Factoring out the greatest common factor
Distributing turns a product into a sum. Given the sum, how do you get the product back?
Where you are 12 in Expressions
When you reach for it
The distributive property multiplies an outside factor into every term of a bracket:
That is the forward direction, from a product to a sum. This lesson runs the same equation from right to left. Given 8a + 20, you find the number that both terms share, 4, pull it out to the front, and write what is left of each term in a bracket. Rewriting a sum as a product this way is called factoring out the greatest common factor, or factoring out a monomial, because the shared factor is a single term such as 4, 5m or 3y^2.
You reach for it whenever a product is more useful than a sum. A product can be read as equal groups: 6x + 9 = 3(2x + 3) says the amount is three copies of 2x + 3. A product shows what a quantity is a multiple of. A product is what solving for one letter needs when that letter sits in two terms, and it is the first move before factoring by grouping or factoring a trinomial. The words term, coefficient and constant come from Terms, coefficients and constants.
This lesson runs over two days. Part 1 finds the greatest common factor of several terms, pulls it to the front, and checks the result by distributing back. Part 2 factors out a negative, factors expressions with fraction coefficients, and reads what the factored form says.
The procedure
Part 1: the greatest common factor of terms
A factor of a term is anything that multiplies into it exactly. The term 12x^3y is 2 \cdot 2 \cdot 3 \cdot x \cdot x \cdot x \cdot y, so 3, 4x, x^2y and 6x^3 are all factors of it. A common factor of several terms is a factor of every one of them, and the greatest common factor (GCF) is the largest one: the number part as large as possible and every letter to as high a power as possible.
For whole numbers, the GCF was found in Greatest common factor and least common multiple, by listing factors or from prime factorizations. For terms with letters, write each term as a product of primes and single letters and keep what every term has. Take 12x^3y and 18x^2y^2:
Both contain one 2, one 3, two x’s and one y, and nothing more is shared. So the GCF is 2 \cdot 3 \cdot x \cdot x \cdot y = 6x^2y. Read off without writing every factor:
- the number part is the GCF of the coefficients, here \operatorname{GCF}(12, 18) = 6;
- each letter that appears in every term goes in to the smallest power it has, here x^2 (from x^3 and x^2) and y (from y and y^2);
- a letter missing from even one term does not go in at all.
The smallest power is the right one because a factor cannot be bigger than the term it divides: x^3 is not a factor of x^2, but x^2 is a factor of both x^2 and x^3 = x^2 \cdot x.
Pulling it to the front. Once the GCF is known, divide each term by it. The distributive property, read from right to left, says
so the GCF goes outside and the quotients go inside, in the same order, with the same signs:
because 12x^3y \div 6x^2y = 2x and 18x^2y^2 \div 6x^2y = 3y. A subtraction is handled as the addition of a negative term, so the minus sign is carried into the bracket with its term. A term that equals the GCF leaves 1 behind, never nothing, since a = a \cdot 1.
Checking. Factoring changes the look of an expression, never its value, so the check is to distribute back: 6x^2y \cdot 2x = 12x^3y and 6x^2y \cdot 3y = 18x^2y^2, which is where we started. Then look inside the bracket. If its terms still share a factor, the factor taken out was common but not the greatest: 12x + 18 = 2(6x + 9) is true, but 6x and 9 still share 3, so take that out too and get 6(2x + 3). Factoring out the GCF is finished when the terms in the bracket have no common factor other than 1.
Part 2: negatives, fractions, and reading the factored form
Factoring out a negative. When the first term is negative, it is usual to take out the negative of the GCF, so the bracket starts with a positive term. Dividing every term by a negative number flips every sign:
since -10x \div (-2) = 5x and 4 \div (-2) = -2. The form 2(-5x + 2) is just as correct, as distributing shows; the two differ only in where the minus sign is kept. A negative factor is worth taking out because the next steps of factoring a trinomial are easier when the leading coefficient is positive.
Fraction coefficients. A common factor need not be a whole number. In \displaystyle \frac{2}{3}x - \frac{5}{6} both coefficients are multiples of \displaystyle \frac{1}{6}, because \displaystyle \frac{2}{3} = \frac{4}{6}, so
The useful factor here is one over the least common denominator, which is the least common multiple of the denominators: it leaves whole numbers in the bracket. “Greatest” stops meaning anything once fractions are allowed (see Boundaries), so the choice is made by what the bracket should look like.
