Factoring means finding the things you multiply together to get a given answer.

You did some work with factoring in grade school. For instance, you found the prime factorization of numbers.

You have also found factors of numbers and their common factors.

When adding or subtracting fractions, you used factors to find the least common denominator.

In algebra we mostly care about factoring polynomials.  We want to find what you need to multiply together to get a given polynomial.  It’s like you’re playing Jeopardy with the distributive property.

Almost all the time we will be factoring quadratic trinomials.

The most common factoring problems look like this: Factor x x + 35 We need to find the quantities we can multiply to get this polynomial.

Factor x x + 35 The answer will have the format (x + ___)(x + ___)

Factor x x + 35 To find the numbers that go in the quantity, find what you can multiply to get 35 that adds up to 12

Factor x x + 35 multiply to get 35 that adds up to 12 The only numbers that do both are 7 and 5. So … (x + 7)(x + 5)

Factor x x + 35 multiply to get 35 that adds up to 12 The only numbers that do both are 7 and 5. So … (x + 7)(x + 5) (x + 5)(x + 7) is also OK.

Factor x x + 36

Factor x x 

Factor x x  There are lots of ways to get 36, like 6  6, 9  4, and 12  3.

Factor x x  There are lots of ways to get 36, like 6  6, 9  4, and 12  3. Only adds up to 13.

Factor x x  There are lots of ways to get 36, like 6  6, 9  4, and 12  3. Only adds up to 13. So the answer is (x + 9)(x + 4)

Factor x x + 40 x x + 24 x x + 9

Factor x x + 40 (x + 8)(x + 5) x x + 24 (x + 6)(x + 4) x x + 9 (x + 9)(x + 1)

Factor x 2 – 16x + 48

Factor x 2 – 16x + 48 The rule is still the same Multiply to get 48 Add to get -16

Factor x 2 – 16x + 48 The rule is still the same -12  -4 = = -16

Factor x 2 – 16x + 48 The rule is still the same -12  -4 = = -16 So it’s (x – 12)(x – 4).

Factor x 2 – 5x + 6 x 2 – 16x + 55 x 2 – 18x + 32

Factor x 2 – 5x + 6 (x – 2)(x – 3) x 2 – 16x + 55 (x – 11)(x – 5) x 2 – 18x + 32 (x – 16)(x – 2)

Factor x 2 – x – 72

Factor x 2 – x – 72 +  This time we need both positive and negative factors because we’re multiplying to get -72. We also need to add to -1.

Factor x 2 – x – 72 +  Consider (x + 9)(x – 8) and (x – 9)(x + 8) Both multiply to -72 Only the 2 nd adds to -1 So … It’s (x – 9)(x + 8)

Factor x 2 + 5x – 24 This time we need to multiply to -24 and add to positive 5

Factor x 2 + 5x – 24 (x + 8)(x – 3)

If you have one positive and one negative factor, the larger factor has the same sign as the middle term in the trinomial. x 2 + 4x – 21 = (x + 7)(x – 3) x 2 – 3x – 18 = (x – 6)(x + 3)

Factor x 2 + 5x – 36 x 2 – 4x – 32 x x – 28

Factor x 2 + 5x – 36 (x – 4)(x + 9) x 2 – 4x – 32 (x – 8)(x + 4) x x – 28 (x + 14)(x – 2)

Factor x 2 + 6x + 9

Factor x 2 + 6x 

Factor x 2 + 6x  (x + 3)(x + 3)

Factor x 2 + 6x  (x + 3)(x + 3) Most books would write this as (x + 3) 2

Factor x x + 64 x 2 – 18x + 81 x x + 36

Factor x x + 64 (x + 8) 2 x 2 – 18x + 81 (x – 9) 2 x x + 36 (x + 6) 2

Factor x 2 – 49

Factor x 2 – 49 We need to multiply to get -49

0x Factor x 2 – 49 We need to multiply to get -49 We need to add to get 0

0x Factor x 2 – 49 We need to multiply to get -49 We need to add to get 0 It’s (x + 7)(x – 7)

Factor x 2 – 100 x 2 – 1 x 2 – 25

Factor x 2 – 100 (x + 10)(x – 10) x 2 – 1 (x – 1)(x + 1) x 2 – 25 (x + 5)(x – 5)

Let’s try a bit of everything.

x 2 + 8x + 12 x 2 – x – 20 x 2 – 16x + 64 x 2 – 12x + 27

x 2 + 8x + 12 (x + 6)(x + 2) x 2 – x – 20 (x – 5)(x + 4) x 2 – 16x + 64 (x – 8) 2 x 2 – 12x + 27 (x – 3)(x – 9)

x 2 – 4x - 45 x 2 – 16 x 2 + 2x + 1 x 2 – 18x + 72

x 2 – 4x - 45 (x – 9)(x + 5) x 2 – 16 (x – 4)(x + 4) x 2 + 2x + 1 (x + 1)(x + 1) x 2 – 18x + 72 (x – 12)(x – 6)

Your book also likes problems like this. Factor a 2 + 2ab – 15b 2

Factor a 2 + 2ab – 15b 2 The rules are still the same, but the answer will have both a and b in it.

