Question

Factorise
${ax}^{2} + bx + c$
where a is not equal to 0 by splitting the
middle term.
​

Answer 1

Step-by-step explanation:

First, factor out all constants which evenly divide all three terms. If a is negative, factor out -1. This will leave an expression of the form d (ax2 + bx + c), where a, b, c, and d are integers, and a > 0. We can now turn to factoring the inside expression.

Here is how to factor an expression ax2 + bx + c, where a > 0:

Write out all the pairs of numbers that, when multiplied, produce a.

Write out all the pairs of numbers that, when multiplied, produce c.

Pick one of the a pairs -- (a1, a2) -- and one of the c pairs -- (c1, c2).

If c > 0: Compute a1c1 + a2c2. If | a1c1 + a2c2| = b, then the factored form of the quadratic is

(a1x + c2)(a2x + c1) if b > 0.

(a1x - c2)(a2x - c1) if b < 0.

If a1c1 + a2c2≠b, compute a1c2 + a2c1. If a1c2 + a2c1 = b, then the factored form of the quadratic is (a1x + c1)(a2x + c2) or (a1x + c1)(a2x + c2). If a1c2 + a2c1≠b, pick another set of pairs.

If c < 0: Compute a1c1 -a2c2. If | a1c1 - a2c2| = b, then the factored form of the quadratic is:

(a1x - c2)(a2x + c1) where a1c1 > a2c2 if b > 0 and a1c1 < a2c2 if b < 0.

Using FOIL, the outside pair plus (or minus) the inside pair must equal b.

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