Question

divide. if the polynomial does not divide evenly, include the remainder as a fraction.
$(2f^{3} + 11f^{2} + 17f + 15) \div (f + 3)$

Explanation:

Step1: Use polynomial long division

Divide the leading term of the dividend \(2f^3\) by the leading term of the divisor \(f\), we get \(2f^2\). Multiply the divisor \((f + 3)\) by \(2f^2\) to get \(2f^3+6f^2\). Subtract this from the dividend:
\[ \begin{align*} &(2f^3 + 11f^2 + 17f + 15)-(2f^3+6f^2)\\ =&(2f^3 - 2f^3)+(11f^2 - 6f^2)+17f + 15\\ =&5f^2+17f + 15 \end{align*} \]

Step2: Divide the new leading term

Divide the leading term of \(5f^2\) by \(f\) to get \(5f\). Multiply the divisor \((f + 3)\) by \(5f\) to get \(5f^2+15f\). Subtract this from \(5f^2+17f + 15\):
\[ \begin{align*} &(5f^2+17f + 15)-(5f^2+15f)\\ =&(5f^2 - 5f^2)+(17f - 15f)+15\\ =&2f + 15 \end{align*} \]

Step3: Divide the new leading term again

Divide the leading term of \(2f\) by \(f\) to get \(2\). Multiply the divisor \((f + 3)\) by \(2\) to get \(2f+6\). Subtract this from \(2f + 15\):
\[ \begin{align*} &(2f + 15)-(2f+6)\\ =&(2f - 2f)+(15 - 6)\\ =&9 \end{align*} \]

Step4: Combine the results

The quotient is \(2f^2 + 5f+2\) and the remainder is \(9\). So we can write the division as:
\[ \frac{2f^3 + 11f^2 + 17f + 15}{f + 3}=2f^2 + 5f + 2+\frac{9}{f + 3} \]
But wait, let's check using synthetic division. The root of \(f + 3 = 0\) is \(f=-3\).
Set up synthetic division:
\[ \begin{array}{r|rrrr} -3 & 2 & 11 & 17 & 15\\ & & -6 & -15 & -6\\ \hline & 2 & 5 & 2 & 9 \end{array} \]
So the quotient is \(2f^2+5f + 2\) and the remainder is \(9\). Wait, but when we do \((2f^3 + 11f^2 + 17f + 15)\div(f + 3)\), using synthetic division, the coefficients of the quotient are \(2,5,2\) and remainder \(9\). Wait, but let's multiply \((f + 3)(2f^2+5f + 2)+9\):
\[ \begin{align*} &(f + 3)(2f^2+5f + 2)+9\\ =&2f^3+5f^2+2f+6f^2+15f+6 + 9\\ =&2f^3+11f^2+17f+15 \end{align*} \]
Wait, but the problem says "if the polynomial does not divide evenly, include the remainder as a fraction". But in this case, when we do the division, the remainder is \(9\), so the result is \(2f^2 + 5f+2+\frac{9}{f + 3}\)? Wait, no, wait, maybe I made a mistake in synthetic division. Wait, let's recalculate synthetic division:

Coefficients of dividend: \(2\) (for \(f^3\)), \(11\) (for \(f^2\)), \(17\) (for \(f\)), \(15\) (constant term).

Bring down the \(2\).

Multiply \(2\) by \(-3\) to get \(-6\). Add to \(11\): \(11-6 = 5\).

Multiply \(5\) by \(-3\) to get \(-15\). Add to \(17\): \(17-15 = 2\).

Multiply \(2\) by \(-3\) to get \(-6\). Add to \(15\): \(15-6 = 9\). Oh! I see, I made a mistake earlier in the last multiplication. So the remainder is \(9\), and the quotient is \(2f^2+5f + 2\). Wait, but when we multiply \((f + 3)(2f^2+5f + 2)\) we get \(2f^3+5f^2+2f+6f^2+15f+6=2f^3+11f^2+17f+6\), then add the remainder \(9\) gives \(2f^3+11f^2+17f+15\), which is correct. So the division is \((2f^3 + 11f^2 + 17f + 15)\div(f + 3)=2f^2 + 5f+2+\frac{9}{f + 3}\)? Wait, no, the quotient is \(2f^2+5f + 2\) and the remainder is \(9\), so the result is \(2f^2 + 5f + 2+\frac{9}{f + 3}\). But wait, let's check with another method, like polynomial long division again.

Divide \(2f^3+11f^2+17f+15\) by \(f + 3\):

1. \(2f^3\div f=2f^2\). Multiply \((f + 3)\) by \(2f^2\): \(2f^3+6f^2\). Subtract from dividend: \(2f^3+11f^2+17f+15-(2f^3+6f^2)=5f^2+17f+15\).

2. \(5f^2\div f = 5f\). Multiply \((f + 3)\) by \(5f\): \(5f^2+15f\). Subtract: \(5f^2+17f+15-(5f^2+15f)=2f+15\).

3. \(2f\div f=2\). Multiply \((f + 3)\) by \(2\): \(2f+6\). Subtract: \(2f+15-(2f+6)=9\).

So the quotient is \(2f^2+5f + 2\) and the remainder is \(9\), so the division is \(2f^2 + 5f+2+\frac{9}{f + 3}\). Wait, but the problem says "if the polynomial does not divide evenly, include the remainder as a fraction". So in this case, the division is not even, so we write it as the quotient plus the remainder over the divisor.

But wait, let's check if \(f=-3\) is a root. Plug \(f=-3\) into the dividend: \(2(-3)^3+11(-3)^2+17(-3)+15=2(-27)+11(9)-51 + 15=-54 + 99-51 + 15=(-54-51)+(99 + 15)=-105+114 = 9\), which is the remainder, so that's correct.

Wait, but maybe the problem expects the answer in the form of quotient with remainder as a fraction, so the result is \(2f^2 + 5f + 2+\frac{9}{f + 3}\)? But let's check the multiplication again:

\((f + 3)(2f^2+5f + 2)=2f^3+5f^2+2f+6f^2+15f+6=2f^3+11f^2+17f+6\). Then add the remainder \(9\): \(2f^3+11f^2+17f+6 + 9=2f^3+11f^2+17f+15\), which matches the dividend. So the division is correct.

But wait, maybe I made a mistake in the synthetic division earlier. The synthetic division gives us that the quotient is \(2f^2+5f + 2\) and the remainder is \(9\), so the division result is \(2f^2 + 5f+2+\frac{9}{f + 3}\).

Answer:

\(2f^2 + 5f + 2+\frac{9}{f + 3}\)