Reading the factored form. A product says something the sum hides. The expression 6x + 9 is six of something and nine ones; the factored form 3(2x + 3) says the same amount splits into three identical groups, each holding two x’s and three ones. The factored form also shows that whenever x is a whole number, 6x + 9 is a multiple of 3. Rewriting an expression to show what it means in a situation has a lesson of its own, Rewriting an expression to show what it means.
Worked examples
Examples 1 to 3 go with Part 1. Examples 4 to 7 go with Part 2.
The greatest common factor of terms
Find the GCF of (a) 16a^2b^3, 24a^4b and 40a^3 \qquad (b) 9n^2 and 10n.
Solution(a) The number part is \operatorname{GCF}(16, 24, 40). In primes, 16 = 2^4, 24 = 2^3 \cdot 3 and 40 = 2^3 \cdot 5; the only shared prime is 2, at most three times, so the number part is 2^3 = 8. The letter a appears in all three terms, as a^2, a^4 and a^3, so it goes in to the smallest power, a^2. The letter b is missing from 40a^3, so it does not go in. The GCF is 8a^2.
Check: 16a^2b^3 = 8a^2 \cdot 2b^3, 24a^4b = 8a^2 \cdot 3a^2b and 40a^3 = 8a^2 \cdot 5a, and the leftovers 2b^3, 3a^2b and 5a share no factor other than 1.
(b) The coefficients 9 and 10 share no prime, so the number part is 1. The letter n appears in both, as n^2 and n, so it goes in once. The GCF is n.
Factoring and checking
Factor (a) 8a + 20 \qquad (b) 15m^2 - 10m, and check by distributing.
Solution(a) \operatorname{GCF}(8, 20) = 4, and a is not in the second term, so the GCF is 4. Dividing, 8a \div 4 = 2a and 20 \div 4 = 5. By the distributive property,
Check: 4 \cdot 2a + 4 \cdot 5 = 8a + 20. The figure shows the same thing as area: two rectangles of height 4, with areas 8a and 20, have widths 2a and 5.
(b) Read the expression as 15m^2 + (-10m). The coefficients give \operatorname{GCF}(15, 10) = 5, and m is in both terms, to the powers 2 and 1, so the GCF is 5m. Dividing, 15m^2 \div 5m = 3m and -10m \div 5m = -2. So
Check: 5m \cdot 3m = 15m^2 and 5m \cdot (-2) = -10m. Inside the bracket, 3m and -2 share nothing, so the factoring is complete.
Three terms, and finishing a partial factoring
(a) Factor 6y^4 - 9y^3 + 3y^2. \qquad (b) Jasper factors 24k^2 + 36k as 4k(6k + 9). Is he finished?
Solution(a) \operatorname{GCF}(6, 9, 3) = 3, and the smallest power of y is y^2, so the GCF is 3y^2. Dividing each term: 6y^4 \div 3y^2 = 2y^2, -9y^3 \div 3y^2 = -3y, and 3y^2 \div 3y^2 = 1. The last term is the GCF itself, and it leaves 1 in the bracket:
Check: 3y^2 \cdot 2y^2 - 3y^2 \cdot 3y + 3y^2 \cdot 1 = 6y^4 - 9y^3 + 3y^2. Writing 3y^2(2y^2 - 3y) instead would distribute back to only two terms, so the 1 cannot be left out.
(b) His statement is true: distributing 4k(6k + 9) gives 24k^2 + 36k. But 6k and 9 still share the factor 3, so 4k was a common factor and not the greatest. Taking the 3 out of the bracket as well, by the associative property of multiplication,
which agrees with finding the GCF directly: \operatorname{GCF}(24, 36) = 12, and k appears to the first power in 36k.
Factoring out a negative
Factor out a negative common factor: (a) -6x - 15 \qquad (b) -2w^2 + 8w - 10.
Solution(a) \operatorname{GCF}(6, 15) = 3, and the first term is negative, so take out -3. Dividing each term by -3 flips its sign: -6x \div (-3) = 2x and -15 \div (-3) = 5. So
Check: -3 \cdot 2x = -6x and -3 \cdot 5 = -15.
