Factorising Quadratic Expressions
A quadratic expression such as \(x^2 - x - 6\) factorises into two brackets by finding two numbers that multiply to the constant and add to the middle coefficient. In NSW Year 9 Advanced maths this begins with monic quadratics (NESA outcome MA5‑ALG‑P‑01) and moves on to non-monic ones.
You will factorise monic trinomials with every sign pattern, take out a common factor first, as in \(3x^2 + 6x - 45\), and deal with a negative \(x^2\) term. You will then factorise non-monic expressions such as \(5x^2 - 13x - 6\), factorise \(12x^2y - 2xy - 4y\) fully, and use factorising to find the width of a rectangle from its area. It is part of Year 9 Advanced Algebra: Expanding and Factorising.
Theory
To factorise a quadratic \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\). This Year 9 Advanced maths page covers a key skill if you are heading for Advanced or Extension maths: reading the signs, taking out a common factor first, non-monic quadratics such as \(6x^2 + 7x - 20\), and area problems.
A quadratic trinomial has the form \(ax^2 + bx + c\). It is monic when \(a = 1\), as in \(x^2 - 5x + 4\), and non-monic otherwise, as in \(4x^2 + 4x - 3\).
Expanding \((x + p)(x + q)\) gives \(x^2 + (p + q)x + pq\). So to factorise a monic quadratic, find \(p\) and \(q\) with product \(c\) and sum \(b\).
The signs tell you where to look. If \(c\) is positive, both numbers have the sign of \(b\). If \(c\) is negative, the numbers have opposite signs and the larger one has the sign of \(b\).
For a non-monic quadratic, find two numbers with product \(ac\) and sum \(b\), use them to split the middle term, then group in pairs.
| \(b\) | \(c\) | The two numbers | Example |
|---|---|---|---|
| \(+\) | \(+\) | both positive | \(x^2 + 7x + 10 = (x + 2)(x + 5)\) |
| \(-\) | \(+\) | both negative | \(x^2 - 7x + 10 = (x - 2)(x - 5)\) |
| \(+\) | \(-\) | opposite signs, larger positive | \(x^2 + 3x - 10 = (x + 5)(x - 2)\) |
| \(-\) | \(-\) | opposite signs, larger negative | \(x^2 - 3x - 10 = (x - 5)(x + 2)\) |
How to factorise a quadratic
- Arrange in descending powers of \(x\). If the \(x^2\) term is negative, take out \(-1\).
- Take out any HCF, including a pronumeral: \(5x^2 y + 10xy - 15y = 5y(x^2 + 2x - 3)\).
- Monic: find two numbers with product \(c\) and sum \(b\), and write \((x + p)(x + q)\).
- Non-monic: find two numbers with product \(ac\) and sum \(b\), split the middle term, group in pairs and take out the common bracket.
- Check by expanding. The middle term is where errors show up.
- Area problems: if area \(=\) quadratic and one side is given, factorise; the other factor is the other side. Then use any extra fact (a perimeter, a length) to find \(x\).
(i) \(x^2 - 13x + 36\)
(ii) \(x^2 + 2x - 48\)
Find two numbers that multiply to the constant and add to the coefficient of \(x\).
(i) Product \(+36\), sum \(-13\): both numbers are negative. \(-4 \times (-9) = 36\) and \(-4 + (-9) = -13\).
| \(x^2 - 13x + 36\) | \(=\) | \(\boldsymbol{(x - 4)(x - 9)}\) |
(ii) Product \(-48\), sum \(+2\): the signs differ, and the larger number is positive. \(8 \times (-6) = -48\) and \(8 + (-6) = 2\).
| \(x^2 + 2x - 48\) | \(=\) | \(\boldsymbol{(x + 8)(x - 6)}\) |
(i) \(4x^2 - 8x - 60\)
(ii) \(30 - x - x^2\)
(i) The HCF is \(4\). Then product \(-15\), sum \(-2\): \(-5\) and \(3\).
| \(4x^2 - 8x - 60\) | \(=\) | \(4(x^2 - 2x - 15)\) |
| \(\) | \(=\) | \(\boldsymbol{4(x - 5)(x + 3)}\) |
(ii) Rearrange in descending powers, then take out \(-1\) so the \(x^2\) term is positive. Product \(-30\), sum \(+1\): \(6\) and \(-5\).
| \(30 - x - x^2\) | \(=\) | \(-x^2 - x + 30\) |
| \(\) | \(=\) | \(-(x^2 + x - 30)\) |
| \(\) | \(=\) | \(\boldsymbol{-(x + 6)(x - 5)}\) |
Multiply \(a \times c = 6 \times (-20) = -120\). Find two numbers with product \(-120\) and sum \(7\): \(15\) and \(-8\).
| \(6x^2 + 7x - 20\) | \(=\) | \(6x^2 + 15x - 8x - 20\) |
| \(\) | \(=\) | \(3x(2x + 5) - 4(2x + 5)\) |
| \(\) | \(=\) | \(\boldsymbol{(2x + 5)(3x - 4)}\) |
Group in pairs and take out \(-4\) from the second pair so the brackets match. Check: \(6x^2 - 8x + 15x - 20 = 6x^2 + 7x - 20\).
(i) Find an expression for its length.
(ii) The length is \(13\) cm. Find the area.
Area \(=\) length \(\times\) width, so factorise the area and look for the factor \((x + 3)\).
(i) \(ac = 6\); \(6 + 1 = 7\).
| \(2x^2 + 7x + 3\) | \(=\) | \(2x^2 + 6x + x + 3\) |
| \(\) | \(=\) | \(2x(x + 3) + 1(x + 3)\) |
| \(\) | \(=\) | \((x + 3)(2x + 1)\) |
| \(\text{Length}\) | \(=\) | \(\boldsymbol{(2x + 1)}\text{ cm}\) |
(ii)
| \(2x + 1\) | \(=\) | \(13\) |
| \(x\) | \(=\) | \(6\) |
| \(\text{Width}\) | \(=\) | \(6 + 3 = 9\text{ cm}\) |
| \(A\) | \(=\) | \(13 \times 9 = \mathbf{117}\text{ cm}^2\) |
Check: \(2(36) + 7(6) + 3 = 117\).
Common pitfalls
Frequently asked questions
How do you factorise x squared plus bx plus c?
Find two numbers that multiply to c and add to b, then write (x + first number)(x + second number). For x squared + 11x + 30, the numbers 5 and 6 give (x + 5)(x + 6).
How do you factorise a quadratic when the x squared coefficient is not 1?
Multiply the x squared coefficient by the constant, then find two numbers with that product that add to the middle coefficient. Split the middle term using them, group the four terms in pairs and take out the common bracket.
How do you know what signs go in the brackets?
If the constant is positive, both signs match the sign of the middle term. If the constant is negative, one bracket has plus and one has minus, and the larger number takes the sign of the middle term.
What if the x squared term is negative?
Take out negative 1 first so the x squared term becomes positive, then factorise what is left in the bracket. Keep the minus sign in front of the answer.
How can I check my factorisation?
Expand your brackets. You should get back the original quadratic exactly, including the middle term and its sign.