Problem 1 3x2 + 8x + 4The first thing we have to do is to find what multiplies toget 3x2. I choose 3x and x, so we put them inparenthesis;(3x + ?)(x + ?)
Problem 1 3x2 + 8x + 4 (3x + ?)(x + ?)The next step is a bit different from what you areexpecting, we will take multiplication problems thatmake 4 and place them in the question mark’s spots,forgetting about the 8 for now. You will do willdifferentiate between the problems like 1(4) and 4(1)giving them separate problems. These will be madeand solved on the next page.Pairs that make 4: 2 and 2, 1 and 4, 4 and 1
Problem 11. (3x + 2)(x + 2) 3x2 + 4 + 8x2. (3x + 1)(x + 4) 3x2 + 4 + 4x3. (3x + 4)(x + 1) 3x2 + 4 + 7xWe will now remember the 8 and our original problem(3x2 + 8x + 4), since the top problem contains an 8 andis exactly the same to our problem, our answer is (3x +2)(x + 2).
Problem 2 2x2 + 13h – 71. For our first step, I will say that to make 2x you need 2xand x.2. Going to our second step we will say that to get -7 youcan use -1 and 7, 1 and -7, 7 and -1, -7 and 1(2x + -1) (x + 7)(2x + 1) (x + -7)(2x + 7) (x +-1)(2x + -7) (x + 1)
Mini LessonDo you remember how in the last slide, when we hadproblems like (2x + 7) (x +-1) & (2x + -7) (x + 1) wherethe negative sign just switched sides? A cheat toanswer that is to write answer of the problem for thefirst one (2x2 – 7 + 5x) and when you are answeringthe next one you can just add a negative to thecoefficient.
Problem 2(2x + -1) (x + 7) is the answer since it multiplies tomake our original problem, 2x2 + 13h – 7.
Problem 3 -x2 + x + 20The way to get around the negative sign on the firstproblem is to change it to -1(x2 – x – 20) and solve theproblem inside the parenthesis ignoring the -1.-1(x – 5)(x + 4)Now you can multiply it all by -1.(-x + 5)(-x - 4)This is your final answer.