.
If no integer pair has product \(ac\) and sum \(b\), the trinomial does not factorise with integers.
Then take a common factor from each pair; the same bracket should appear twice.
| \(ac\) | \(b\) | Signs of \(m\) and \(n\) |
|---|---|---|
| positive | positive | both positive |
| positive | negative | both negative |
| negative | either | opposite signs; the larger takes the sign of \(b\) |
The ac method
- Tidy up: write in descending powers; take out any HCF, and \(-1\) if the \(x^2\) term is negative.
- Multiply \(a \times c\).
- Find the pair \(m, n\) with product \(ac\) and sum \(b\).
- Split \(bx\) into \(mx + nx\).
- Group in pairs and take out the common factor of each pair. The brackets must match; if they do not, check the signs.
- Take out the common bracket, then check by expanding.
Variations
- Two pronumerals: \(ax^2 + bxy + cy^2\) works the same way; the split terms carry \(xy\).
- A repeated bracket: let \(u\) be the bracket, factorise in \(u\), substitute back and simplify each bracket.
- Letters as coefficients: the pair may involve the letter, for example \(3x^2 + (3p + 2)x + 2p\) splits as \(3px + 2x\).
- Volumes and areas: factorise the expression; each factor is a length.
(i) \(3x^2 + 10x + 8\)
(ii) \(8x^2 - 2x - 15\)
(i) \(ac = 3 \times 8 = 24\). Two numbers with product \(24\) and sum \(10\): \(6\) and \(4\). Split \(10x\) into \(6x + 4x\) and group.
| \(3x^2 + 10x + 8\) | \(=\) | \(3x^2 + 6x + 4x + 8\) |
| \(\) | \(=\) | \(3x(x + 2) + 4(x + 2)\) |
| \(\) | \(=\) | \(\boldsymbol{(3x + 4)(x + 2)}\) |
(ii) \(ac = 8 \times (-15) = -120\). Product \(-120\), sum \(-2\): \(-12\) and \(10\).
| \(8x^2 - 2x - 15\) | \(=\) | \(8x^2 - 12x + 10x - 15\) |
| \(\) | \(=\) | \(4x(2x - 3) + 5(2x - 3)\) |
| \(\) | \(=\) | \(\boldsymbol{(4x + 5)(2x - 3)}\) |
Check (ii): \(8x^2 - 12x + 10x - 15\) \(= 8x^2 - 2x - 15\ \checkmark\). The order of the split does not matter: \(10x - 12x\) gives the same brackets.
(i) \(12 + 5x - 2x^2\)
(ii) \(30x^3 - 35x^2 - 15x\)
(i) Write it in descending powers and take out \(-1\). Then \(ac = 2 \times (-12) = -24\), sum \(-5\): \(-8\) and \(3\).
| \(12 + 5x - 2x^2\) | \(=\) | \(-(2x^2 - 5x - 12)\) |
| \(2x^2 - 5x - 12\) | \(=\) | \(2x^2 - 8x + 3x - 12\) |
| \(\) | \(=\) | \(2x(x - 4) + 3(x - 4)\) |
| \(\) | \(=\) | \((2x + 3)(x - 4)\) |
| \(12 + 5x - 2x^2\) | \(=\) | \(-(2x + 3)(x - 4)\) |
| \(\) | \(=\) | \(\boldsymbol{(2x + 3)(4 - x)}\) |
(ii) Take out the HCF \(5x\) first. Then \(ac = 6 \times (-3) = -18\), sum \(-7\): \(-9\) and \(2\).
| \(30x^3 - 35x^2 - 15x\) | \(=\) | \(5x(6x^2 - 7x - 3)\) |
| \(6x^2 - 7x - 3\) | \(=\) | \(6x^2 - 9x + 2x - 3\) |
| \(\) | \(=\) | \(3x(2x - 3) + 1(2x - 3)\) |
| \(\) | \(=\) | \((3x + 1)(2x - 3)\) |
| \(30x^3 - 35x^2 - 15x\) | \(=\) | \(\boldsymbol{5x(3x + 1)(2x - 3)}\) |
When grouping gives \(+1(2x - 3)\), write the \(1\): it becomes the \(+1\) in \((3x + 1)\).
