Weekly Quiz 8 will be given after today’s lecture, during the last 25 minutes of class. 1 Please CLOSE YOUR LAPTOPS, and turn off and put away your cell phones, and get out your notetaking materials. Section 6.5 Factoring Binomials Review of factoring techniques: 1. ALWAYS look for a GCF first with any polynomial you are trying to factor. 2. After that, count how many terms are in the polynomial. If there are – FOUR terms: Use factoring by grouping. – THREE terms ( a trinomial): • If the leading coefficient is 1 (if it looks like x2 + bx + c), – Find factors (x + t)(x + v) such that t + v = b and t*v = c. • If the leading coefficient is not 1 (if it looks like ax2 + bx + c), – Use the British method, which starts with multiplying a*c, and then finding factors of that number that add up to b. Today we are going to learn techniques for factoring two-term polynomials (binomials). Case one: Binomials in which both terms are perfect squares. Previously, we 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 this equation to factor the difference of 2 squares. a2 – b2 = (a – b)(a + b) Examples: a2 – b2 = (a – b)(a + b) Factor x2 – 16. Since this polynomial can be written as (x)2 – (4)2, x2 – 16 = (x – 4)(x + 4). Factor 9x2 – 4. Since this polynomial can be written as (3x)2 – (2)2, 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). Examples a2 – b2 = (a – b)(a + b) 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. Example: a2 – b2 = (a – b)(a + b) 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!) 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) 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) [You can prove that these formulas work by multiplying out the two factors on the right side of each equation. If you do this, you’ll get six products at the first stage, but you’ll find that everything cancels out except the a3 and b3 terms.] NOTE: These formulas are on the formula sheet that you can use on each test and quiz, so you don’t have to memorize them, but you do need to know how to apply them.) Formulas: Sum of cubes: a3 + b3 = (a + b)(a2 – ab + b2) Diff. of cubes: a3 – b3 = (a – b)(a2 + ab + b2) Examples HINT: It will help you a lot when doing cube problems if you always start by writing blank parentheses, then figure out what goes in them. Those will be the “a” and “b” in your formula. 1. Factor x3 + 1. HINT: This one looks like ( )3 + ( )3. Since this polynomial can be written as (x)3 + (1)3, 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. HINT: This one looks like ( y )3 - ( 4 )3. What goes in the parentheses? Therefore we can use the difference of cubes formula with a = y and b = 4. Answer: y3 – 64 = (y – 4)(y2 + 4y + 16). Examples: a3 + b3 = (a + b)(a2 – ab + b2) a3 – b3 = (a – b)(a2 + ab + b2) 3. Factor 8t3 + s6. HINT: This one looks like ( )3 + ( )3. What goes in the parentheses? This polynomial can be written as (2t)3 + (s2)3, so 8t3 + s6 = (2t + s2)((2t)2 – (2t)(s2) + (s2)2) = (2t + s2)(4t2 – 2s2t + s4). 4. Factor x3y6 – 27z3. HINT: This one looks like ( )3 - ( )3. What goes in the parentheses? This polynomial can be written as (xy2)3 – (3z)3, so x3y6 – 27z3 = (xy2 – 3z)((xy2)2 + (3z)(xy2) + (3z)2) = (xy2 – 3z)(x2y4 + 3xy2z + 9z2). Remember to ALWAYS check to see if you can factor out any common factors before attempting to use any other factoring techniques or formulas. Example 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). The assignment on this material (HW 35) is due at the start of class on Monday. Please open your laptops, log in to the MyMathLab course web site, and open Weekly Quiz 8. • You have 25 minutes to finish this 10-question quiz. • Put away cell phones, graphing calculators, notes, headphones, etc. • You can use a scientific calculator and a pink formula sheet. • After you submit your quiz, turn your worksheet in to the TA, and then you are free to leave. Your scratch work will be reviewed for possible partial credit points.