Question

use synthetic division to rewrite the following fraction in the form $q(x) + \frac{r(x)}{d(x)}$, where $d(x)$ is the denominator of the original fraction, $q(x)$ is the quotient, and $r(x)$ is the remainder.\
$\frac{2x^5 - 16x^4 + 37x^3 - 28x^2 - 35x}{x - 5}$

Explanation:

Step1: Identify the root of the denominator

For the denominator \(x - 5\), the root is \(x = 5\) (since \(x - 5=0\) gives \(x = 5\)). The coefficients of the numerator polynomial \(2x^{5}-16x^{4}+37x^{3}-28x^{2}-35x + 0\) (we add the constant term \(0\) as there is no constant term in the original numerator) are \(2, - 16,37,-28,-35,0\).

Step2: Perform synthetic division

Set up synthetic division with root \(5\) and coefficients \(2, - 16,37,-28,-35,0\):

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• Bring down the leading coefficient \(2\).
• Multiply \(2\) by \(5 = 10\), add to \(-16\): \(-16 + 10=-6\).
• Multiply \(-6\) by \(5=-30\), add to \(37\): \(37-30 = 7\).
• Multiply \(7\) by \(5 = 35\), add to \(-28\): \(-28 + 35=7\).
• Multiply \(7\) by \(5 = 35\), add to \(-35\): \(-35+35 = 0\).
• Multiply \(0\) by \(5 = 0\), add to \(0\): \(0 + 0=0\).

The quotient polynomial \(q(x)\) has coefficients \(2, - 6,7,7,0\) (the last non - zero remainder is the constant term of the quotient? Wait, no. The degree of the quotient is one less than the degree of the dividend. The dividend is degree \(5\), so the quotient is degree \(4\). So \(q(x)=2x^{4}-6x^{3}+7x^{2}+7x+0\) (we can write \(q(x)=2x^{4}-6x^{3}+7x^{2}+7x\)) and the remainder \(r(x) = 0\) (since the last entry in the synthetic division result is \(0\)).

Step3: Rewrite the fraction

The original fraction is \(\frac{2x^{5}-16x^{4}+37x^{3}-28x^{2}-35x}{x - 5}\). Using the division algorithm for polynomials \(f(x)=d(x)q(x)+r(x)\), where \(f(x)\) is the numerator, \(d(x)=x - 5\), \(q(x)=2x^{4}-6x^{3}+7x^{2}+7x\) and \(r(x) = 0\). So \(\frac{f(x)}{d(x)}=q(x)+\frac{r(x)}{d(x)}=2x^{4}-6x^{3}+7x^{2}+7x+\frac{0}{x - 5}\) or simply \(2x^{4}-6x^{3}+7x^{2}+7x\) (since \(\frac{0}{x - 5}=0\)).

Answer:

\(2x^{4}-6x^{3}+7x^{2}+7x+\frac{0}{x - 5}\) (or \(2x^{4}-6x^{3}+7x^{2}+7x\) as \(\frac{0}{x - 5}=0\))