Factorising Monic Quadratic Trinomials
A monic quadratic \(x^2 + bx + c\) factorises as \((x + m)(x + n)\), where \(m\) and \(n\) multiply to \(c\) and add to \(b\). This NSW Year 9 Advanced maths subtopic factorises monic trinomials (NESA outcome MA5‑ALG‑P‑01).
You will factorise trinomials with every sign pattern, take out a common factor first, handle a negative \(x^2\) term, and recognise a trinomial that does not factorise with integers. You will factorise expressions in two pronumerals, with a bracket as the variable or with a constant \(a\), find every \(k\) that allows factorising, and use factors to evaluate a large expression without a calculator. It is part of Year 9 Advanced Quadratics.
Every question with a fully worked solution.
- Factorising Monic Quadratic Trinomials - Video - Factorising Monic Quadratics Watch
Theory
A monic quadratic such as \(x^2 + 7x + 10\) factorises into \((x + a)(x + b)\), where \(a\) and \(b\) multiply to the constant and add to the middle coefficient. This Year 9 Advanced maths page shows the product and sum method, how to handle the signs, and how to spot quadratics in disguise. It leads on to Advanced and Extension maths.
A quadratic trinomial has three terms and highest power \(2\): \(ax^2 + bx + c\). It is monic when \(a = 1\), so it starts with \(x^2\).
Factorising is the reverse of expanding: writing the expression as a product. Since \((x + a)(x + b) = x^2 + (a + b)x + ab\), the two numbers must have product \(c\) and sum \(b\).
Factorise fully means take out any highest common factor (HCF) first, and keep going until no bracket can be factorised again.
Some trinomials cannot be factorised with integers: \(x^2 + 4x + 2\) has no integer pair with product \(2\) and sum \(4\).
| Constant \(c\) | Middle \(b\) | The two numbers | Example |
|---|---|---|---|
| positive | positive | both positive | \(x^2 + 9x + 14 = (x + 2)(x + 7)\) |
| positive | negative | both negative | \(x^2 - 9x + 14 = (x - 2)(x - 7)\) |
| negative | positive | opposite signs, bigger one positive | \(x^2 + 5x - 14 = (x + 7)(x - 2)\) |
| negative | negative | opposite signs, bigger one negative | \(x^2 - 5x - 14 = (x - 7)(x + 2)\) |
How to factorise \(x^2 + bx + c\)
- HCF first. Take out any common factor, including \(-1\) or a negative HCF if the \(x^2\) term is negative.
- Write the target: product \(= c\), sum \(= b\).
- List factor pairs of \(c\) and use the signs to choose the pair with the right sum.
- Write the brackets \((x + a)(x + b)\) with the signs included.
- Check by expanding: the middle term must come back.
Harder forms
- A repeated bracket such as \((x + 3)^2 + \ldots(x + 3) + \ldots\): let \(u = x + 3\), factorise, then substitute back and simplify.
- \(x^4\) and \(x^2\): let \(u = x^2\). Then check each bracket for a difference of two squares.
- Unknown coefficient: list every integer pair with the given product; each pair's sum is a possible coefficient.
- Word problems: if a total is (rows) \(\times\) (seats per row), factorise the total and match one factor to the known one.
(i) \(x^2 + 13x + 40\)
(ii) \(x^2 - 2x - 35\)
(iii) \(x^2 - 15x + 56\)
For \(x^2 + bx + c\), find two numbers with product \(c\) and sum \(b\). The signs of \(c\) and \(b\) tell you the signs of the numbers.
| \(\text{(i) product } 40,\ \text{sum } 13\) | \(:\) | \(5 \text{ and } 8\) |
| \(x^2 + 13x + 40\) | \(=\) | \(\boldsymbol{(x + 5)(x + 8)}\) |
(ii) The product is negative, so the numbers have opposite signs; the sum is negative, so the bigger one is negative.
| \(\text{product } {-35},\ \text{sum } {-2}\) | \(:\) | \({-7} \text{ and } 5\) |
| \(x^2 - 2x - 35\) | \(=\) | \(\boldsymbol{(x - 7)(x + 5)}\) |
(iii) Positive product, negative sum: both numbers are negative.
| \(\text{product } 56,\ \text{sum } {-15}\) | \(:\) | \({-7} \text{ and } {-8}\) |
| \(x^2 - 15x + 56\) | \(=\) | \(\boldsymbol{(x - 7)(x - 8)}\) |
Check (ii) by expanding: \(x^2 + 5x - 7x - 35\) \(= x^2 - 2x - 35\ \checkmark\)
(i) \(28 + 3x - x^2\)
(ii) \(4x^3 - 8x^2 - 60x\)
(iii) \(a^2 - ab - 12b^2\)
(i) Rewrite in descending powers and take out \(-1\), so the \(x^2\) term is positive.
