Chapter 1: Factoring and Quadratic Equations Section 1.4: Factoring Trinomials in the Form ax2 + bx + c where a ≠ 1 #1 – 18: Rewrite as a polynomial with 4 terms (if possible) then factor by grouping and check your answer, state if a polynomial is prime. 1) 5x2 + 11x + 6 First multiply the 5*6 = 30 Factors of 30 Choice to break up middle term What middle column simplifies to 1 * 30 1x + 30x 31x 2 * 15 2x + 15x 17x 3 * 10 3x + 10x 13x 5*6 5x + 6x 11x = 5x2 + 5x + 6x + 6 = (5x2 + 5x) + (6x + 6) = 5x(x+1) + 6(x+1) Solution: (x+1)(5x+6) 3 Chapter 1: Factoring and Quadratic Equations 3) 5x2 – 11x + 6 First multiply the 5*6 = 30 Factors of 30 Choice to break up middle term What middle column simplifies to -1 * -30 -1x + -30x -31x -2 * -15 -2x + -15x -17x -3 * -10 -3x + -10x -13x -5 * -6 -5x + -6x -11x Factors of 30 Choice to break up middle term What middle column simplifies to 1 * 12 1x + 12x 13x 2*6 2x + 6x 8x 3*4 3x + 4x 7x = 5x2 - 5x - 6x + 6 = (5x2 - 5x) + (-6x + 6) = 5x(x-1) + -6(x-1) = 5x(x -1) – 6(x -1) Solution: (x-1)(5x-6) 5) 4x2 + 7x + 3 First multiply the 4*3 = 12 = 4x2 + 4x + 3x + 3 = (4x2 + 4x) + (3x + 3) = 4x(x+1) + 3(x+1) Solution: (x+1)(4x+3) 4 Chapter 1: Factoring and Quadratic Equations 7) 4x2 – 7x + 3 First multiply the 4*3 = 12 Factors of 30 Choice to break up middle term What middle column simplifies to -1*-12 -1x + - 12x -13x -2*-6 -2x + - 6x -8x -3*-4 -3x + - 4x -7x Factors of 30 Choice to break up middle term What middle column simplifies to -1*14 -1z + 14z 13z 1*-14 1z + -14z -13z -2*7 -2z + 7z 5z 2*-7 2z + -7z -5z = 4x2 + -3x + -4x + 3 = x(4x - 3) + (-1)(4x – 3) = x(4x - 3) - 1(4x – 3) Solution: (4x – 3)(x – 1) 9) 2z2 + 5z – 7 First multiply the 2*-7= -14 = 2z2 + (-2z) + 7z – 7 = 2z(z-1) + 7(z-1) Solution: (z-1)(2z+7) 5 Chapter 1: Factoring and Quadratic Equations 11) 2z2 – 5z – 7 First multiply the 2*-7= -14 Factors of -14 Choice to break up middle term What middle column simplifies to -1*14 -1z + 14z 13z 1*-14 1z + -14z -13z -2*7 -2z + 7z 5z 2*-7 2z + -7z -5z Factors of 42 Choice to break up middle term What middle column simplifies to 1 * 42 1x + 42x 43x 2 * 21 2x + 21x 23x 3 * 14 3x + 14x 17x 6*7 6x + 7x 13x = 2z2 + 2z + - 7z -7 = 2z(z+1) + -7(z+1) Solution: (z+1)(2z – 7) 13) 6x2 + 23x + 7 First multiply the 6*7 = 42 = 6x2 + 2x + 21x + 7 = (6x2 + 2x) + (21x + 7) = 2x(3x+1) + 7(3x+1) Solution: (3x+1)(2x+7) 6 Chapter 1: Factoring and Quadratic Equations 15) 3x2 + 10x + 7 First multiply the 3*7 = 21 Factors of 21 Choice to break up middle term What middle column simplifies to 1 * 21 1x + 21x 23x 3*7 3x + 7x 10x Factors of 30 Choice to break up middle term What middle column simplifies to 1 * 30 1x + 30x 31x 2 * 15 2x + 15x 17x 3 * 10 3x + 10x 13x 5*6 5x + 6x 11x 3x2 + 3x + 7x + 7 = 3x(x + 1) + 7(x +1) Solution: (x+1)(3x + 7) 17) 5x2 + 13x + 6 First multiply the 5*6 = 30 = 5x2 + 3x + 10x + 6 = (5x2 + 3x) + (10x + 6) = x(5x+3) + 2(5x+3) Solution: (5x+3)(x+2) 7 Chapter 1: Factoring and Quadratic Equations #19-44: Factor, using bottoms up or the guess and check method, state if a polynomial is prime (notice #19 – 30 are the same as #1-12, and you should get the same answer regardless of the technique you use to factor.) 