Question

the area of a rectangle is $5x^3 + 19x^2 + 6x - 18$ with length $x + 3$. using synthetic division, what is the width of the rectangle?\
\\(\bigcirc\\) $5x^2 + 4x - 6$\
\\(\bigcirc\\) $5x^2 + 34x + 108 + \frac{306}{x + 3}$\
\\(\bigcirc\\) $5x^3 + 4x^2 - 6x$\
\\(\bigcirc\\) $5x^2 + 34x + 108 + \frac{306}{x - 3}$

Explanation:

Step1: Identify the dividend and divisor

The area of the rectangle (dividend) is \(5x^3 + 19x^2 + 6x - 18\) and the length (divisor) is \(x + 3\). For synthetic division, we use the root of the divisor, which is \(x=-3\) (since \(x + 3=0\) gives \(x=-3\)).

Step2: Set up synthetic division

Write the coefficients of the dividend: \(5\) (for \(x^3\)), \(19\) (for \(x^2\)), \(6\) (for \(x\)), and \(-18\) (constant term). Bring down the leading coefficient \(5\).

Step3: Multiply and add

• Multiply \(5\) by \(-3\) to get \(-15\). Add to the next coefficient: \(19 + (-15)=4\).
• Multiply \(4\) by \(-3\) to get \(-12\). Add to the next coefficient: \(6 + (-12)= - 6\).
• Multiply \(-6\) by \(-3\) to get \(18\). Add to the last coefficient: \(-18+18 = 0\) (remainder is \(0\), which makes sense as \(x + 3\) is a factor).

Step4: Form the quotient

The coefficients of the quotient (width) are \(5\) (for \(x^2\)), \(4\) (for \(x\)), and \(-6\) (constant term). Wait, no, wait the area is \(5x^3+19x^2 + 6x - 18\)? Wait, maybe I misread the area. Wait the user's image says "the area of a rectangle is \(5x^3 + 19x^2 + 6x - 18\) with length \(x + 3\)". Wait but the options have \(5x^2+34x + 108+\frac{306}{x + 3}\) etc. Wait, maybe the area is \(5x^3+19x^2+6x - 18\)? No, maybe the area is \(5x^3+19x^2+6x - 18\)? Wait no, let's check the options. Wait the third option is \(5x^2+34x + 108+\frac{306}{x + 3}\)? Wait, maybe I made a mistake. Wait, let's re - examine the problem. Wait the area is \(5x^3+19x^2+6x - 18\)? No, maybe the area is \(5x^3+19x^2+6x - 18\)? Wait, no, let's do synthetic division correctly. Wait, the length is \(x + 3\), so divisor is \(x+3\), root \(x=-3\). Coefficients of area: \(5\), \(19\), \(6\), \(-18\).

Bring down \(5\).

\(5\times(-3)=-15\), \(19-15 = 4\).

\(4\times(-3)=-12\), \(6-12=-6\).

\(-6\times(-3)=18\), \(-18 + 18=0\).

So the quotient is \(5x^2+4x - 6\). Wait, but that's the first option? Wait no, the first option is \(5x^2+4x - 6\). Wait, but maybe the area was misread. Wait, maybe the area is \(5x^3+19x^2+6x - 18\)? Then the width is \(5x^2 + 4x-6\), which is the first option. Wait, but the user's options: first option is \(5x^2+4x - 6\), second is \(5x^3+4x^2-6x\), third is \(5x^2+34x + 108+\frac{306}{x + 3}\), fourth is \(5x^2+34x + 108+\frac{306}{x - 3}\). Wait, maybe I misread the area. Wait, maybe the area is \(5x^3+19x^2+6x - 18\)? No, maybe the area is \(5x^3+19x^2+6x - 18\), and length \(x + 3\). Then synthetic division gives \(5x^2+4x - 6\), which is the first option.

Wait, but let's check again. Let's multiply \((x + 3)(5x^2+4x - 6)=5x^3+4x^2-6x+15x^2+12x - 18=5x^3+19x^2+6x - 18\), which matches the area. So the width is \(5x^2+4x - 6\).

Answer:

A. \(5x^2 + 4x-6\)