The area of a rectangle is $(x^4+4x^3+ 3x^2-4x-4)$ , and the length of the rectangle is $(x^3 +5x^2 +8x+ 4)$ . If area length x width, what is the width of the rectangle?

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Answer 1 · 2016-12-28T09:52:39

$x-1$

The width is given by the rational expression:

$(x^4+4x^3+3x^2-4x-4)/(x^3+5x^2+8x+4)$

Let us see if we can simplify the expression:

Note that $x=1$ is a zero of the numerator, since the sum of the coefficients is zero. That is:

$1+4+3-4-4 = 0$

So $(x-1)$ is a factor:

$x^4+4x^3+3x^2-4x+4 = (x-1)(x^3+5x^2+8x+4)$

So:

$(x^4+4x^3+3x^2-4x-4)/(x^3+5x^2+8x+4) = x-1$

The area of a rectangle is (x^4+4x^3+ 3x^2-4x-4)(x^4+4x^3+ 3x^2-4x-4), and the length of the rectangle is (x^3 +5x^2 +8x+ 4)(x^3 +5x^2 +8x+ 4). If area length x width, what is the width of the rectangle?