Any questions on the Section 5.6 homework?

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Section 5.7 Factoring by Special Products

Previously, we learned a shortcut for squaring a binomial. (a + b) 2 = a 2 + 2ab + b 2 (a – b) 2 = a 2 – 2ab + b 2 So if the first and last terms of our polynomial to be factored can be written as expressions squared, and the middle term of our polynomial is twice the product of those two expressions, then we can use these two previous equations to easily factor the polynomial. a 2 + 2ab + b 2 = (a + b) 2 a 2 – 2ab + b 2 = (a – b) 2 Factoring by Special Products

Example Factor the polynomial 16x 2 – 8xy + y 2. Since the first term, 16x 2, can be written as (4x) 2, and the last term, y 2 is obviously a square, we check the middle term. 8xy = 2(4x)(y) (twice the product of the expressions that are squared to get the first and last terms of the polynomial) Therefore 16x 2 – 8xy + y 2 = (4x – y) 2. Note: You can use FOIL method to verify that the factorization for the polynomial is accurate. (Multiply (4x – y)(4x – y) and show you get 16x 2 – 8xy + y 2.)

Note that this would be a pretty tough one to do by the “factoring by grouping” method, since 36×121=HUGE NUMBER. But if we notice that 36 = 6 2 and 121=11 2, then we can check and see if (6x + 11) 2 might be the correct factoring. And sure enough, when we multiply out (6x + 11)(6x + 11), we do get the correct trinomial after we simplify: 36x x + 66x + 11 = 36x x Example from the online homework:

Previously, we also discovered a formula for finding the product of the sum and difference of two terms (a – b)(a + b) = a 2 – b 2 We can use the reverse of the previous equation to see how to quickly factor the difference of 2 squares. a 2 – b 2 = (a – b)(a + b) This formula can really save you some time!

Examples: Factor x 2 – 16. Since this polynomial can be written as x 2 – 4 2, x 2 – 16 = (x – 4)(x + 4). Factor 9x 2 – 4. Since this polynomial can be written as (3x) 2 – 2 2, 9x 2 – 4 = (3x – 2)(3x + 2). Factor 16x 2 – 9y 2. Since this polynomial can be written as (4x) 2 – (3y) 2, 16x 2 – 9y 2 = (4x – 3y)(4x + 3y).

Examples Factor x 8 – y 6. Since this polynomial can be written as (x 4 ) 2 – (y 3 ) 2, x 8 – y 6 = (x 4 – y 3 )(x 4 + y 3 ). Factor x This one is the sum of two squares, not the difference of squares, so it can’t be factored. This polynomial is a prime polynomial.

Example: Remember that you should always factor out any common factors first, before you start any other technique. Step 1: Factor out the GCF, which in this case is 4. 36x 2 – 64 = 4(9x 2 – 16) Step 2: Factor the polynomial 9x 2 – 16 The polynomial can be written as (3x) 2 – (4) 2, so (9x 2 – 16) = (3x – 4)(3x + 4). Our final result is 36x 2 – 64 = 4(3x – 4)(3x + 4). ↑ (Don’t forget to write in 4 (the GCF) as part of your final answer!) Factor 36x 2 – 64.

Example from the online homework: Note that 4096 is a HUGE NUMBER. But if it is a square, then we’re in business. How do you tell if it’s a perfect square? Take the square root (on your calculator…) Answer: 4096 = So s 4 – 4096 = (s 2 ) 2 – 64 2 = (s 2 – 64)(s ) Are we done yet? No, because s 2 – 64 factors further into (s + 8)(s – 8) Final answer: (s + 8)(s – 8)(s )

Factor x 4 – 13x (Make sure you always factor completely!) x 4 is (x 2 ) 2, so our factors will look like (x 2 + __ )(x 2 + __ ) Step 1: Find two factors of 36 that add up to -13 Answer: -9 and -4 So our factors are (x 2 – 9)(x 2 – 4) (Note that both of the factors are differences of squares, so you’re not done yet!) So this further factors into (x – 3)(x + 3)(x – 2)(x + 2) Example

There are two additional types of binomials that can be factored easily by remembering a formula. We have not studied these special products previously, as they involve cubes of terms, rather than just squares. a 3 + b 3 = (a + b)(a 2 – ab + b 2 ) a 3 – b 3 = (a – b)(a 2 + ab + b 2 ) NOTE: These formulas will be on the pink formula sheet that we hand out at each test and quiz, so you don’t have to memorize them, but you do need to know how to apply them.) (If you don’t yet have a copy of this formula sheet in your notebook, raise your hand and we’ll give you a yellow copy to keep.)