(b) \operatorname{GCF}(2, 8, 10) = 2, and w is not in the last term, so take out -2: -2w^2 \div (-2) = w^2, 8w \div (-2) = -4w, and -10 \div (-2) = 5. So
Check: -2 \cdot w^2 = -2w^2, -2 \cdot (-4w) = 8w, -2 \cdot 5 = -10. Every sign in the bracket is the opposite of the sign in the original, which is what dividing by a negative does. Taking out 2 instead gives 2(-w^2 + 4w - 5), which is equally true.
Fraction coefficients
Factor so that the bracket has whole-number coefficients: (a) \displaystyle \frac{3}{4}t + \frac{1}{4} \qquad (b) \displaystyle \frac{2}{5}x - \frac{3}{10}.
Solution(a) Both coefficients are quarters, so take out \displaystyle \frac{1}{4}: \displaystyle \frac{3}{4}t \div \frac{1}{4} = 3t and \displaystyle \frac{1}{4} \div \frac{1}{4} = 1. So
Check: \displaystyle \frac{1}{4} \cdot 3t = \frac{3}{4}t and \displaystyle \frac{1}{4} \cdot 1 = \frac{1}{4}.
(b) The denominators are 5 and 10, with least common multiple 10, so write both coefficients in tenths: \displaystyle \frac{2}{5} = \frac{4}{10}. Taking out \displaystyle \frac{1}{10} leaves 4x and -3:
Check: \displaystyle \frac{1}{10} \cdot 4x = \frac{4}{10}x = \frac{2}{5}x and \displaystyle \frac{1}{10} \cdot (-3) = -\frac{3}{10}. Taking out \displaystyle \frac{1}{5} instead gives \displaystyle \frac{1}{5}\left(2x - \frac{3}{2}\right), which is also true but leaves a fraction in the bracket.
Equal rows of chairs
Kalinda sets out 28 chairs on the floor of a hall and 4 balcony rows of n chairs each, so 4n + 28 chairs in all. (a) Factor 4n + 28 and say what the factored form means. (b) For which numbers of equal groups can the chairs be split, each group holding the same mix of balcony and floor chairs, whatever n is?
Solution(a) \operatorname{GCF}(4, 28) = 4, and 4n \div 4 = n, 28 \div 4 = 7, so
The chairs make 4 identical groups, each one balcony row of n chairs together with 7 floor chairs.
(b) A split into d equal groups of the same mix needs d to divide both 4 (the balcony rows) and 28 (the floor chairs), so d is a common factor of 4 and 28: 1, 2 or 4. For example 4n + 28 = 2(2n + 14) is two groups of two balcony rows and 14 floor chairs. No other number works for every n, even without asking for the same mix: when n = 1 there are 32 chairs and when n = 2 there are 36, and \operatorname{GCF}(32, 36) = 4.
A fraction factor and a negative factor
(a) Three friends split a bill for a pizza costing a dollars, drinks costing b dollars and a cake costing c dollars. Splitting item by item, each pays \displaystyle \frac{1}{3}a + \frac{1}{3}b + \frac{1}{3}c dollars. Factor this and say what the factored form means. \qquad (b) The temperature falls 2 degrees an hour for h hours, then drops another 6 degrees when the wind changes, a change of -2h - 6 degrees. Factor out -2 and say what the bracket means.
Solution(a) All three coefficients are \displaystyle \frac{1}{3}, so take out \displaystyle \frac{1}{3}. Each term divided by \displaystyle \frac{1}{3} leaves its letter:
The bracket is the whole bill, so each friend pays one third of the total. Splitting every item three ways and splitting the total three ways come to the same share. Check with a = 18, b = 6 and c = 12: item by item, 6 + 2 + 4 = 12; from the total, \displaystyle \frac{1}{3} \cdot 36 = 12.
(b) Dividing each term by -2 flips its sign: -2h \div (-2) = h and -6 \div (-2) = 3. So
The factored form is a fall of 2 degrees an hour for h + 3 hours: the drop when the wind changes is worth three more hours of the steady fall. Check with h = 4: -8 - 6 = -14 and -2 \cdot 7 = -14.
