Any questions on the Section 5.6 homework?

1. Any questions on the Section 5.6 homework?

2. Now please CLOSE YOUR LAPTOPS and turn off and put away your cell phones. Sample Problems Page Link (Dr. Bruce Johnston)

3. Section 5.7 Factoring by Special Products

4. Factoring by Special Products • Previously, we learned a shortcut for squaring a binomial. • (a + b)2 = a2 + 2ab + b2 • (a – b)2 = a2 – 2ab + b2 • 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. • a2 + 2ab + b2 =(a + b)2 • a2 – 2ab + b2 = (a – b)2

5. Example Factor the polynomial 16x2 – 8xy + y2. Since the first term, 16x2, can be written as (4x)2, and the last term, y2 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 16x2 – 8xy + y2 = (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 16x2 – 8xy + y2 .)

6. Example from the online homework: 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 = 62 and 121=112, 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: 36x2 + 66x + 66x + 11 = 36x2 + 132x + 121.

7. Previously, we also discovered a formula for finding the product of the sum and difference of two terms (a – b)(a + b) = a2 – b2 We can use the reverse of the previous equation to see how to quickly factor the difference of 2 squares. a2 – b2 = (a – b)(a + b) This formula can really save you some time!

8. Examples: Factor x2 – 16. Since this polynomial can be written as x2 – 42, x2 – 16 = (x – 4)(x + 4). Factor 9x2 – 4. Since this polynomial can be written as (3x)2 – 22, 9x2 – 4 = (3x – 2)(3x + 2). Factor 16x2 – 9y2. Since this polynomial can be written as (4x)2 – (3y)2, 16x2 – 9y2 = (4x – 3y)(4x + 3y).

9. Examples Factor x8 – y6. Since this polynomial can be written as (x4)2 – (y3)2, x8 – y6 = (x4 – y3)(x4 + y3). Factor x2 + 4. 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.

10. Example: Factor 36x2 – 64. 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. 36x2 – 64 = 4(9x2 – 16) Step 2: Factor the polynomial 9x2 – 16 The polynomial can be written as (3x)2 – (4)2, so (9x2 – 16) = (3x – 4)(3x + 4). Our final result is 36x2 – 64 = 4(3x – 4)(3x + 4). ↑ (Don’t forget to write in 4 (the GCF) as part of your final answer!)

11. 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 = 642. So s4 – 4096 = (s2)2 – 642 = (s2 – 64)(s2 + 64) Are we done yet? No, because s2 – 64 factors further into (s + 8)(s – 8) Final answer: (s + 8)(s – 8)(s2 + 64)

12. Example Factor x4 – 13x2 + 36. (Make sure you always factor completely!) • x4 is (x2)2, so our factors will look like (x2 + __ )(x2 + __ ) • Step 1: Find two factors of 36 that add up to -13 • Answer: -9 and -4 • So our factors are (x2 – 9)(x2 – 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)

13. 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. a3 + b3 = (a + b)(a2 – ab + b2) a3 – b3 = (a – b)(a2 + ab + b2) 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.)

14. Examples • Formulas: • Sum of cubes: a3 +b3 = (a+b)(a2–ab + b2) • Diff. of cubes:a3–b3 = (a–b)(a2+ab + b2) 1. Factor x3 + 1. Since this polynomial can be written as x3 + 13, we can use the sum of cubes formula, with a = x and b = 1 Answer: x3 + 1 =(x + 1)(x2 – x + 1). 2. Factor y3 – 64. Since this polynomial can be written as y3 – 43, we can use the difference of cubes formula with a = y and b = 4. Answer: y3 – 64 =(y – 4)(y2 + 4y + 16).

15. Examples: • a3 + b3 = (a + b)(a2 – ab + b2) • a3 – b3 = (a – b)(a2 + ab + b2) 3. Factor 8t3 + s6. This polynomial can be written as (2t)3 + (s2)3, 8t3 + s6 = (2t + s2)((2t)2 – (2t)(s2) + (s2)2) = (2t + s2)(4t2 – 2s2t + s4). 4. Factor x3y6 – 27z3. This polynomial can be written as (xy2)3 – (3z)3, x3y6 – 27z3 = (xy2 – 3z)((xy2)2 + (3z)(xy2) + (3z)2) = (xy2 – 3z)(x2y4 + 3xy2z + 9z2).

16. Example Remember to ALWAYS check to see if you can factor out any common factors before attempting to use any other factoring techniques or formulas. Factor 375y6 – 24y3. 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.) 375y6 – 24y3 = 3y3(125y3 – 8) Since the second part can be written as(5y)3 – 23, 125y3 – 8 = (5y – 2)((5y)2 + (5y)(2) + 22) = (5y – 2)(25y2 + 10y + 4). Final answer: 3y3(5y – 2)(25y2 + 10y + 4).

17. Review of techniques from sections 5.5-5.7:Choosing a Factoring Strategy Steps for factoring a polynomial: • Factor out any common factors. (Always check this first, before doing any other factoring method.) • Look at number of termsin 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. • See if any factors can be further factored. • Check by multiplying all factors out to make sure you get back to the original polynomial.

18. Example Factor x2y – y3 • Factor out the GCF. x2y – y3 = y(x2 – y2) • Since the remaining polynomial x2 – y2 has TWO TERMS, check the squares and cubes formulas. You’ll find that this is a difference of squares, which fits the formula a2 – b2 = (a + b)(a – b) y(x2 – y2) = y(x – y)(x + y). • Check by multiplying (with 3 factors, multiply 2 at a time.) y(x – y)(x + y) = y(x2 + xy – xy – y2)(FOIL) = y(x2 – y2)(simplify) = x2y – y3 (distributive property)

19. Example Factor x3 – 2x2 – 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.) x3 – 2x2– 36x + 72= x2(x – 2) – 36(x – 2) = (x – 2)(x2 – 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.

20. Example Factor 2a3 – 90 – 18a + 10a2 • Factor out the GCF. 2a3 – 90 – 18a + 10a2 = 2(a3 – 45 – 9a + 5a2) • Although the remaining polynomial has four terms, there is no common factor in the first pair to factor out, so we rearrange the terms. a3– 45 – 9a + 5a2 =a3 + 5a2– 9a– 45 now factor each group: = a2(a + 5) – 9(a + 5) = (a + 5)(a2 – 9) • But the last polynomial can be further factored, so our final answer is 2(a + 5)(a – 3)(a + 3). • Check by multiplying.

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

22. You may now OPEN your LAPTOPS and begin working on the homework assignment.