(Thomason – Fall 2015)

Techniques for Factoring Polynomials

"To factor" means "to write as an indicated product." The following is a list of the techniques for factoring polynomials that you are expected to know when you begin a college credit math course such as MATH 1314 – College Algebra. Each technique is accompanied by an example that illustrates the technique. First, factor out the greatest common factor (GCF) from all of the terms. Example: Factor 15x2 − 35x . Solution: Both terms contain factors of 5 and x, and have no other factors in common. Therefore the GCF is 5x. So 15x2 − 35x = 5x ⋅ 3x − 5x ⋅7= 5x(3x–7). The other factoring techniques depend on how many terms there are. If there are two terms, use the Difference of Two Squares pattern. Difference of Two Squares Pattern: a2 − b2 = (a + b)(a − b) Example: Factor 25x2 − 49y2 . Solution: This binomial is the difference of two squares. The first term is the square of 5x and the second term is the square of 7y. So 25x2 − 49y2= (5x)2 − (7y)2 = (5x +7y)(5x – 7y). Note: The sum of two squares cannot be factored using real numbers. Example: 81v2 +16z2 is the sum of two squares and cannot be factored using real numbers. If there are three terms in which the first and third terms are squares of numbers or expressions, it may be a perfect square trinomial. There are two Perfect Square Trinomial Patterns: Square of a Sum Pattern: a2 + 2ab + b2 = (a + b)2 Square of a Difference Pattern : a2 − 2ab + b2 = (a − b)2 Notice that in the first of these patterns, the middle term of the trinomial is plus twice the product of the quantities being squared in the first and third terms while in the second of these patterns, the middle term of the trinomial is minus twice the product of the quantities being squared in the first and third terms. It is this distinction that determines whether the perfect square trinomial factors into the square of a sum or into the square of a difference. Example: Factor 9x2 + 30xy + 25y2 . Solution: Notice that the first term of this trinomial is the square of 3x and that the third term is the square of 5y. Furthermore, the middle term is plus twice the product of 3x and 5y. That is +2(3x)(5y) = +30xy. So using the square of a sum pattern we get 9x2 + 30xy + 25y2 =

€

(3x)2 + 2(3x)(5y) + (5y)2 = (3x + 5y)2 . Example: Factor 4t2 − 36tw + 81w2 . Solution: Notice that the first term of this trinomial is the square of 2t and that the third term is the square of 9w. Furthermore, the middle term is minus twice the product of 2t and 9w. That is –2(2t)(9w) = –36tw. So using the square of a difference pattern we get 4t2 − 36tw + 81w2 =

€

(2t)2 − 2(2t)(9w) + (9w)2 (2t − 9w)2 .

(Thomason – Spring 2015) p. 2 of 3

If there are three terms but it's not a perfect square trinomial, use one of the following methods depending on whether the first coefficient is 1 or not. To factor ax2 + bx + c , when a=1, find two numbers, call them m and n, whose product is c and whose sum is b. Then

€

x2 + bx + c = (x +m)(x + n) . Example: Factor x2 − 2x −15 . Solution: This is a trinomial with a first coefficient of 1, so we need to find factors of –15 that add up to –2. Those factors are –5 and 3. (If you aren't particularly good with doing arithmetic in your head, you may want to write down a list of the different ways of factoring –15 as the product of two numbers.) Then x2 − 2x −15 = (x − 5)(x + 3). To factor ax2 + bx + c , when a≠1, we can use a procedure that is frequently called the ac Method (also called the Master Product Method). This method consists of four steps: Step 1: Determine the product of the first coefficient, a, and the constant term, c. This product, a ⋅ c , is called the Master Product. Step 2: Find two numbers whose product is the Master Product and whose sum is the middle coefficient, b. Step 3: Use the two number determined in Step 2 to write the middle term of the original trinomial as two terms. Step 4: Then there will be four terms which can be factored by grouping. Example: Factor 6x2 − 23x + 20 . Solution: This is a trinomial with a first coefficient that is not 1, so we can use the ac Method. Step 1: ac = (6)(20), which is 120. Step 2: Now we must find two numbers whose product is 120 and whose sum is –23. Those numbers are –15 and –8. (Even if you are pretty good with doing arithmetic in your head, you will probably need to write down a list of the different ways of factoring 120 in order to determine the pair of factors that adds up to –23.) Step 3: We use –15 and –8 to write the middle term of the given trinomial as two terms: 6x2 − 23x + 20 = 6x2 −15x −8x + 20 Step 4: Now that we have four terms we can factor by grouping: 6x2 −15x −8x + 20 = 3x(2x − 5) − 4(2x − 5) = (2x − 5)(3x − 4) . If there are four terms, factor by grouping (The Grouping Method). Example: Factor 4x3 − 7x2 − 20x + 35 . Solution: There is no factor in common to all four terms. However, the first two terms have a common factor of x2 and the last two terms have a common factor of –5. So we group the first two terms together and group the last two terms together. 4x3 − 7x2 − 20x + 35 =

