1.  We visited this idea earlier, and will build on it now  How would we factor 3x 2 + 11x + 6? ◦ Recall, to factor x 2 + 2x + 3, we put one ‘x’ in each bracket, since x*x = x 2 ◦ Now we need 2 ‘things’ whose products is 3x 2  We can use 3x*x = 3x 2  So, (3x + a)(x + b) – we need ‘a’ and ‘b’ such that when multiplied by 3x and x, gives us the original quadratic  3x 2 + 11x + 6 = (3x + 2)(x + 3)

2.  Now that we can factor a quadratic like this: 3x 2 + 11x + 6 = (3x + 2)(x + 3), we can determine the x-intercepts If we had y = a*b, what are the x-intercepts? ◦ Recall, x-intercepts occurs when y = 0 ◦ What would make y=0 in y=a*b?  Either a = 0, or b = 0, right? ◦ If y=(3x + 2)(x + 3), then either (3x+2)=0 or (x+3)=0 ◦ We solve for x in each expression, and that gives us the x-intercepts

3.  (x+3)=0  x + 3 = 0  x + 3 – 3 = 0 – 3  x = -3  Therefore, the x- intercepts are -2/3 and -3.

4.  Factor 4x 2 – 8x – 5:  Either (4x + a)(x + b) OR (2x + a)(2x + b) It’s usually the one where ‘x’ is not by itself in one bracket  (2x + a)(2x + b)  What multiplies to -5? ◦ -5 and +1 or -1 and +5 – try them out!  (2x – 5)(2x + 1) ◦ = 4x 2 + 2x – 10x -5 ◦ = 4x 2 – 8x – 5 ◦ We found the right solution! No need to try the other possibilities.

5.  Factor 12x 2 – 25x + 12:  Either (12x + a)(x + b) OR (4x + a)(3x + b) It’s usually the one where ‘x’ is not by itself in one bracket  (4x + a)(3x + b)  What multiplies to +12? ◦ 3 x 4, -3 x -4, 12 x 1, -12 x -1  (4x + 3)(3x + 4) ◦ = 12x 2 + 16x + 9x + 12 ◦ = 12x 2 + 25x + 12 ◦ Not the right solution! Try the other possibilities.

6.  What multiplies to +12? ◦ 3 x 4, -3 x -4, 12 x 1, -12 x -1  (4x - 3)(3x - 4) ◦ =12x 2 – 16x – 9x + 12 ◦ =12x 2 – 25x + 12  We found the right solution! So the factors are (4x-3) and (3x-4)  Now we need to find the x-intercepts…

7.  We determined: 12x 2 – 25x + 12 = (4x - 3)(3x - 4)  Recall, x-intercepts occur when y = 0  Here, that means when (4x – 3) = 0 & (3x – 4) = 0