Non-monic quadratic equations: factorisation practice | Year 10

Build from an already arranged equation to a rearrangement and then a repeated root. Each time, check the factors before solving.

Split a middle term using a pair with product ac and sum b.
Rearrange a quadratic to equal zero before applying the zero-product rule.
Find fractional roots and recognise when two factors give the same root.

The idea

A non-monic quadratic has a coefficient other than one on its squared term. For ax² + bx + c, splitting the middle term works when two numbers multiply to ac and add to b. Factor by grouping, check by expansion, and use the zero-product rule only after the equation equals zero.

Before you begin: Expand two brackets, factor out a common factor and solve a linear equation.

Worked examples

01 Start with the method

Solve the equation, showing both factors and both roots.

2 x^{2} + 7 x + 3 = 0

Step 1

The product ac is 2 × 3 = 6. The numbers 6 and 1 have product 6 and sum 7, so split 7x into 6x + x.

2 x^{2} + 6 x + x + 3 = 0

Step 2

Factor each pair. The repeated bracket is x + 3.

2 x (x + 3) + (x + 3) = (2 x + 1) (x + 3) = 0

Step 3

At least one factor must equal zero. Solve each resulting linear equation.

2 x + 1 = 0 or x + 3 = 0

\Rightarrow x = - \frac{1}{2} or x = - 3

Step 4

Expand the factors to check the original coefficients. Substituting either root also makes the original left side zero.

(2 x + 1) (x + 3) = 2 x^{2} + 7 x + 3

A mistake to avoid

From 2x + 1 = 0 the root is −1/2, not −1. A factor gives a linear equation to solve; its constant is not automatically the root.

02Rearrange before factorising

Solve this equation.

3 x^{2} + 5 x = 2

Step 1

Subtract 2 from both sides so the product will equal zero after factorising.

3 x^{2} + 5 x - 2 = 0

Step 2

The required product is −6 and sum is 5. Split 5x as 6x − x, then factor by grouping.

3 x (x + 2) - (x + 2) = (3 x - 1) (x + 2) = 0

Step 3

Solve each factor equation. The first root requires division by 3.

3 x - 1 = 0 or x + 2 = 0

Result

x = \frac{1}{3} or x = - 2

03Recognise a repeated root

Solve, stating how many distinct roots there are.

4 x^{2} - 12 x + 9 = 0

Step 1

The first and last terms are squares; the middle term is twice the product with a minus sign.

4 x^{2} - 12 x + 9 = {(2 x - 3)}^{2}

Step 2

A squared expression is zero only when the expression itself is zero.

{(2 x - 3)}^{2} = 0

\Rightarrow 2 x - 3 = 0

Step 3

Both identical factors give the same solution. Count it once as a distinct root.

x = \frac{3}{2}

Result

x = \frac{3}{2} (one distinct, repeated root)

Practice questions with answers

Work these out first, then open each solution to compare your method. These questions are untimed.

Question 1

Factorise and solve. Include the fractional root.

2 x^{2} - 5 x - 3 = 0

Show solution 1

(2 x + 1) (x - 3) = 0

\Rightarrow x = - \frac{1}{2} or x = 3

Split −5x as −6x + x. The factors expand to 2x² − 5x − 3; each factor gives one root.

Question 2

Factorise and solve.

3 x^{2} - x - 2 = 0

Show solution 2

(3 x + 2) (x - 1) = 0

\Rightarrow x = - \frac{2}{3} or x = 1

Split −x as −3x + 2x. Expanding the two factors checks every coefficient.

Question 3

Rearrange, then solve.

2 x^{2} + x = 6

Show solution 3

(2 x - 3) (x + 2) = 0

\Rightarrow x = \frac{3}{2} or x = - 2

Subtract 6 first. The factors expand to 2x² + x − 6.

Question 4

Find both roots.

6 x^{2} + 11 x + 3 = 0

Show solution 4

(3 x + 1) (2 x + 3) = 0

\Rightarrow x = - \frac{1}{3} or x = - \frac{3}{2}

The cross terms 9x and 2x add to 11x. Solve both linear factors, retaining the denominators.

Question 5

Find the distinct root.

9 x^{2} - 12 x + 4 = 0

Show solution 5

{(3 x - 2)}^{2} = 0

\Rightarrow x = \frac{2}{3}

This is a perfect square. Both factors are 3x − 2, so they give one distinct root.

Common questions

What makes a quadratic non-monic?

Its squared term has a coefficient other than one. For example, 2x² + 7x + 3 is non-monic, while x² + 7x + 3 is monic.

What if I cannot find an integer factor pair?

First check that you have rearranged correctly and removed any common factor. Not every quadratic factorises into brackets with rational coefficients. Completing the square or the quadratic formula gives another method.

Can I divide by a factor containing x?

Doing so can lose a solution if that factor could equal zero. Keep the equation as a zero product and solve each factor equation instead.