Boundaries
Every term must contain the factor. The move rewrites ab + ac as a(b + c), which needs a in every term. In 3x^2 + 5 the second term has no x and 3 does not divide 5, so the GCF is 1. Writing \displaystyle x\left(3x + \frac{5}{x}\right) is dividing by x, which fails when x = 0 and does not leave a polynomial in the bracket; it is not factoring out a monomial.
“Greatest” needs whole-number coefficients. The GCF of terms is defined from the integer factors of the coefficients and the whole-number powers of the letters. Once fractions are allowed, any nonzero number is a factor of any other (\displaystyle 5 = 2 \cdot \frac{5}{2}), so there is no greatest one, and the factor taken out is chosen for a purpose, such as whole numbers in the bracket.
The GCF is fixed only up to sign. -2(5x - 2) and 2(-5x + 2) are both correct. Taking out the negative is a convention, not a rule of algebra.
A GCF of 1 does not mean the expression cannot be factored. It means only that this move has nothing to take out. The expression x^2 + 5x + 6 has GCF 1, yet it equals (x + 2)(x + 3), which Factoring when the leading coefficient is one finds. A single term, such as 12x^3, has no sum to turn into a product at all.
Where this goes next
Factoring out the GCF is the first step in every later factoring. In factoring by grouping the common factor is itself a bracket: in x(y + 3) + 2(y + 3) both terms contain y + 3, and it comes out exactly as a monomial does. In Factoring when the leading coefficient is one and Factoring trinomials, taking out the GCF first makes the rest smaller: 2x^2 + 10x + 12 = 2(x^2 + 5x + 6). Once a quadratic is a product, the Solving quadratics by factoring solves it.
The same move solves a formula for one letter whenever that letter sits in more than one term: kx + 5x = 20 becomes x(k + 5) = 20, with x in one place. It also prepares a function for reading: the sideways shift of a graph can be read only after the coefficient of the variable is factored out, as in 2x - 6 = 2(x - 3), which is the subject of Phase shift. And it is one of the ways to rewrite an expression to show what it means in a situation.
Exercises
Part 1: the greatest common factor of terms
- Find the GCF of 10x^2 and 25x.
- Find the GCF of 9a^3b and 15a^2b^2.
- Find the GCF of 8m^4, 12m^2 and 20m^3.
- Find the GCF of 7p and 9q.
Part 1: factor and check
- Factor 6y + 30.
- Factor 14c - 21.
- Factor 9z^2 + 4z.
- Factor 18h^3 - 12h^2.
- Factor 10a^2b + 15ab^2.
- Factor 4r^3 - 8r^2 + 4r.
- Factor 16x^4 + 24x^3 - 8x^2.
- Isidora writes 30k^2 + 20k = 5k(6k + 4). Is that true? Is she finished? If not, finish it.
Part 2: factoring out a negative
- Factor out a negative common factor: -12w + 18.
- Factor out a negative common factor: -x^3 + 4x^2 - 7x.
Part 2: fraction coefficients
- Factor so that the bracket has whole-number coefficients: \displaystyle \frac{1}{5}a + \frac{4}{5}.
- Factor so that the bracket has whole-number coefficients: \displaystyle \frac{5}{6}u - \frac{1}{3}.
- Factor so that the bracket has whole-number coefficients: \displaystyle \frac{3}{8}v^2 + \frac{3}{4}v.
Part 2: reading and using the factored form
- Nikodem plants a vegetable bed with 8 trays of s seedlings each and 12 loose seedlings, so 8s + 12 seedlings in all. He plants them in equal rows, every row the same mix of tray seedlings and loose ones. (a) Write 8s + 12 as 4 equal rows; what is in each row? (b) Write it as 2 equal rows. (c) Can he make 3 such rows?
- Factor 10n + 15, and use the factored form to show that for every whole number n the value is a multiple of 5. Is it always odd?
- A rectangle has area 6x^2 + 15x and width 3x. Find its length.
- Factor out \displaystyle -\frac{1}{2} from \displaystyle -\frac{1}{2}x + 4. Use the factored form to explain why the expression is negative whenever x is greater than 8.
