Question

the area of a triangle is given by the expression: 14x² + 20x - 15 and the width is 14x - 1. find the length of the rectangle:

Explanation:

Step1: Recall the formula for the area of a rectangle

The area formula for a rectangle is \(A = length\times width\). So, \(length=\frac{A}{width}\). Here, \(A = 14x^{2}+20x - 15\) and \(width=14x - 1\).

Step2: Perform polynomial long - division

Divide \(14x^{2}+20x - 15\) by \(14x - 1\).

• First, divide the leading term of the dividend \(14x^{2}\) by the leading term of the divisor \(14x\): \(\frac{14x^{2}}{14x}=x\).
• Multiply the divisor \(14x - 1\) by \(x\): \(x(14x - 1)=14x^{2}-x\).
• Subtract this from the dividend: \((14x^{2}+20x - 15)-(14x^{2}-x)=14x^{2}+20x - 15 - 14x^{2}+x = 21x-15\).
• Then divide the leading term of the new dividend \(21x\) by the leading term of the divisor \(14x\): \(\frac{21x}{14x}=\frac{3}{2}\).
• Multiply the divisor \(14x - 1\) by \(\frac{3}{2}\): \(\frac{3}{2}(14x - 1)=21x-\frac{3}{2}\).
• Subtract this from \(21x - 15\): \((21x-15)-(21x-\frac{3}{2})=21x-15 - 21x+\frac{3}{2}=-\frac{27}{2}\).

Since the remainder \(-\frac{27}{2}\) is non - zero, there is an error. Let's use the formula \(A = lw\), so \(l=\frac{14x^{2}+20x - 15}{14x - 1}\).
We can rewrite \(14x^{2}+20x - 15\) as \(14x^{2}-x+21x - 15\).
\(\frac{14x^{2}-x+21x - 15}{14x - 1}=\frac{x(14x - 1)+ \frac{3}{2}(14x - 1)-\frac{27}{2}}{14x - 1}\).
If we assume the problem is a rectangle (maybe a mis - label of the figure as a triangle, since for a rectangle \(A = lw\)), and if we consider the remainder as a mistake in the problem setup (assuming it's a perfect division), we can also use synthetic division (for \(14x-1 = 0\), \(x=\frac{1}{14}\)). But a better way is to use the formula:

$$
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$$

Check: \((x+\frac{3}{2})(14x - 1)=14x^{2}-x + 21x-\frac{3}{2}=14x^{2}+20x-\frac{3}{2}\). If we assume the area formula was a typo and the area is \(14x^{2}+20x-\frac{3}{2}\), then the length is \(x + \frac{3}{2}\).

Answer:

\(x+\frac{3}{2}\)