Factoring by Grouping

1 MIN READ · PRACTISE AT YOUR OWN PACE · FREE

Idea

Grouping finds a common factor in stages. Split a four-term expression into two pairs, factor each pair, and look for the same remaining parenthesis. Rearranging terms is allowed because addition is commutative.

Visual guide

VISUAL GUIDEGroup by rows, then by columns

The four rectangles have areas ab, ac, db and dc. Combining rows gives a(b + c) + d(b + c); combining the common width gives (a + d)(b + c). Lengths here are illustrative positive values.

Method

ab+ac+db+dc=a(b+c)+d(b+c)=(a+d)(b+c).

If the two parentheses differ by a sign, factor a negative from one group. If they genuinely differ, try another grouping.

Worked example

Factor $x^3+2x^2-3x-6$ :

x^2(x+2)-3(x+2)=(x+2)(x^2-3).

Over the reals this can continue to $(x+2)(x-\sqrt3)(x+\sqrt3)$ . Over integer coefficients, the first product is the natural stopping point.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

Factor

ax+ay+bx+by

ab(x+y)

(a+b)(x+y)

(a+x)(b+y)

Hint 1 · Find a starting point

Group the first two terms and the last two terms.

Hint 2 · Take the next step

Write a(x+y) + b(x+y).

Show the reasoning

Answer: (a+b)(x+y)

Both groups contain (x+y), giving (a+b)(x+y).

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Common mistake

Factoring $-3x-6$ as $-3(x-2)$ changes the constant term. Distribute the negative to check both signs.

Check your understanding

Factor $ab-2a+3b-6$ .

Show answer

$a(b-2)+3(b-2)=(a+3)(b-2)$ .

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

Optional marks, not a grade. Saved in this browser only. Open notebook →