| \(28 + 3x - x^2\) | \(=\) | \(-(x^2 - 3x - 28)\) |
| \(\) | \(=\) | \(-(x - 7)(x + 4)\) |
| \(\) | \(=\) | \(\boldsymbol{(7 - x)(x + 4)}\) |
The last step moves the minus sign into the first bracket: \(-(x - 7) = 7 - x\).
(ii) The HCF of all three terms is \(4x\). Take it out first.
| \(4x^3 - 8x^2 - 60x\) | \(=\) | \(4x(x^2 - 2x - 15)\) |
| \(\) | \(=\) | \(\boldsymbol{4x(x - 5)(x + 3)}\) |
(iii) Work with the coefficients \(1, -1, -12\): product \(-12\), sum \(-1\) gives \(-4\) and \(3\). Each number carries a \(b\).
| \(a^2 - ab - 12b^2\) | \(=\) | \(\boldsymbol{(a - 4b)(a + 3b)}\) |
(i) \((x - 2)^2 - 3(x - 2) - 10\)
(ii) \(x^4 - 10x^2 + 9\)
(i) The bracket \((x - 2)\) appears twice. Let \(u = x - 2\).
| \(u^2 - 3u - 10\) | \(=\) | \((u - 5)(u + 2)\) |
| \(\) | \(=\) | \((x - 2 - 5)(x - 2 + 2)\) |
| \(\) | \(=\) | \(\boldsymbol{x(x - 7)}\) |
(ii) \(x^4 = (x^2)^2\), so let \(u = x^2\): product \(9\), sum \(-10\) gives \(-1\) and \(-9\).
| \(u^2 - 10u + 9\) | \(=\) | \((u - 1)(u - 9)\) |
| \(x^4 - 10x^2 + 9\) | \(=\) | \((x^2 - 1)(x^2 - 9)\) |
| \(\) | \(=\) | \(\boldsymbol{(x - 1)(x + 1)(x - 3)(x + 3)}\) |
Each bracket in the second line is a difference of two squares, so it factorises again. 'Fully' means keep going until no bracket factorises.
(i) Find every integer \(k\) for which \(x^2 + kx - 8\) factorises as \((x + a)(x + b)\) with \(a\) and \(b\) integers.
(ii) Factorise \(x^2 - 5x + 6\). Hence evaluate \(2003^2 - 5 \times 2003 + 6\) without a calculator.
(i) We need \(ab = -8\) and \(k = a + b\). List every integer pair with product \(-8\).
| \(a, b\) | \(1, -8\) | \(-1, 8\) | \(2, -4\) | \(-2, 4\) |
|---|---|---|---|---|
| \(k = a + b\) | \(-7\) | \(7\) | \(-2\) | \(2\) |
| \(k\) | \(=\) | \(\boldsymbol{{-7},\ {-2},\ 2,\ 7}\) |
(ii) Product \(6\), sum \(-5\): \(-2\) and \(-3\).
| \(x^2 - 5x + 6\) | \(=\) | \((x - 2)(x - 3)\) |
| \(2003^2 - 5 \times 2003 + 6\) | \(=\) | \((2003 - 2)(2003 - 3)\) |
| \(\) | \(=\) | \(2001 \times 2000 = \boldsymbol{4\,002\,000}\) |
Common pitfalls
Frequently asked questions
What is a monic quadratic?
A quadratic whose \(x^2\) term has coefficient \(1\), such as \(x^2 - 5x + 6\).
How do you factorise a monic quadratic trinomial?
Find two numbers that multiply to the constant term and add to the coefficient of \(x\). Write them in the brackets \((x + a)(x + b)\).
How do the signs work when factorising?
If the constant is positive, both numbers have the sign of the middle term. If the constant is negative, the numbers have opposite signs and the larger one takes the sign of the middle term.
What if the x squared term is negative?
Take out \(-1\) (or a negative common factor) first, factorise the monic quadratic left inside, and keep the minus sign in front.
Can every quadratic trinomial be factorised?
No. If no pair of integers has the right product and sum, it cannot be factorised with integers, for example \(x^2 + 3x + 1\).
What does factorise fully mean?
Take out the highest common factor first, then factorise until no bracket can be factorised any further.