Year 10A - Factorising non-monic quadratics - Free practice

Year 10A Factorising Non-Monic Quadratics Worksheets & Practice

Master factorising and solving quadratics with a leading coefficient, ready for senior Methods maths.

Quick answer

How do you factorise non-monic quadratics and solve quadratic equations?

A non-monic quadratic has a leading coefficient other than 1, e.g. 6x² + 7x − 3. Factorise it using the ac method: multiply a and c, find two numbers that multiply to ac and add to b, split the middle term, then factor by grouping. Quadratic equations can also be solved with the quadratic formula or by completing the square.

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Worked examples

Year 10A factorising non-monic quadratics: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

Example 1 · Factorising a non-monic quadratic using the ac method
Factorise $6x^2 + 11x + 3$ .
Show worked solution
Key idea: For a non-monic quadratic $ax^2 + bx + c$ , find two numbers that multiply to $ac$ and add to $b$ , then split the middle term and factorise by grouping.
1. Identify $a = 6$ , $b = 11$ , $c = 3$ , so $ac = 6 \times 3 = 18$ .
2. Find two numbers that multiply to $18$ and add to $11$ : these are $2$ and $9$ .
3. Split the middle term: $6x^2 + 2x + 9x + 3$ .
4. Factorise by grouping in pairs: $2x(3x + 1) + 3(3x + 1)$ .
5. Factor out the common bracket: $(3x + 1)(2x + 3)$ .
Answer: $(3x + 1)(2x + 3)$
Example 2 · Solving a non-monic quadratic equation by factorising
Solve $4x^2 - 4x - 3 = 0$ .
Show worked solution
Key idea: Factorise the left-hand side using the ac method, then apply the null factor law.
1. Identify $a = 4$ , $b = -4$ , $c = -3$ , so $ac = 4 \times (-3) = -12$ .
2. Find two numbers that multiply to $-12$ and add to $-4$ : these are $-6$ and $2$ .
3. Split and group: $4x^2 - 6x + 2x - 3 = 2x(2x - 3) + 1(2x - 3) = (2x - 3)(2x + 1)$ .
4. Apply the null factor law: $2x - 3 = 0$ or $2x + 1 = 0$ .
5. Solve each: $x = \frac{3}{2} \text{ or } x = -\frac{1}{2}$
Answer: $x = \dfrac{3}{2}$ or $x = -\dfrac{1}{2}$
Example 3 · Solving a quadratic equation in a real-life context
A rectangular garden has a length that is $3$ metres more than twice its width. The area of the garden is $27$ square metres. Find the width of the garden.
Show worked solution
Key idea: Write an equation for the area using the given information, rearrange to standard form, then solve by factorising.
1. Let the width be $w$ metres. Then the length is $(2w + 3)$ metres.
2. Write the area equation: $w(2w + 3) = 27$ , which expands to $2w^2 + 3w = 27$ .
3. Rearrange to standard form: $2w^2 + 3w - 27 = 0$ .
4. Use the ac method: $ac = 2 \times (-27) = -54$ . Two numbers that multiply to $-54$ and add to $3$ : these are $9$ and $-6$ . Split and group: $2w^2 + 9w - 6w - 27 = w(2w + 9) - 3(2w + 9) = (2w + 9)(w - 3)$ .
5. Apply the null factor law: $2w + 9 = 0$ gives $w = -\dfrac{9}{2}$ (reject, as width must be positive), or $w - 3 = 0$ gives $w = 3$ .
Answer: The width of the garden is $3$ metres.

Common mistakes

Where students usually lose marks

Forgetting to rearrange to standard form before factorising

For $3x^2 + 5x = 2$ , student writes $x(3x + 5) = 2$ and concludes $x = 2$ or $3x + 5 = 2$ .

The null factor law only works when the product equals zero. Always rearrange to $ax^2 + bx + c = 0$ first before factorising.

Wrong signs when splitting the middle term

For $ac = -12$ and $b = 4$ , student chooses $-6$ and $-2$ because $(-6) \times (-2) = 12$ .

Check both conditions: the two numbers must multiply to $ac$ (with its sign) and add to $b$ . Write out factor pairs of $|ac|$ and test signs carefully until both conditions are met.

Sign error when factorising a group with a negative factor

$6x^2 - 9x + 4x - 6 = 3x(2x - 3) + 2(2x - 3)$ , but the second group gives $+2(2x - 3) = 4x - 6$ which is incorrect when it should be $+2(2x - 3) = 4x - 6$ student writes $3x(2x-3) - 2(2x + 3)$ .

After splitting, factor out a negative from the second pair when needed. Check by expanding your factored group back to make sure it matches the original terms exactly.

Discarding valid negative solutions without checking the context

For a quadratic with solutions $x = 5$ and $x = -5$ , student automatically discards $x = -5$ in every problem.

Only reject a solution if it is physically impossible given the context (for example, a negative length). In purely algebraic problems, both solutions are valid unless you have a specific reason to exclude one.

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Your turn

Solve $6x^2 + 7x - 3 = 0$ using any method. Enter the larger solution, correct to 2 decimal places.

Got a different factorising non-monic quadratics question wrong? Upload it to Tuterly and get the worked solution and a fresh one to practise, free.

About this worksheet

Why we built it

Non-monic quadratics are expressions like 6x² + 11x + 3, where the coefficient of x² is something other than 1. This is one of the trickier skills in Year 10A and a genuine foundation for senior Methods maths. These worksheets step through factorising by grouping (the ac method), completing the square and the quadratic formula, with fully worked solutions so students can see exactly where their reasoning should go.

What's covered

Sub-skills your student will practise

✓Identifying monic vs non-monic quadratic expressions
✓Factorising non-monic quadratics using the ac (splitting the middle) method
✓Factorising by grouping in pairs after splitting
✓Solving non-monic quadratic equations by factorising
✓Solving quadratic equations using the quadratic formula
✓Solving quadratic equations by completing the square

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Frequently asked

Questions parents ask about factorising non-monic quadratics

What makes a quadratic 'non-monic' and why does it matter?

A quadratic is non-monic when the coefficient of x² is not 1, for example 3x² + 10x + 8. It matters because the simple factorising shortcuts from Year 9 no longer work directly, so students need a reliable method like the ac approach.

What is the ac method and how does it work?

For ax² + bx + c, multiply a × c, then find two numbers that multiply to give ac and add to give b. Split the middle term using those two numbers, then factorise by grouping in pairs. For example, 6x² + 11x + 3 has ac = 18; the pair 9 and 2 works, giving (2x + 3)(3x + 1).

When should a student use the quadratic formula instead of factorising?

If no whole-number pair satisfies the ac condition, the quadratic may not factorise over the integers. The quadratic formula x = (-b ± √(b² - 4ac)) ÷ 2a always works and is the go-to method for Year 10A and senior maths when factorising fails or feels slow.

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