Factorise a quadratic with a non-unit leading coefficient

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Factorise a quadratic with a non-unit leading coefficient

For ax²+bx+c, the first coefficient affects both brackets. One reliable method splits the middle term into two whose coefficients multiply to ac and add to b. Grouping then exposes a common binomial factor.

For 2x²+7x+3, ac=6 and the useful pair is 6 and 1. Rewrite 7x as 6x+x, then factorise the two groups. Expanding the final brackets checks the leading, middle and constant coefficients together.

Worked example: Factorise 2x²+7x+3.

Split the middle: 2x²+6x+x+3.

Factorise the groups: 2x(x+3)+1(x+3).

Take out the shared bracket.

Answer: (2x+1)(x+3).

Check your understanding: Expand (3x−2)(x+1).

Answer: 3x²+x−2.

Why: The four products give 3x²+3x−2x−2.