Question

(3x^2+26x+16)/(3x^2-7x-6)×(x^2+2x-15)/2x^2+9x-5) can someone help me solve this step by step please.

Answer 1

So firstly, we are going to complete the square for each polynomial. To do this, we need to split the middle term of each polynomial into two terms that add up to the middle term and multiply to the product of the first and third term.

Starting with 3x^2+26x+16, the two terms that fit are 24x and 2x since they add up to 26x and multiply to 48x^2. Now replace 26x with 24x + 2x: $3x^2+24x+2x+16$

Next, you are going to factor 3x^2 + 24x and 2x + 16 separately. Make sure they have the same quantity in the parentheses: $3x(x+8)+2(x+8)$

Next, thanks to distributive property, we can rewrite this as (3x + 2)(x + 8).

The process is going to be the same with the other 3 polynomials, so I'll brush through them real quickly:

$3x^2-7x-6\\ 3x^2-9x+2x-6\\ 3x(x-3)+2(x-3)\\ (3x+2)(x-3)$

$x^2+2x-15\\ x^2-3x+5x-15\\ x(x-3)+5(x-3)\\ (x+5)(x-3)$

$2x^2+9x-5\\ 2x^2+10x-x-5\\ 2x(x+5)-1(x+5)\\ (2x-1)(x+5)$

So right now, just for reminder our fraction looks like this: $\frac{(3x+2)(x+8)(x+5)(x-3)}{(3x+2)(x-3)(2x-1)(x+5)}$

Next, cancel out (x+5), (x-3), and (3x+2) and your answer will be: $\frac{x+8}{2x-1}$