Factor a 2 + 2ab – 15b 2 Multiply to get -15 Add up to 2

Factor a 2 + 2ab – 15b 2 Multiply to get -15 Add up to 2 It’s (a + 5b)(a – 3b)

Factor x 2 + 6xy + 8y 2 n 2 – 2np – 35p 2

Factor x 2 + 6xy + 8y 2 (x + 4y)(x + 2y) n 2 – 2np – 35p 2 (n + 5p)(n – 7p)

Your book also likes problems like this. Factor x x

Factor x x What’s different this time is that the first part has x 10. This means the answer will have the form (x 5 + __)(x 5 + __)

Factor x x  Everything else is the same. So, the answer is … (x 5 + 9)(x 5 + 7)

Factor x 4 – 26x completely.

Factor x 4 – 26x completely. (x 2 – 25)(x 2 – 1)

Factor x 4 – 26x completely. (x 2 – 25)(x 2 – 1) … BUT, we’re not done.  Both parts can be factored again.

Factor x 4 – 26x completely. (x 2 – 25)(x 2 – 1) (x + 5)(x – 5)(x + 1)(x – 1)

There are two more things that can complicate factoring.

First … Quadratic coefficients

If there’s a coefficient that makes the problem look like ax 2 + bx + c The answer will usually have the form (ax + __)(x + __)

ax 2 + bx + c You still want to find numbers that will multiply to “c”. (ax + __)(x + __)

ax 2 + bx + c Unfortunately, they WON’T just add up to b.

ax 2 + bx + c Remember FOIL. Outside + Inside needs to add to b

Factor 3x x + 14

Factor 3x x + 14 The answer will have the form (3x + __)(x + __)

Factor 3x x + 14 The answer will have the form (3x + __)(x + __) Since 7  2 = 14, it might be (3x + 7)(x + 2) or (3x + 2)(x + 7)

Factor 3x x + 14 Which is right? (3x + 7)(x + 2) (3x + 2)(x + 7) Check outside + inside

Factor 3x x + 14 Which is right? (3x + 7)(x + 2) = 13  (3x + 2)(x + 7) = 23

Factor 3x x + 14 The answer is (3x + 2)(x + 7)

Factor 5x 2 + 2x – 3

Factor 5x 2 + 2x – 3 Could be (5x + __)(x – __) or (5x – __)(x + __)

Factor 5x 2 + 2x – 3 The numbers at the end will be 3 and 1 (one + and one –)

Factor 5x 2 + 2x – 3 Consider (5x + 3)(x – 1) (5x + 1)(x – 3) (5x – 3)(x + 1) (5x – 1)(x + 3) Check outside + inside

Factor 5x 2 + 2x – 3 Consider (5x + 3)(x – 1) -5+3= -2 (5x + 1)(x – 3) -15+1= -14 (5x – 3)(x + 1) 5–3 = 2 (5x – 1)(x + 3) 15–1 = 14 Check outside + inside

Factor 5x 2 + 2x – 3 The answer is (5x – 3)(x + 1)

Factor 2x x x 2 – 37x x 2 – x – 10

Factor 2x x + 24 (2x + 3)(x + 8) 7x 2 – 37x + 10 (7x – 2)(x – 5) 3x 2 – x – 10 (3x + 5)(x – 2)

The other possible complication is common factors.

Common factor problems usually involve binomials, like this one: Factor 6x x 6

The answer typically has the form ___( __ + __ )  The common factor goes outside the parentheses.  Divide the original problem by the common factor to get what stays in the parentheses.

Factor 6x x 6

Factor 6x x 6 To find the common factor…  Find the biggest number that goes into both 6 and 15 (the GCF)  Choose the smaller exponent … Here it’s 3x 6

Factor 6x x 6 So our answer has the form 3x 6 ( __ + __ )

Factor 6x x 6 So our answer has the form 3x 6 ( __ + __ ) Now divide both terms by 3x 6  Divide coefficients.  Subtract exponents.

Factor 6x x 6 The final answer is … 3x 6 (2x + 5)

Factor 12x 4 y 2 – 8xy 3

Factor 12x 4 y 2 – 8xy 3 Common factor is 4xy 2 So answer is 4xy 2 (__ + __)

Factor 12x 4 y 2 – 8xy 3 4xy 2 (3x 3 – 4y)

Factor 7x x 4 18x 3 – 27x 4 30a 5 b a 3 b 3

Factor 7x x 4 7x 4 (x + 3) 18x 3 – 27x 4 9x 3 (2 – 3x) 30a 5 b a 3 b 3 5a 3 b 2 (6a 2 + 5b)

Factor 2x x x 2 completely.

Factor 2x x x 2 completely. First take out a common factor. Here it’s 2x 2

Factor 2x x x completely. 2x 2 (x 2 + 8x + 15) Now factor what’s inside the parentheses.

Factor 2x x x completely. 2x 2 (x 2 + 8x + 15) = 2x 2 (x + 5)(x + 3)