Factoring Sums and Differences of Squares

Section 1.5: Factoring Sums and Differences of Squares #1- 42: Completely factor the binomials, remember to factor out the GCF first when applicable (if a problem is prime say so). 1) x2 – 9 Divide the exponent by 2 and take the square root of the number. Then create two basically identical parenthesis, except for opposite signs. Solution: (x + 3)(x – 3) 2) x2 + 9 This is a sum of squares and it is prime. Solution: Prime 5) y2 – 36 Divide the exponent by 2 and take the square root of the number. Then create two basically identical parenthesis, except for opposite signs. Solution: (y + 6)(y – 6) 7) y2 + 36 This is a sum of squares and it is prime. Solution: Prime 9) 25a2 – 81 Divide the exponent by 2 and take the square root of the numbers. Then create two basically identical parenthesis, except for opposite signs. Solution: (5a + 9)(5a – 9) 11) 25a2 + 81 This is a sum of squares and it is prime. Solution: Prime 13) 49x2 – 36 Divide the exponent by 2 and take the square root of the numbers. Then create two basically identical parenthesis, except for opposite signs. Solution: (7x + 6)(7x – 6) 15) 49x2 + 36 This is a sum of squares and it is prime. Solution: Prime 17) x3 – 64x First factor out the common factor of x. = x(x2 – 64) Now factor the inside of the parenthesis. Solution: x(x + 8)(x – 8) 19) x3 + 64x First factor out the common factor of x. This will be the answer, as what is left inside the parenthesis doesn’t factor. Solution: x(x2 + 64) 21) 3x2 – 27 First factor out the common factor of 3. = 3(x2 – 9) Now factor the inside of the parenthesis. Solution: 3(x+3)(x – 3) 23) 3x2 + 27 First factor out the common factor of 3. This will be the answer, as what is left inside the parenthesis doesn’t factor. Solution: 3(x2 + 9) 25) 9 – 25x2 This could be rewritten with the -25x2 first. If you do this your answer would be –(5x + 3)(5x-3) I will leave the problem written in the order it is given. So the answer will look a bit backwards. Take the square root of the 9 to get 3, the square root of 25 gives 5 and divide the exponent by 2. Solution: (3 + 5x)(3 – 5x) 27) 81 – 16x2 This could be rewritten with the -16x2 first. If you do this your answer would be –(4x + 9)(4x-9) I will leave the problem written in the order it is given. So the answer will look a bit backwards. Take the square root of the 81 to get 9, the square root of 16 gives 4 and divide the exponent by 2. Solution: (9 – 4x)(9 + 4x) 29) x4 – 9 You need to divide the exponent of the x by 2. Solution: (x2 + 3)(x2 – 3) 31) 16x4 – 25 Divide the exponent by 2 and square root the numbers. Solution: (4x2 + 5)(4x2 – 5) 33) 98y2 – 2x4 First factor out the GCF of 2. = 2(49y2 – x4) Next divide the exponents by 2 and square root the number. Solution: 2(7y + x2)(7y – x2) 35) x4 – 16 Trick problem************ this problem is a trick, because one of the two parenthesis factors after factor the first time. Step 1 answer: = (x2 + 4)(x2 – 4) While this is correct, it is not completely factored. The x2 + 4 parenthesis is prime, but the x2 – 4 factors more. Solution: (x2 + 4)(x + 2)(x – 2) 37) 2x4 – 512 First factor out the GCF of 2. = 2(x4 – 256) Now divide the exponent by 2 and square root the 256. = 2(x2 + 16)(x2 – 16) Again this is tricky. The above line is correct, but not completely factored. I need to factor the x2 – 16 parenthesis to get the best final answer. Solution: 2(x2 + 16)(x + 4)(x – 4) 39) y4 – 2401 I needed to use my calculator to find the square root of 2401. = (y2 + 49)(y2 – 49) Now factor the second parenthesis. Solution: (y2 + 49)(y + 7)(y – 7) 41) x4 + 4 This is a sum of squares, so it is prime. Solution: Prime