Question

hw 3.5: polynomial division
score: 3/15 answered: 3/15
question 4
divide: \\(\frac{6x^3 - 12x^2 - 17x + 21}{3x^2 - 7}\\). give your answer in the form \\(\text{quotient} + \frac{\text{remainder}}{\text{divisor}}\\)
\\(\frac{6x^3 - 12x^2 - 17x + 21}{3x^2 - 7} = \\)
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Explanation:

Step1: Divide the leading terms

Divide the leading term of the numerator \(6x^3\) by the leading term of the denominator \(3x^2\). So, \(\frac{6x^3}{3x^2} = 2x\). This is the first term of the quotient.

Step2: Multiply the divisor by the first term of the quotient

Multiply the divisor \(3x^2 - 7\) by \(2x\): \(2x(3x^2 - 7)=6x^3 - 14x\).

Step3: Subtract this product from the numerator

Subtract \(6x^3 - 14x\) from the numerator \(6x^3 - 12x^2 - 17x + 21\):
\[

$$\begin{align*} &(6x^3 - 12x^2 - 17x + 21)-(6x^3 - 14x)\\ =&6x^3 - 12x^2 - 17x + 21 - 6x^3 + 14x\\ =& - 12x^2 - 3x + 21 \end{align*}$$

\]

Step4: Divide the new leading term

Now, divide the leading term of the new polynomial \(-12x^2\) by the leading term of the divisor \(3x^2\). So, \(\frac{-12x^2}{3x^2}=-4\). This is the next term of the quotient.

Step5: Multiply the divisor by the second term of the quotient

Multiply the divisor \(3x^2 - 7\) by \(-4\): \(-4(3x^2 - 7)=-12x^2 + 28\).

Step6: Subtract this product from the polynomial obtained in Step 3

Subtract \(-12x^2 + 28\) from \(-12x^2 - 3x + 21\):
\[

$$\begin{align*} &(-12x^2 - 3x + 21)-(-12x^2 + 28)\\ =&-12x^2 - 3x + 21 + 12x^2 - 28\\ =& - 3x - 7 \end{align*}$$

\]

Step7: Write the final form

The quotient is \(2x - 4\) and the remainder is \(-3x - 7\), and the divisor is \(3x^2 - 7\). So, we write the expression as:
\(2x - 4+\frac{-3x - 7}{3x^2 - 7}\)

Answer:

\(2x - 4+\frac{-3x - 7}{3x^2 - 7}\)