(i) \(4a^2 - 4ab - 3b^2\)
(ii) \(3(x - 1)^2 + 4(x - 1) - 4\)
(i) Use the coefficients \(4, -4, -3\): \(ac = -12\), sum \(-4\): \(-6\) and \(2\). Each split term carries \(ab\).
| \(4a^2 - 4ab - 3b^2\) | \(=\) | \(4a^2 - 6ab + 2ab - 3b^2\) |
| \(\) | \(=\) | \(2a(2a - 3b) + b(2a - 3b)\) |
| \(\) | \(=\) | \(\boldsymbol{(2a + b)(2a - 3b)}\) |
(ii) Let \(u = x - 1\). Then \(3u^2 + 4u - 4\): \(ac = -12\), sum \(4\): \(6\) and \(-2\).
| \(3u^2 + 4u - 4\) | \(=\) | \(3u^2 + 6u - 2u - 4\) |
| \(\) | \(=\) | \(3u(u + 2) - 2(u + 2)\) |
| \(\) | \(=\) | \((3u - 2)(u + 2)\) |
| \(\) | \(=\) | \(\big(3(x - 1) - 2\big)\big((x - 1) + 2\big)\) |
| \(\) | \(=\) | \(\boldsymbol{(3x - 5)(x + 1)}\) |
Watch the \(3\) when substituting back: \(3u = 3(x - 1) = 3x - 3\).
(i) \(3x + 2\) is a factor of \(12x^2 + kx - 10\). Find \(k\) and the other factor.
(ii) A rectangle has area \((6x^2 + 13x + 6)~\text{cm}^2\) and one side \((2x + 3)\) cm. Find the other side, and both lengths when \(x = 4\).
(i) The other factor is \((px + q)\). The \(x^2\) terms: \(3x \times px = 12x^2\), so \(p = 4\). The constants: \(2 \times q = -10\), so \(q = -5\).
| \((3x + 2)(4x - 5)\) | \(=\) | \(12x^2 - 15x + 8x - 10\) |
| \(\) | \(=\) | \(12x^2 - 7x - 10\) |
| \(k\) | \(=\) | \(\boldsymbol{-7}\) |
| \(\text{other factor}\) | \(=\) | \(\boldsymbol{4x - 5}\) |
(ii) Area \(=\) length \(\times\) width, so factorise the area. \(ac = 36\), sum \(13\): \(9\) and \(4\).
| \(6x^2 + 13x + 6\) | \(=\) | \(6x^2 + 9x + 4x + 6\) |
| \(\) | \(=\) | \(3x(2x + 3) + 2(2x + 3)\) |
| \(\) | \(=\) | \((2x + 3)(3x + 2)\) |
| \(\text{other side}\) | \(=\) | \(\boldsymbol{(3x + 2)}~\text{cm}\) |
| \(2(4) + 3\) | \(=\) | \(\boldsymbol{11}~\text{cm}\) |
| \(3(4) + 2\) | \(=\) | \(\boldsymbol{14}~\text{cm}\) |
Check: \(11 \times 14 = 154\) and \(6(16) + 13(4) + 6 = 154\ \checkmark\)
Common pitfalls
Frequently asked questions
What is a non-monic quadratic?
A quadratic \(ax^2 + bx + c\) in which \(a \ne 1\), such as \(3x^2 + 10x + 8\).
What is the ac method?
Multiply \(a\) by \(c\), find two numbers with that product that add to \(b\), split the middle term using them, then group the four terms in pairs and take out the common bracket.
Does it matter which way round I split the middle term?
No. Either order gives the same factors, although the common factors taken out of each pair will look different.
What if the x squared coefficient is negative?
Take out \(-1\) first, factorise what is left, and keep the minus sign in front. You can then move it into one bracket: \(-(x - 4) = (4 - x)\).
How do I know if a non-monic quadratic cannot be factorised?
If no pair of integers multiplies to \(ac\) and adds to \(b\), it does not factorise with integers.