how do i divide factors ax^2+bx+c

Question

Thwe question is "The area of a rectangle is 20x^2-27x-8. The length is 4x+1. What is the width?"

Answer 1

64÷1.40

Answer 2

Let S=20x²-27x-8: the area, L=4x+1: the length, and W=the width, so you have:

S=LW  Substitute the equation given for S and L, getting

20x²-27x-8=(4x+1)W*   The left side is quadratic, but (4x+1) is linear, so W must be linear.

Let W=ax+b, so 20x²-27x-8=(4x+1)(ax+b)=4ax²+(4b+a)x+b.   We have:

4a=20 and b=-8, so we have: a=5 and b=-8   Thus, W=5x-8  

CK: 4b+a=4x(-8)+(5)=-27   CKD.

Therefore the width is: 5x-8 

* We can factor this using long division: 20x²-27x-8=5x(4x+1)-5x-27x-8=5x(4x+1)-32x-8

=5x(4x+1)-8(4x+1)=(4x+1)(5x-8)   We have: W=5x-8