Reasoning
- Explain why the GCF of x^7 and x^4 is x^4, and not x^7 or x^3.
- Leander says 2x + 5 cannot be factored, because 2 and 5 share no factor. Marisela writes \displaystyle 2x + 5 = 2\left(x + \frac{5}{2}\right). Who is right?
- Someone factors 12x^2 - 8x as 4(3x^2 - 2) and checks by putting x = 1 into both. The check passes. Explain why it passed, and give a check that catches the error.
- Explain why, after the GCF has been taken out, the terms left in the bracket have no common factor other than 1.
Answers
- 5x. \operatorname{GCF}(10, 25) = 5, and x appears to the powers 2 and 1.
- 3a^2b. \operatorname{GCF}(9, 15) = 3; the smallest powers are a^2 and b.
- 4m^2. \operatorname{GCF}(8, 12, 20) = 4; the smallest power of m is m^2.
- 1. 7 and 9 share no prime, and neither letter is in both terms.
- 6(y + 5). \operatorname{GCF}(6, 30) = 6; check: 6y + 30.
- 7(2c - 3). 14c \div 7 = 2c and -21 \div 7 = -3.
- z(9z + 4). The coefficients share only 1, but z is in both terms.
- 6h^2(3h - 2). \operatorname{GCF}(18, 12) = 6 and the smallest power is h^2.
- 5ab(2a + 3b). 10a^2b \div 5ab = 2a and 15ab^2 \div 5ab = 3b.
- 4r(r^2 - 2r + 1). The last term 4r is the GCF itself and leaves 1.
- 8x^2(2x^2 + 3x - 1). \operatorname{GCF}(16, 24, 8) = 8; -8x^2 \div 8x^2 = -1.
- True, but not finished: 6k and 4 still share 2. Finished, 10k(3k + 2), since \operatorname{GCF}(30, 20) = 10.
- -6(2w - 3). -12w \div (-6) = 2w and 18 \div (-6) = -3.
- -x(x^2 - 4x + 7). Each term divided by -x: x^2, -4x, 7.
- \displaystyle \frac{1}{5}(a + 4). \displaystyle \frac{4}{5} \div \frac{1}{5} = 4.
- \displaystyle \frac{1}{6}(5u - 2). \displaystyle \frac{1}{3} = \frac{2}{6}, so the bracket holds 5u and -2.
- \displaystyle \frac{3}{8}v(v + 2). Both coefficients are multiples of \displaystyle \frac{3}{8}, since \displaystyle \frac{3}{4} = 2 \cdot \frac{3}{8}, and v is in both terms.
- (a) 4(2s + 3): each row is 2 trays and 3 loose seedlings. (b) 2(4s + 6). (c) No: 3 does not divide 8, so the 8 trays cannot be shared equally among 3 rows.
- 5(2n + 3), which is 5 times a whole number, so a multiple of 5. Yes, always odd: 2n + 3 is odd, and 5 times an odd number is odd.
- 2x + 5. 6x^2 + 15x = 3x(2x + 5), and area is width times length.
- \displaystyle -\frac{1}{2}(x - 8), since \displaystyle 4 \div \left(-\frac{1}{2}\right) = -8. When x > 8 the bracket x - 8 is positive, and a negative times a positive is negative. At x = 10, for instance, \displaystyle -\frac{1}{2} \cdot 2 = -1.
- x^4 is a factor of x^7 = x^4 \cdot x^3 and of x^4 itself. x^7 is not a factor of x^4, which has only four x’s, and x^3 is a common factor but smaller than x^4.
- Both, in different senses. With whole-number coefficients nothing but 1 comes out, which is Leander’s point. Marisela’s equation is true, since any nonzero number can be factored out once fractions are allowed; that is why “greatest” needs whole numbers.
- At x = 1 every power of x equals 1, so 12x^2 - 8x and 12x^2 - 8 both give 4. Distributing back gives 4(3x^2 - 2) = 12x^2 - 8, not 12x^2 - 8x; or put x = 2: 32 against 40. The correct factoring is 4x(3x - 2).
- If the bracket’s terms shared a factor d other than 1, then d times the GCF would divide every original term, a common factor bigger than the greatest one, which is impossible.