Suppose the width of a rectangle is $2 b - 1$ $2b - 1$ and the area is $4 b^{3} + 5 b - 3$ $4b^3 + 5b - 3$ . What is the length of the rectangle?

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Answer 1 · 2016-06-20T14:36:55

The formula for area of a rectangle is $A = L \times W$ $A = L xx W$ .

Therefore $\frac{A}{W} = L$ $A/W = L$ , and by continuation $2 b - 1$ $2b - 1$ must be a factor of $4 b^{3} + 5 b - 3$ $4b^3+ 5b - 3$ .

By synthetic division:

$\frac{1}{2}_|4 0 5 -3$ $1/2"_|4 0 5 -3"$
$2 1 3$ $" 2 1 3"$
$-------------------------$ $"-------------------------"$
$4 2 6 0$ $" 4 2 6 0"$

Hence, $\frac{4 b^{3} + 5 b - 3}{2 b - 1}$ $(4b^3 + 5b - 3)/(2b - 1)$ gives a quotient of $4 b^{2} + 2 b + 6$ $4b^2 + 2b+ 6$ .

$:.$ The length of the rectangle measures $(4b^2 + 2b+ 6)$ inches.

Hopefully this helps!

Suppose the width of a rectangle is 2b - 12b−12b - 1 and the area is 4b^3 + 5b - 34b3+5b−34b^3 + 5b - 3. What is the length of the rectangle?