€

(4x3 − 7x2) − (20x − 35) =

€

x2(4x − 7) − 5(4x − 7) Now the two resulting terms have a common factor of (4x − 7) so we can factor it out and get 4x3 − 7x2 − 20x + 35 =

€

x2(4x − 7) − 5(4x − 7) =

€

(4x − 7)(x2 − 5) .

(Thomason – Spring 2015) p. 3 of 3

Warning: To factor a polynomial completely, we frequently have to apply more than one technique. After factoring by any method, always take a close look at each of your factors to see if any of them can factored. Example: Factor

€

−6x5 + 28x4 −16x3. Solution: As always, we first factor out the greatest common factor, which in this case is

€

−2x3. (Note: When the highest power term’s coefficient is negative, it is best to include the negative in the common factor.) So we have

€

−6x5 + 28x4 −16x3 =

€

−2x3(3x2 −14x + 8) But we aren't finished. It’s possible the trinomial in parentheses,

€

3x2 −14x + 8 , can be factored using the ac Method. Step 1: ac = (3)(8), which is 24. Step 2: Now we try to find two numbers whose product is 24 and whose sum is –14. Those numbers are –12 and –2. Step 3: We use –12 and –2 to write the middle term of the trinomial as two terms:

€

3x2 −14x + 8 =

€

3x2 −12x − 2x + 8 Step 4: Now that we have four terms we can factor by grouping:

€

3x2 −12x − 2x + 8 =

€

3x(x − 4) − 2(x − 4) =

€

(x − 4)(3x − 2). So

€

3x2 −14x + 8 =

€

(x − 4)(3x − 2). Then back to our original problem, we have

€

−6x5 + 28x4 −16x3 =

€

−2x3(3x2 −14x + 8) =

€

−2x3(x − 4)(3x − 2). Example: Factor

€

6x3 − 5x2 − 54x + 45. Solution:

€

6x3 − 5x2 − 54x + 45 =

€

x2(6x2 − 5) − 9(6x2 − 5) =

€

(6x − 5)(x2−9) factoring by grouping =

€

(6x − 5)((x − 3)(x + 3) using the difference of two squares pattern None of the factors in the last result is factorable so we are finished. Example: Factor 2b6 −162b2 . Solution: 2b6 −162b2 = 2b2(b4 −81) by factoring out the GCF = 2b2(b2 + 9)(b2 − 9) by using the difference of two squares pattern =2b2(b2 + 9)(b + 3)(b − 3) by using the difference of two squares pattern again None of the factors in the last result is factorable so we are finished. Example: Factor 3x3 −30x 2y + 48xy2 . Solution: The GCF for this trinomial is 3x. So we get3x3 −30x 2y + 48xy2 = 3x(x2 −10xy +16y2) . Now we must see if the trinomial we obtained can be factored. Its first term is the square of x and its third term is the square of 4y so it looks like we might have a perfect square trinomial. However, to be a perfect square trinomial the middle term has to be plus or minus twice the product of the quantities being squared in the first and third terms, which would be ±2(x)(4y), or ±8xy, neither of which is the –10xy in our trinomial. Our trinomial does have a leading coefficient of 1, so we try that factoring method. We need factors of 16 that add up to –10. Those factors are –2 and –8 so we get 3x3 −30x 2y + 48xy2 = 3x(x2 −10xy +16y2)=3x(x − 2y)( x − 8y) .