MPM2D Common Factors and Grouping: Notes and Practice

Try this without a calculator: $17 \times 23 + 17 \times 77$ . Multiplying both products and adding is slow. Notice that 17 appears in both products, so pull it out: $17 \times 23 + 17 \times 77 = 17 (23 + 77) = 17 \times 100 = 1700 .$ You just factored: you rewrote a sum as a product by taking out something both parts share. Factoring is expanding run backwards. Expanding turns $2 x (x + 2)$ into $2 x^{2} + 4 x$ ; factoring turns $2 x^{2} + 4 x$ back into $2 x (x + 2)$ . Why bother? A product is often more useful than a sum. If a rectangle has area $2 x^{2} + 4 x$ , the factored form $2 x (x + 2)$ suggests possible side lengths. In the next unit, factored form will tell you where a parabola crosses the $x$ -axis. Every factoring method in this unit starts with the one in this lesson.

The greatest common factor

A factor of an expression is something that divides into it with no remainder. The greatest common factor (GCF) of several terms is the largest factor they all share. Find it in two parts. For the numbers, take the GCF of the coefficients: the GCF of 12 and 18 is 6, because 6 is the largest number that divides both. For each variable, take the lowest power that appears in every term: $x^{3}$ and $x^{2}$ share $x^{2}$ . A variable that is missing from even one term is not part of the GCF. So the GCF of $6 x^{2} y$ and $9 x y^{2}$ is $3 x y$ : 3 from the coefficients, $x$ (the lower power of $x$ ), and $y$ (the lower power of $y$ ).

Find the GCF of all the terms.
Write the GCF outside a bracket.
Divide each term by the GCF and write the quotients inside the bracket, with their signs.
Check by expanding: the GCF times the bracket must give back the original expression, and the terms left in the bracket should have no common factor other than 1.

12 a^{2} b - 18 a b^{2} = 6 ab (2 a) + 6 ab (- 3 b) = 6 ab (2 a - 3 b)

Fig 1Left: a rectangle of area 2x² + 4x has height 2x and length x + 2, so 2x² + 4x = 2x(x + 2). Right: factoring 2x − 2y + ax − ay by grouping into pairs that share the bracket (x − y).

The area model from expanding also works backwards. The terms $2 x^{2}$ and $4 x$ are the areas of two rectangles that share a side. The longest side they can share is the GCF, $2 x$ . Dividing gives the other sides: $2 x^{2} \div 2 x = x$ and $4 x \div 2 x = 2$ . So the whole rectangle is $2 x$ by $x + 2$ , and $2 x^{2} + 4 x = 2 x (x + 2)$ . If you take out only part of the GCF, say $2 (x^{2} + 2 x)$ , the answer is equal but not fully factored, because the bracket still has a common factor $x$ . Fully factored means no bracket can be factored any further.

The 1 does not disappear

When the GCF is a whole term, dividing that term by itself leaves 1, not 0. For

6 x^{3} - 3 x^{2} + 3 x

the GCF is

3 x

, and

3 x \div 3 x = 1

, so the answer is

3 x (2 x^{2} - x + 1)

. Writing

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

would expand to only two terms, so it cannot be right.

When the first term is negative, it is usually tidier to take out a negative common factor, so the bracket starts with a positive term. Dividing by a negative changes every sign inside: $- 4 x^{2} + 8 x = - 4 x (x - 2)$ , because $- 4 x^{2} \div (- 4 x) = x$ and $8 x \div (- 4 x) = - 2$ . Expand to check: $- 4 x \cdot x = - 4 x^{2}$ and $- 4 x \cdot (- 2) = 8 x$ .

VideoSee how to find the greatest common factor of every term and divide it out to factor a polynomial.Watch on YouTube

A common factor that is a binomial

A common factor does not have to be a single term. In $3 x (x - 2) + 5 (x - 2)$ both parts are multiplied by the same bracket $(x - 2)$ . Treat that bracket as one object, the way you treated 17 above, and take it out: $3 x (x - 2) + 5 (x - 2) = (x - 2) (3 x + 5) .$ What remains in the second bracket is what was multiplying $(x - 2)$ in each part. Sometimes the brackets are opposites, like $(x - 3)$ and $(3 - x)$ . Since $3 - x = - (x - 3)$ , you can rewrite one to match: $5 x (x - 3) - 2 (3 - x) = 5 x (x - 3) + 2 (x - 3) = (x - 3) (5 x + 2) .$

Factoring by grouping

An expression with four terms may have no factor common to all four, yet still factor. In $2 x - 2 y + a x - a y$ the first two terms share 2 and the last two share $a$ . Factor each pair separately: $2 (x - y) + a (x - y)$ . Now the bracket $(x - y)$ is common to both parts, so take it out: $(x - y) (2 + a)$ . This is factoring by grouping. If the brackets do not match after the first step, try pairing the terms differently: $a x + b y + a y + b x$ has no luck as written, but reordered as $a x + a y + b x + b y$ it gives $a (x + y) + b (x + y) = (x + y) (a + b)$ . If no pairing gives matching brackets, the expression cannot be factored by grouping.

Expression	Group in pairs	Factored
$2 x - 2 y + a x - a y$	$2 (x - y) + a (x - y)$	$(x - y) (2 + a)$
$x y + 3 x + 2 y + 6$	$x (y + 3) + 2 (y + 3)$	$(y + 3) (x + 2)$
$x^{2} + 3 x - 2 x - 6$	$x (x + 3) - 2 (x + 3)$	$(x + 3) (x - 2)$
$10 x y - 15 y + 4 x - 6$	$5 y (2 x - 3) + 2 (2 x - 3)$	$(2 x - 3) (5 y + 2)$

Grouping works when both pairs leave the same bracket

Watch the sign of the second pair

x^{2} + 3 x - 2 x - 6

the second pair is

- 2 x - 6

. Take out

- 2

, not

2

, so the bracket matches:

- 2 x - 6 = - 2 (x + 3)

. Taking out

+ 2

gives

2 (- x - 3)

, which does not match

(x + 3)

. The third row of the table is exactly how the next lesson factors trinomials such as

2 x^{2} + 11 x + 5

Common Factors and Grouping

Key things to remember

The greatest common factor

A common factor that is a binomial

Factoring by grouping

Worked examples

Where marks get lost

Flashcards 7

Practice