Factorising Monic Quadratic Trinomials
Every question with a fully worked solution.
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Theory
A monic quadratic trinomial such as \(x^2 + 7x + 12\) factorises into two brackets, \((x + 3)(x + 4)\). This Year 9 maths page shows how to find the two numbers that multiply to the number term and add to the \(x\) coefficient, how the signs work, and what to do when there is a common factor or a negative \(x^2\) term.
A quadratic expression has \(x^2\) as its highest power, such as \(x^2 - 5x + 6\). A trinomial has three terms.
A quadratic is monic when the coefficient of \(x^2\) is \(1\). In \(x^2 + bx + c\), \(b\) is the coefficient of \(x\) and \(c\) is the constant (number) term.
Expanding \((x + p)(x + q)\) gives \(x^2 + (p + q)x + pq\). So to factorise \(x^2 + bx + c\), look for two numbers \(p\) and \(q\) with
Not every trinomial factorises with whole numbers. \(x^2 + x + 1\) has no two whole numbers that work.
The general pattern:
Two special cases:
How to factorise \(x^2 + bx + c\)
- Take out any common factor first, and take out \(-1\) if the \(x^2\) term is negative.
- Write down the product \(c\) and the sum \(b\).
- Use the signs. If \(c\) is positive, both numbers have the sign of \(b\). If \(c\) is negative, the numbers have opposite signs and the bigger one has the sign of \(b\).
- List factor pairs of \(c\) until one pair has the right sum (or difference). For large numbers, use a table.
- Write the two brackets, \((x + p)(x + q)\).
- Check by expanding. The middle term is the one most often wrong.
(i) \(x^2 + 9x + 14\)
(ii) \(x^2 - 9x + 18\)
Find two numbers that multiply to the number term and add to the \(x\) coefficient. The number term is positive, so both numbers have the sign of the middle term.
(i) Product \(14\), sum \(9\): \(2 \times 7 = 14\) and \(2 + 7 = 9\).
| \(x^2 + 9x + 14\) | \(=\) | \(\boldsymbol{(x + 2)(x + 7)}\) |
(ii) Product \(18\), sum \(-9\): both numbers are negative. \((-3) \times (-6) = 18\) and \(-3 + (-6) = -9\).
| \(x^2 - 9x + 18\) | \(=\) | \(\boldsymbol{(x - 3)(x - 6)}\) |
(i) \(x^2 + 2x - 24\)
(ii) \(x^2 - 4x - 60\)
The number term is negative, so the two numbers have opposite signs. Their difference is the size of the \(x\) coefficient.
(i) Product \(-24\), sum \(2\): \(6 \times (-4) = -24\) and \(6 + (-4) = 2\).
| \(x^2 + 2x - 24\) | \(=\) | \(\boldsymbol{(x + 6)(x - 4)}\) |
(ii) Product \(-60\), sum \(-4\). List the factor pairs of \(60\) and look for a difference of \(4\):
| Pair | \(1, 60\) | \(2, 30\) | \(3, 20\) | \(4, 15\) | \(5, 12\) | \(6, 10\) |
|---|---|---|---|---|---|---|
| Difference | \(59\) | \(28\) | \(17\) | \(11\) | \(7\) | \(4\) |
The sum is negative, so the bigger number is negative: \(-10\) and \(6\).
| \(x^2 - 4x - 60\) | \(=\) | \(\boldsymbol{(x - 10)(x + 6)}\) |
Every term is divisible by \(5\). Take it out first, so that the bracket is monic (starts with \(x^2\)).
| \(5x^2 + 15x - 50\) | \(=\) | \(5(x^2 + 3x - 10)\) |
Product \(-10\), sum \(3\): \(5\) and \(-2\).
| \(\) | \(=\) | \(\boldsymbol{5(x + 5)(x - 2)}\) |
Check: \(5(x^2 - 2x + 5x - 10) = 5x^2 + 15x - 50\). ✓
Rewrite in descending powers and take out \(-1\), so that the bracket starts with \(+x^2\).
| \(12 + x - x^2\) | \(=\) | \(-(x^2 - x - 12)\) |
Product \(-12\), sum \(-1\): \(-4\) and \(3\).
| \(\) | \(=\) | \(-(x - 4)(x + 3)\) |
Multiply the \(-1\) into the first bracket: \(-(x - 4) = 4 - x\).
| \(\) | \(=\) | \(\boldsymbol{(4 - x)(x + 3)}\) |
Both \(-(x - 4)(x + 3)\) and \((4 - x)(x + 3)\) are correct answers.
Common pitfalls
Frequently asked questions
How do you factorise a quadratic trinomial?
For \(x^2 + bx + c\), find two numbers that multiply to \(c\) and add to \(b\). If the numbers are \(p\) and \(q\), the answer is \((x + p)(x + q)\).
What does monic mean?
A monic quadratic has \(1\) as the coefficient of \(x^2\). For example, \(x^2 - 5x + 6\) is monic, but \(2x^2 - 5x + 6\) is not.
How do I know which number is negative?
If the number term is negative, one number is positive and one is negative, and the bigger one takes the sign of the \(x\) coefficient. For \(x^2 + 4x - 12\) the numbers are \(6\) and \(-2\).
What if the quadratic has a common factor?
Take the common factor out first. Then factorise the monic quadratic left inside the bracket, and keep the common factor at the front.
Does every quadratic factorise?
No. If no two whole numbers have the right product and sum, the quadratic does not factorise this way. Other methods, such as the quadratic formula, are met later.