Question

simplify the rational expression \\(\frac{6x^4 - 5x^2 + 120x - 121}{3x - 3}\\) in the form \\(q(x) + \frac{r(x)}{b(x)}\\). use the keyboard to type your answer in the box. you may use geogebra cas calculator or any cas calculator. the rational expression \\(\frac{6x^4 - 5x^2 + 120x - 121}{3x - 3}\\) in the form \\(q(x) + \frac{r(x)}{b(x)}\\) is \\(square\\).

Explanation:

Step1: Perform Polynomial Long Division

We divide the numerator \(6x^4 - 5x^2 + 120x - 121\) by the denominator \(3x - 3\).

First, divide the leading term of the numerator \(6x^4\) by the leading term of the denominator \(3x\): \(\frac{6x^4}{3x}=2x^3\). Multiply \(3x - 3\) by \(2x^3\) to get \(6x^4 - 6x^3\). Subtract this from the numerator:

\[ \begin{align*} &(6x^4 - 5x^2 + 120x - 121)-(6x^4 - 6x^3)\\ =&6x^4 - 5x^2 + 120x - 121 - 6x^4 + 6x^3\\ =&6x^3 - 5x^2 + 120x - 121 \end{align*} \]

Step2: Continue Long Division

Now, divide the leading term of the new numerator \(6x^3\) by \(3x\): \(\frac{6x^3}{3x}=2x^2\). Multiply \(3x - 3\) by \(2x^2\) to get \(6x^3 - 6x^2\). Subtract this from the new numerator:

\[ \begin{align*} &(6x^3 - 5x^2 + 120x - 121)-(6x^3 - 6x^2)\\ =&6x^3 - 5x^2 + 120x - 121 - 6x^3 + 6x^2\\ =&x^2 + 120x - 121 \end{align*} \]

Step3: Continue Long Division

Divide the leading term of the new numerator \(x^2\) by \(3x\): \(\frac{x^2}{3x}=\frac{1}{3}x\). Multiply \(3x - 3\) by \(\frac{1}{3}x\) to get \(x^2 - x\). Subtract this from the new numerator:

\[ \begin{align*} &(x^2 + 120x - 121)-(x^2 - x)\\ =&x^2 + 120x - 121 - x^2 + x\\ =&121x - 121 \end{align*} \]

Step4: Continue Long Division

Divide the leading term of the new numerator \(121x\) by \(3x\): \(\frac{121x}{3x}=\frac{121}{3}\). Multiply \(3x - 3\) by \(\frac{121}{3}\) to get \(121x - 121\). Subtract this from the new numerator:

\[ \begin{align*} &(121x - 121)-(121x - 121)\\ =&121x - 121 - 121x + 121\\ =&0 \end{align*} \]

Step5: Combine the Results

Putting it all together, the quotient \(q(x)\) is \(2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3}\) and the remainder \(r(x)\) is \(0\). So the rational expression simplifies to:

\[ 2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3}+\frac{0}{3x - 3} \]

Simplifying the quotient terms (we can also write the polynomial part with integer coefficients by factoring out \(\frac{1}{3}\) if needed, but since the problem allows the form \(q(x)+\frac{r(x)}{b(x)}\) and the remainder is \(0\), we can present the quotient as is or simplify the polynomial):

First, let's rewrite the quotient with a common denominator to combine terms:

\(2x^3+2x^2+\frac{1}{3}x+\frac{121}{3}=\frac{6x^3 + 6x^2 + x + 121}{3}\)

But since the division gave us a remainder of \(0\), the simplified form is the quotient itself (because \(\frac{r(x)}{b(x)} = 0\)). Wait, actually, when we did the long division, we can also check by multiplying back:

\((3x - 3)(2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3})=3x(2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3})-3(2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3})\)

\[ \begin{align*} &=6x^4 + 6x^3 + x^2 + 121x-6x^3 - 6x^2 - x - 121\\ &=6x^4 - 5x^2 + 120x - 121 \end{align*} \]

Which matches the numerator. So the remainder \(r(x) = 0\), so the expression simplifies to \(2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3}+\frac{0}{3x - 3}\), or we can write the polynomial part with integer coefficients by factoring:

\(2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3}=\frac{6x^3 + 6x^2 + x + 121}{3}\)

But another way is to note that when we did the long division, we can also factor the denominator \(3x - 3 = 3(x - 1)\) and see if we can factor the numerator. Let's try factoring the numerator \(6x^4 - 5x^2 + 120x - 121\). Let's try \(x = 1\): \(6 - 5 + 120 - 121 = 0\), so \((x - 1)\) is a factor. Then we can perform polynomial division by \((x - 1)\) first, then by \(3\) (since the denominator is \(3(x - 1)\)).

Divide \(6x^4 - 5x^2 + 120x - 121\) by \(x - 1\) using synthetic division:

Coefficients: \(6, 0, -5, 120, -121\) (for \(x^4, x^3, x^2, x, constant\))

Using synthetic division with root \(1\):

\[ \begin{array}{r|rrrrr} 1 & 6 & 0 & -5 & 120 & -121\\ & & 6 & 6 & 1 & 121\\ \hline & 6 & 6 & 1 & 121 & 0 \end{array} \]

So the quotient is \(6x^3 + 6x^2 + x + 121\), so \(6x^4 - 5x^2 + 120x - 121=(x - 1)(6x^3 + 6x^2 + x + 121)\). Then the original expression is \(\frac{(x - 1)(6x^3 + 6x^2 + x + 121)}{3(x - 1)}=\frac{6x^3 + 6x^2 + x + 121}{3}\) (since \(x eq1\) to avoid division by zero).

Simplifying \(\frac{6x^3 + 6x^2 + x + 121}{3}=2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3}\)

Since the remainder is \(0\), the expression in the form \(q(x)+\frac{r(x)}{b(x)}\) is \(2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3}+\frac{0}{3x - 3}\), which simplifies to \(2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3}\) (because the remainder term is zero).

Answer:

\(2x^3 + 2x^2+\frac{1}{3}x+\frac{121}{3}\) (or equivalently \(\frac{6x^3 + 6x^2 + x + 121}{3}\))