19) 5x2 + 11x + 6 First multiply the 5*6 = 30 Rewrite as x2 + 11x + 30 and factor this Factors of 30 Sum of factors 1 * 30 1+30 = 31 2 * 15 2 + 15 = 17 3 * 10 3 + 10 = 13 5*6 5 + 6 = 11 (x + 5)(x+6) Finish divide by 5 / reduce / bottoms up 5 6 (𝑥 + 5) (𝑥 + 5) Solution: (x+1)(5x+6) 8 Chapter 1: Factoring and Quadratic Equations 21) 5x2 – 11x + 6 First multiply the 5*6 = 30 Rewrite as x2 - 11x + 30 and factor this Factors of 30 Sum of factors -1 * -30 -1+-30 = -31 -2 * -15 -2 + -15 = -17 -3 * -10 -3 + -10 = -13 -5 * -6 -5 + -6 = -11 (x - 5)(x-6) Finish divide by 5 / reduce / bottoms up 5 6 (𝑥 − 5) (𝑥 − 5) Solution: (x-1)(5x-6) 23) 4x2 + 7x + 3 First multiply 4*3 = 12 and rewrite the problem x2 + 7x + 12 Factor this (x+3)(x+4) Now divide by 4 / reduce and bottoms up 3 4 (𝑥 + 4) (𝑥 + 4) Solution: (4x+ 3)(x+1) 9 Chapter 1: Factoring and Quadratic Equations 25) 4x2 – 7x + 3 First multiply 4*3 = 12 and rewrite the problem x2 - 7x + 12 Factor this (x-3)(x-4) Now divide by 4 / reduce and bottoms up 3 4 (𝑥 − 4) (𝑥 − 4) Solution: (4x- 3)(x-1) 27) 2z2 + 5z – 7 First multiply the 2*-7 = -14 and rewrite the problem z2 + 5z – 14 factor this (z+7)(z-2) 7 2 now divide by 2 / reduce and bottoms up (𝑧 + 2) (𝑧 − 2) Solution: (2z + 7)(z – 1) 29) 2z2 – 5z – 7 First multiply the 2*-7 = -14 and rewrite the problem z2 + 5z – 14 factor this (z-7)(z+2) 7 2 now divide by 2 / reduce and bottoms up (𝑧 − 2) (𝑧 + 2) Solution: (2z - 7)(z + 1) 10 Chapter 1: Factoring and Quadratic Equations 31) 3x2 – 11x + 10 First multiply the 3*10 = 30 and rewrite the problem x2 – 11x + 30 factor this (x – 5)(x – 6) 5 6 now divide by 3 / reduce and bottoms up (𝑥 − 3) (𝑥 − 3) Solution: (3x – 5)(x – 2) 33) 2b2 – 15b + 7 First multiply 2 * 7 = 14 and rewrite the problem b2 – 15b + 14 factor this (b – 14)(b – 1) Now divide by 2 / reduce and bottoms up = (𝑏 − 14 2 1 ) (𝑏 − 2) Solution: (b – 7)(2b – 1) 35) 6y2 – 7y – 5 First multiply the 6*-5 = -30 and rewrite the problem x2 – 7x - 30 factor this (x + 3)(x - 10) 3 now divide by 6 / reduce and bottoms up (𝑥 + 6) (𝑥 − 1 10 6 ) 5 (𝑥 + 2) (𝑥 − 3) Solution: (2x + 1)(3x – 5) 11 Chapter 1: Factoring and Quadratic Equations 37) 8a2 + a – 7 First multiply 8* -7 = -56 and rewrite the problem a2 + a – 56 and factor this (a+8)(a-7) Now divide by 8 / reduce and bottoms up 8 7 (𝑎 + 8) (𝑎 − 8) Solution: (a + 1)(8a – 7) 39) 2x2 – 5x – 7 First multiply 2* -7 = -14 and rewrite the problem x2 - 5x - 14 and factor this (x + 2)(x – 7) Now divide by 2 / reduce and bottoms up 2 7 (𝑥 + 2) (𝑥 − 2) Solution: (x + 1)(2x – 7) 12 Chapter 1: Factoring and Quadratic Equations 41) 3x2 + 5x + 6 First multiply 3*6 = 18 and rewrite the problem x2 + 5x + 18 my choices for the lasts are 1*9 or 2*9 or 3*6 none of these gives me the 5 that I need so this is prime Solution: Prime 43) 2x2 + 5x – 8 First multiply 2*-8 = -16 and rewrite the problem x2 + 5x - 16 This is prime, so the original problem is also prime. Solution: prime #45-64: Factor out the GCF and then factor by bottoms up or the guess and check method. 