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

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Question 375317: The perimeter of a rectangle is (4x^2+10x+20)inches and its length is (x^2-4x+8. Find the width.
Answer by mananth(16946) About Me  (Show Source):
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(4x^2+10x+20)inches and its length is (x^2-4x+8). Find the width.
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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
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m.ananth@hotmail.ca