1. Factor x Since this polynomial can be written as x , we can use the sum of cubes formula, with a = x and b = 1 Answer: x = (x + 1)(x 2 – x + 1). 2. Factor y 3 – 64. Since this polynomial can be written as y 3 – 4 3, we can use the difference of cubes formula with a = y and b = 4. Answer: y 3 – 64 = (y – 4)(y 2 + 4y + 16). Examples Formulas: Sum of cubes: a 3 + b 3 = (a + b)(a 2 – ab + b 2 ) Diff. of cubes:a 3 – b 3 = (a – b)(a 2 + ab + b 2 )

3. Factor 8t 3 + s 6. This polynomial can be written as (2t) 3 + (s 2 ) 3, 8t 3 + s 6 = (2t + s 2 )((2t) 2 – (2t)(s 2 ) + (s 2 ) 2 ) = (2t + s 2 )(4t 2 – 2s 2 t + s 4 ). 4. Factor x 3 y 6 – 27z 3. This polynomial can be written as (xy 2 ) 3 – (3z) 3, x 3 y 6 – 27z 3 = (xy 2 – 3z)((xy 2 ) 2 + (3z)(xy 2 ) + (3z) 2 ) = (xy 2 – 3z)(x 2 y 4 + 3xy 2 z + 9z 2 ). Examples: a 3 + b 3 = (a + b)(a 2 – ab + b 2 ) a 3 – b 3 = (a – b)(a 2 + ab + b 2 )

Remember to ALWAYS check to see if you can factor out any common factors before attempting to use any other factoring techniques or formulas. Step 1: Factor out the GCF. (Tip: Since 375 is such a big number, start by factoring the smaller number 24, then see if any of its factors will divide into 375. You will find that the number 3 is a divisor of both 24 and 375.) 375y 6 – 24y 3 = 3y 3 (125y 3 – 8) Since the second part can be written as (5y) 3 – 2 3, 125y 3 – 8 = (5y – 2)((5y) 2 + (5y)(2) ) = (5y – 2)(25y y + 4). Final answer: 3y 3 (5y – 2)(25y y + 4). Example Factor 375y 6 – 24y 3.

Review of techniques from sections : Choosing a Factoring Strategy Steps for factoring a polynomial: 1)Factor out any common factors. (Always check this first, before doing any other factoring method.) 2)Look at number of terms in polynomial If 2 terms, look for difference of squares, difference of cubes or sum of cubes. (Use the formula sheet for these!) If 3 terms, use techniques for factoring into 2 binomials. If 4 or more terms, try factoring by grouping. 3)See if any factors can be further factored. 4)Check by multiplying all factors out to make sure you get back to the original polynomial.

Factor out the GCF. x 2 y – y 3 = y(x 2 – y 2 ) Since the remaining polynomial x 2 – y 2 has TWO TERMS, check the squares and cubes formulas. You’ll find that this is a difference of squares, which fits the formula a 2 – b 2 = (a + b)(a – b) y(x 2 – y 2 ) = y(x – y)(x + y). Check by multiplying (with 3 factors, multiply 2 at a time.) y(x – y)(x + y) = y(x 2 + xy – xy – y 2 ) (FOIL) = y(x 2 – y 2 ) (simplify) = x 2 y – y 3 (distributive property) Example Factor x 2 y – y 3

Factor x 3 – 2x 2 – 36x + 72 There are no factors common to all the terms to factor out. Since there are 4 terms, we try factoring by grouping (Group the first 2 terms and last 2 terms, then factor out common terms.) x 3 – 2x 2 – 36x + 72 = x 2 (x – 2) – 36(x – 2) = (x – 2)(x 2 – 36) Notice that our second polynomial can be further factored, since it is the difference of squares. Final answer: (x – 2)(x + 6)(x – 6). Check by multiplying the factors. Example

Factor 2a 3 – 90 – 18a + 10a 2 Factor out the GCF. 2a 3 – 90 – 18a + 10a 2 = 2(a 3 – 45 – 9a + 5a 2 ) Although the remaining polynomial has four terms, there is no common factor in the first pair to factor out, so we rearrange the terms. a 3 – 45 – 9a + 5a 2 = a 3 + 5a 2 – 9a – 45 now factor each group: = a 2 (a + 5) – 9(a + 5) = (a + 5)(a 2 – 9) But the last polynomial can be further factored, so our final answer is 2(a + 5)(a – 3)(a + 3). Check by multiplying. Example

Reminder: This homework assignment on Section 5.7 is due at the start of next class period.

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