Factoring and Quadratic Equations. Section 1.2 Factoring by Grouping

Chapter 1: Factoring and Quadratic Equations Section 1.2: Factoring by Grouping #1 – 36: Factor by Grouping, state if a polynomial is prime 1) x2 + 5x + 2x + 10 Think of this as (x2 + 5x) + (2x + 10) Factor out an x from the first parenthesis and a 2 from the second parenthesis. = x(x+5) + 2(x+5) Now factor out a (x+5) Solution: (x+5)(x+2) 3) x2 – 5x – 2x + 10 This is a bit trickier. Think of this as (x2 – 5x) + (-2x + 10) notice I put a plus in front of the second parenthesis. This does equal the original problem. Factor out an x from the first parenthesis and -2 from the second =x(x-5) + (-2)(x – 5) (notice what I did with the signs in the second parenthesis when I factored out a negative) This can be written better. = x(x-5) – 2(x-5) Now factor out a (x-5) Solution: (x – 5)(x – 2) 5) x2 + 9x + 4x + 36 7) x2 – 9x – 4x + 36 Think of this as Chapter 1: Factoring and Quadratic Equations (x2 – 9x) + (-4x + 36) notice I put a plus in front of the second parenthesis. This does equal the original problem. Factor out an x from the first parenthesis and -4 from the second =x(x-9) + (-4)(x-9) (notice what I did with the signs in the second parenthesis when I factored out a negative) This can be written better. = x(x-9) – 4(x-9) Now factor out a (x-9) Solution: (x-9)(x-4) 9) y2 + 10y + 3y + 30 Think of this as (y2 + 10y) + (3y + 30) Factor out a y from the first parenthesis and a 3 from the second parenthesis. = y(y+10) + 3(y + 10) Now factor out a (y+10) Solution: (y+10)9y+3) 11) y2 – 10y – 3y + 30 13) 5x2 + 5x + 6x + 6 Think of this as (5x2 + 5x) + (6x + 6) Factor out an 5x from the first parenthesis and a 6 from the second parenthesis. = 5x(x+1) + 6(x+1) Now factor out a (x+1) Solution: (x+1)(5x+6) 15) 5x2 – 5x – 6x+ 6 17) 4y2 + 3y + 4y + 3 Chapter 1: Factoring and Quadratic Equations 19) 4y2 – 3y – 4y + 3 Think of this as (4y2 – 3y) + (-4y + 3) notice I put a plus in front of the second parenthesis. This does equal the original problem. Factor out an x from the first parenthesis and -1 from the second =y(4y – 3) + (-1)(4y – 3) (notice what I did with the signs in the second parenthesis when I factored out a negative) This can be written better. = y(4y – 3) – 1(4y – 3) Now factor out a (4y – 3) Solution: (4y – 3)(y – 1) 21) 2z2 – 2z + 7z – 7 23) 2z2 + 7z – 2z – 7 Think of this as (2z2 + 7z) + (-2z – 7) notice I put a plus in front of the second parenthesis. This does equal the original problem. Factor out an z from the first parenthesis and -1 from the second = z(2z+7) + (-1)(2z + 7) (notice what I did with the signs in the second parenthesis when I factored out a negative) This can be written better. = z(2z+7) – 1(2z +7) Now factor out a (2z + 7) Solution: (2z+7)(z-1) Chapter 1: Factoring and Quadratic Equations 25) x3 + 2x2 + 6x + 12 27) x3 + 6x + 3x2 + 18x Think of this as (x3 + 6x) + (3x2 + 18x) Factor out an x2 from the first parenthesis and a 3x from the second parenthesis. = x2(x+6) + 3x(x+6) Now factor out a (x+6) Okay Solution: (x+6)(x2+3x) this can be written better, notice the second parenthesis has a common factor of an x. I should factor that out and write it first. Best Solution: x(x+6)(x+3) 29) y3 + 4y2 + y + 4 31) z3 + 5z2 – z – 5 Think of this as (z3 + 5z2) + (-z – 5) notice I put a plus in front of the second parenthesis. This does equal the original problem. Factor out an z2 from the first parenthesis and -1 from the second = z2(z+5) + (-1)(z+5) (notice what I did with the signs in the second parenthesis when I factored out a negative) This can be written better. = z2(z+5) – 1(z+5) Now factor out a (z+5) Solution: (z+5)(z2 – 1) if you factored before, you might know how to simplify this more. If you wrote (z+5)(z+1)(z-1) it is also correct, but too advanced for now. Chapter 1: Factoring and Quadratic Equations 33) 5x3 + 4x2 + 10x + 8 35) 7t3 + 5t2 + 21t + 15