the area of a rectangle is $(x^{4}+4x^{3}+3x^{2}-4x - 4)$, and the length of the rectangle is…

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 × width, what is the width of the rectangle?\n$x + 1$\n$x - 9$\n$x + 4$\n$x - 1$

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 × width, what is the width of the rectangle?\n$x + 1$\n$x - 9$\n$x + 4$\n$x - 1$

Answer

Explanation:

Step1: Recall width formula

Since area = length × width, then width = $\frac{\text{area}}{\text{length}}$. So we need to divide $(x^{4}+4x^{3}+3x^{2}-4x - 4)$ by $(x^{3}+5x^{2}+8x + 4)$.

Step2: Perform polynomial long - division

We divide the leading term of the dividend $x^{4}+4x^{3}+3x^{2}-4x - 4$ (which is $x^{4}$) by the leading term of the divisor $x^{3}+5x^{2}+8x + 4$ (which is $x^{3}$). The result is $x$. Multiply $x^{3}+5x^{2}+8x + 4$ by $x$: $x(x^{3}+5x^{2}+8x + 4)=x^{4}+5x^{3}+8x^{2}+4x$. Subtract this from the dividend: [ \begin{align*} &(x^{4}+4x^{3}+3x^{2}-4x - 4)-(x^{4}+5x^{3}+8x^{2}+4x)\ =&x^{4}+4x^{3}+3x^{2}-4x - 4 - x^{4}-5x^{3}-8x^{2}-4x\ =&-x^{3}-5x^{2}-8x - 4 \end{align*} ] Now divide the leading term of the new dividend $-x^{3}-5x^{2}-8x - 4$ (which is $-x^{3}$) by the leading term of the divisor $x^{3}$ (which is $x^{3}$), the result is $- 1$. Multiply $x^{3}+5x^{2}+8x + 4$ by $-1$: $-(x^{3}+5x^{2}+8x + 4)=-x^{3}-5x^{2}-8x - 4$. Subtract this from the new dividend: $(-x^{3}-5x^{2}-8x - 4)-(-x^{3}-5x^{2}-8x - 4)=0$.

Answer:

D. $x - 1$