Factoring and Quadratic Equations. Section 1.1 The Greatest Common Factor

Chapter 1: Factoring and Quadratic Equations Section 1.1: The Greatest Common Factor #1-16: Factor out the GCF. 1) 2y – 10 Both 2 and 10 are divisible by 2, so 2 is part of the common factor. Both terms don’t have a y, so there is no letter in the GCF. Just write a 2 to the left of a parenthesis, and divide the numbers by 2. This is an okay answer 2(1y – 5) , however it isn’t necessary to leave a 1 in front of the y. Solution: 2(y – 5) 3) 14x3 – 7x2 + 7x Each number is divisible by 7, so 7 is part of the common factor. Each term has an x, and the smallest power of x that occurs is x1, so x1 is also part of the GCF. The GCF is 7x Divide each number by 7, and take one x away from each term to get your answer. Solution: 7x(2x2 – 7x + 1) 5) b5 – 3b4 + b3 Each term has a b and the smallest power of b that occurs in the problem is b3, hence b3 is the GCF. I will write a b3 to the left of a parenthesis, and take away 3 b’s from each term inside the parenthesis. Solution: b2 – 3b + 1 7) 12x4 – 3x3 9) 4x3y + 12x2y3 4 divides evenly into both numbers and it is part of the GCF. Each term has an x and the smallest power of x that occurs is x2, so x2 is part of the GCF Each term has a y and the smallest power of y that occurs is y, so y is part of the GCF. The GCF is 4x2y I will divide each number by 4, take 2 x’s and 1 y away from each term. Chapter 1: Factoring and Quadratic Equations Solution: 4x2y(x+3y2) 11) 16a4b2 – 18ab3 13) 12xyz3 – 14x2y3 – 2xz 15) 16r2st3 – 4r3st2 + 12rst #17-26: Factor out a (-1) from each polynomial. 17) -x + 2 I can either write a (-1) or just a – in front of a parenthesis, then switch the signs inside the parenthesis. Solution: -1(x – 2) or -(x – 2) the second answer is considered better 19) -x – 3 21) -5x + 9 I can either write a (-1) or just a – in front of a parenthesis, then switch the signs inside the parenthesis. Solution: -1(5x – 9) or -(5x – 9) 23) -3x + 6y – 7 I can either write a (-1) or just a – in front of a parenthesis, then switch the signs inside the parenthesis. Solution: -1(3x – 6y + 7) or –(3x – 6y + 7) 25) -4x + 6z + 11s Chapter 1: Factoring and Quadratic Equations #27-38: Factor each polynomial by factoring out the opposite of the GCF. 27) -4x3 – 12x2 29) -12x3 + 4x2 – 8x I can either write a (-1) or just a – in front of a parenthesis, then switch the signs inside the parenthesis. Solution: -1(12x3 – 4x2 + 8x) or -(12x3 – 4x2 + 8x) 31) -3z + 6z3 33) -14a4b2 – 6a2b 35) -8xyz3 – 4x2y3 + 2xyz 37) -16r2st3 – 4r3st2 – 12rst2 Section 1.1: The Greatest Common Factor #39 – 52: Factor out the GCF 39) x(x-4) + 3(x-4) There is an (x – 4) on each side of the plus sign. It is the common factor. When I take the (x-4) away from the problem I am left with x + 3, and that is what I will put inside a parenthesis. Solution: (x – 4)(x + 3) 41) x2(y – 2) – 3(y – 2) Chapter 1: Factoring and Quadratic Equations There is an (y – 2) on each side of the minus sign. It is the common factor. When I take the (y 2) away from the problem I am left with x2 - 3, and that is what I will put inside a parenthesis. Solution: (y – 2)(x2 – 3) 43) 3y(z + 1) – 4(z + 1) 45) x(3x – 4) – 2(3x – 4) 47) 3x(2x – 7) + 4(2x – 7) 49) 2x2(3x – 5y) – 5x(3x – 5y) There is an (3x – 5y) on each side of the minus sign. It is the common factor. When I take the (3x-5y) away from the problem I am left with 2x2 - 5x, and that is what I will put inside a parenthesis. Solution: (3x – 5y)(2x2 – 5x) 51) 8y(y – 5) – 9(y – 5)