Factorising Monic Quadratic Trinomials
A monic quadratic trinomial has the form \(x^2+bx+c\), with a leading coefficient of 1. It factorises as \((x+p)(x+q)\) when \(p\) and \(q\) multiply to \(c\) and add to \(b\), so \(x^2-9x+14=(x-2)(x-7)\). In NSW Year 10 Advanced maths the trinomials include common factors, negative leading terms and two pronumerals (NESA outcome MA5‑ALG‑P‑01).
You will factorise trinomials with positive and negative terms, take out a common factor or \(-1\) first, and factorise expressions in two pronumerals such as \(a^2+2ab-15b^2\). You will treat a bracket as a single term, factorise a quartic as a quadratic in \(a^2\), find all values of a coefficient that make a trinomial factorise, and use factors to prove that an expression is never prime. It is part of Year 10 Advanced Quadratics.
Every question with a fully worked solution.
- Factorising Monic Quadratic Trinomials - Video - Factorising Monic Quadratics Watch
Theory
A monic quadratic trinomial \(x^2 + bx + c\) factorises as \((x + m)(x + n)\), where \(m\) and \(n\) multiply to \(c\) and add to \(b\). This Year 10 Advanced maths page covers the sign rules, common factors first, two-variable and substituted forms, and using factors in reasoning. It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
A quadratic trinomial has three terms and highest power \(2\): \(ax^2 + bx + c\). It is monic when \(a = 1\), as in \(x^2 - 7x + 10\).
Factorising is the reverse of expanding. Since \((x + m)(x + n) = x^2 + (m + n)x + mn\), you need two numbers with product \(c\) and sum \(b\).
Factorise fully means keep going until no bracket can be factorised further: take out any common factor first, and check whether a bracket is itself a difference of two squares.
Not every trinomial factorises with integers. \(x^2 + 3x + 5\) has no pair of integers with product \(5\) and sum \(3\).
How to factorise \(x^2 + bx + c\)
- Tidy: write in descending powers, take out any common factor, and take out \(-1\) if the \(x^2\) term is negative.
- Find two numbers with product \(c\) and sum \(b\).
- Write \((x + m)(x + n)\), putting back any common factor in front.
- Check by expanding: the middle term must come back exactly.
Using factorised forms
- Areas: a rectangle of area \(x^2 + bx + c\) has sides equal to the two factors. Subtract them to compare length and width.
- Unknown coefficients: expand the factorised form and match coefficients, then use any extra condition given.
- Which values work: list the factor pairs of the constant (Example 4).
- Prime or not: if an expression in \(n\) factorises into two integers that are both greater than \(1\), the value is not prime.
- Long products: multiply the brackets in pairs whose \(x^2\) and \(x\) terms match, then substitute \(u\) for the repeated part and factorise in \(u\).
(i) \(x^2 + 13x + 42\)
(ii) \(x^2 - 5x - 84\)
(iii) \(y^2 - 15y + 54\)
Find two numbers that multiply to the constant and add to the coefficient of the middle term.
(i) Product \(42\), sum \(13\): \(6\) and \(7\).
| \(x^2 + 13x + 42\) | \(=\) | \(\boldsymbol{(x + 6)(x + 7)}\) |
(ii) Product \(-84\), so opposite signs; sum \(-5\), so the bigger one is negative. Pairs of \(84\): \(1 \times 84, 2 \times 42, 3 \times 28, 4 \times 21, 6 \times 14, 7 \times 12\). The pair differing by \(5\) is \(7\) and \(12\).
| \(x^2 - 5x - 84\) | \(=\) | \(\boldsymbol{(x - 12)(x + 7)}\) |
(iii) Product \(54\), sum \(-15\): both negative, \(-6\) and \(-9\).
| \(y^2 - 15y + 54\) | \(=\) | \(\boldsymbol{(y - 6)(y - 9)}\) |
Check (ii) by expanding: \(x^2 + 7x - 12x - 84 = x^2 - 5x - 84\).
(i) \(3x^2 + 6x - 72\)
(ii) \(45 + 4x - x^2\)
(i) Take out the highest common factor first.
| \(3x^2 + 6x - 72\) | \(=\) | \(3(x^2 + 2x - 24)\) |
| \(\) | \(=\) | \(\boldsymbol{3(x + 6)(x - 4)}\) |
(ii) Rewrite in descending powers and take out \(-1\) so the \(x^2\) term is positive.
| \(45 + 4x - x^2\) | \(=\) | \(-(x^2 - 4x - 45)\) |
| \(\) | \(=\) | \(-(x - 9)(x + 5)\) |
| \(\) | \(=\) | \(\boldsymbol{(9 - x)(x + 5)}\) |
Both \(-(x - 9)(x + 5)\) and \((9 - x)(x + 5)\) are correct: the minus sign has been put inside the first bracket.
(i) \(p^2 + 3pq - 28q^2\)
(ii) \((x + 2)^2 + 4(x + 2) - 21\)
(iii) \(a^4 - 10a^2 + 9\)
(i) Treat it as \(p^2 + 3p - 28\) with \(q\) attached to each number: \(7\) and \(-4\).
| \(p^2 + 3pq - 28q^2\) | \(=\) | \(\boldsymbol{(p + 7q)(p - 4q)}\) |
(ii) Let \(u = x + 2\). Then factorise in \(u\) and substitute back.
| \(u^2 + 4u - 21\) | \(=\) | \((u + 7)(u - 3)\) |
| \(\) | \(=\) | \((x + 2 + 7)(x + 2 - 3)\) |
| \(\) | \(=\) | \(\boldsymbol{(x + 9)(x - 1)}\) |
(iii) It is a quadratic in \(a^2\). Then each bracket is a difference of two squares.
| \(a^4 - 10a^2 + 9\) | \(=\) | \((a^2 - 1)(a^2 - 9)\) |
| \(\) | \(=\) | \(\boldsymbol{(a - 1)(a + 1)(a - 3)(a + 3)}\) |
Expanding \((x + m)(x + n)\) gives \(x^2 + (m + n)x + mn\). So \(mn = 24\) and \(b = m + n\). For \(b\) to be positive, take \(m\) and \(n\) both positive.
| \(1 \times 24\) | \(:\) | \(b = 25\) |
| \(2 \times 12\) | \(:\) | \(b = 14\) |
| \(3 \times 8\) | \(:\) | \(b = 11\) |
| \(4 \times 6\) | \(:\) | \(b = 10\) |
So \(b = \boldsymbol{10}\), \(\boldsymbol{11}\), \(\boldsymbol{14}\) or \(\boldsymbol{25}\).
Negative pairs such as \(-4 \times -6\) give negative values of \(b\), which the question rules out.
Common pitfalls
Frequently asked questions
How do you factorise x squared + bx + c?
Find two numbers that multiply to \(c\) and add to \(b\). If they are \(m\) and \(n\), the factorised form is \((x + m)(x + n)\). Check by expanding.
What does monic mean in maths?
A monic polynomial has a leading coefficient of 1. So x squared - 5x + 6 is a monic quadratic, but 2x squared - 5x + 3 is not.
How do you factorise when the constant is negative?
A negative constant means the two numbers have opposite signs. Look for the factor pair that differs by the middle coefficient, and give the bigger number the same sign as the middle term.
How do you factorise a quadratic with a negative x squared term?
Take out -1 first so the x squared term is positive, factorise what is left, then keep the -1 in front or multiply it into one bracket.
Can every quadratic be factorised?
No. If no pair of integers has the right product and sum, the trinomial does not factorise over the integers. It may still factorise with surds, which completing the square or the quadratic formula will show.