45) 4m2 + 34m – 18 First factor out the GCF of 2 = 2(2m2 + 17m – 9) Now bottoms up factor the inside of the parenthesis 2(m2 + 17m – 18) 2(m + 18)(m – 1) 2 (𝑚 + 18 1 ) (𝑚 − 2) 2 Solution: 2(m+9)(2m-1) 13 Chapter 1: Factoring and Quadratic Equations 47) 4z3 – 13z2 + 3z First factor out the GCF of z = z(4z2 – 13z + 3) Now bottoms up factor the inside of the parenthesis z(z2 – 13z + 12) z(z – 1)(z – 12) 1 𝑧 (𝑧 − 4) (𝑧 − 12 4 ) Solution: z(4z – 1)(z – 3) 49) 20x3 – 18x2 + 4x First factor out the GCF of 2x = 2x(10x2 – 9x + 2) Now I will bottom up what is left inside the parenthesis. 2x(x2 – 9x + 20) 2x(x – 5)(x – 4) 5 4 2𝑥 (𝑥 − 10) (𝑥 − 10) 1 2 2𝑥 (𝑥 − 2) (𝑥 − 5) Solution: 2x(2x – 1)(5x – 2) 14 Chapter 1: Factoring and Quadratic Equations 51) -16x2 + 44x – 10 First factor out the GCF of -2 = -2(8x2 – 22x + 5) Now bottoms up factor what’s left in the parenthesis. -2(x2 – 22x + 40) -2(x – 20)(x – 2) −2 (𝑥 − 20 2 ) (𝑥 − 8) 8 5 1 −2 (𝑥 − 2) (𝑥 − 4) Solution: -2(2x – 5)(4x – 1) 53) 18x2 – 21x – 15 First factor out the GCF of 3 = 3(6x2 – 7x – 5) Now bottoms up factor the inside of the parenthesis 3(x2 – 7x – 30) 3(x+ 3)(x-10) 3 (𝑥 + 6) (𝑥 − 3 10 1 5 6 ) 3 (𝑥 + 2) (𝑥 − 3) Solution: 3(2x+1)(3x-5) 15 Chapter 1: Factoring and Quadratic Equations 55) 18x3 – 21x2 – 15x First factor out the GCF of 3x = 3x(6x2 – 7x – 5) Now bottoms up factor the inside of the parenthesis 3x(x2 – 7x – 30) 3x(x+ 3)(x-10) 3 3𝑥 (𝑥 + 6) (𝑥 − 1 10 6 ) 5 3𝑥 (𝑥 + 2) (𝑥 − 3) Solution: 3x(2x+1)(3x-5) 57) -18x2 + 21x + 15 First factor out the GCF of -3 = -3(6x2 – 7x – 5) Now bottoms up factor the inside of the parenthesis -3(x2 – 7x – 30) -3(x+ 3)(x-10) 3 −3 (𝑥 + 6) (𝑥 − 1 10 6 ) 5 −3 (𝑥 + 2) (𝑥 − 3) Solution: -3(2x+1)(3x-5) 16 Chapter 1: Factoring and Quadratic Equations 59) 12x2 + 10x + 12 First factor out GCF of 2 = 2(6x2 + 5x + 6) Now bottoms up what is left inside the parenthesis. 2(x2 + 5x + 36) the inside of the parenthesis is prime, so I can’t factor more. It would be wrong to write prime as my answer as I was able to factor out a GCF of 2. Solution: 2(6x2 + 5x + 6) 61) 3x3 + 5x2 + 6x First factor out the GCF of x = x(3x2 + 5x + 6) Now bottoms up factor what is left x(x2 + 5x + 18) the inside of the parenthesis is prime, so I can’t factor more. It would be wrong to write prime as my answer as I was able to factor out a GCF of x. Solution: x(3x2 + 5x + 6) 63) 4b3 + 6b2 - 6b factor out GCF of 2b = 2b(2b2 + 3b – 3) now bottoms up factor the inside of the parenthesis 2b(b2 + 3b – 6) the inside of the parenthesis is prime, so I can’t factor more. It would be wrong to write prime as my answer as I was able to factor out a GCF of 2b. Solution: 2b(2b2 + 3b – 3) 17
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