SOLUTION: The perimeter of a rectangle is (4x^2+ 10x+20) inches and its lenght is (x^2+4x+8) inches. Find the width.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The perimeter of a rectangle is (4x^2+ 10x+20) inches and its lenght is (x^2+4x+8) inches. Find the width.      Log On


   

Question 375314: The perimeter of a rectangle is (4x^2+ 10x+20) inches and its lenght is (x^2+4x+8) inches. Find the width.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Let width be w
2*(l+w)= perimeter
2*((x^2-4x+8)+w)=(4x^2+10x+20)
2x^2-8x+16+2w=4x^2+10x+20
reorganize the terms
4x^2-2x^2+10x+8x+20-16=2w
2x^2+18x+4=2w
/2
x^2+9x+2=w
...
m.ananth@hotmail.ca