# Finding a polynomial $f(x)$ that when divided by $x+3$ yields quotient $2x^2-x+7$ and remainder $10$

I'm struggling to grasp this particular question:

When a polynomial $$f(x)$$ is divided by $$x+3$$, the quotient is $$2x^2-x+7$$ and the remainder is $$10$$. What is $$f(x)$$?

This is what I did:

\begin{align} f(x) &= (x+3)(2x^2-x+7) \\ &= 2x^3-x^2+7x+6x^2-3x+21 \\ &= 2x^3+5x^2+4x+21 \end{align}

Making sure it is correct:

Then, I realized that the remainder has to be 10 not 0...I have no idea how to do that

• If you were doing ordinary division with numbers, you could divide $437$ by $23$, say, and get $19$ evenly. Then if you wanted a number where the quotient was also $19$, but with a remainder of $10$, you would just add $10$ to the original number: $447$ divided by $23$ is $19$, with a remainder of $10$. Mar 5 at 21:22
• Oh, ok that makes much more sense, thank you so much! Mar 5 at 21:24

Just add $$10$$ to your $$f(x)$$...it becomes the remainder when you do the division.

In general, when you divide a function $$p(x)$$ by $$q(x)$$ and you get $$f(x)$$ as the quotient with a remainder of $$r(x)$$, you can alyways represent the answer as: $$\frac{r(x)}{q(x)} + f(x)$$

So you can rewrite your question algebraically as $$\frac{10}{x+3} + 2x^2-x+7$$ and then can simplify as follows:

$$\frac{10}{x+3} + 2x^2-x+7 = \frac{10 + (x+3)(2x^2-x+7)}{x+3} = \frac{10 + 2x^3 -x^2+7x+6x^2-3x+21}{x+3} = \frac{2x^3 +5x^2 + 4x + 31}{x+3}$$

"Then, I realized that the remainder has to be 10 not 0...I have no idea how to do that"

Don't you?

If $$f(x) = (x+3)(2x^2 -x + 7)$$ then

$$f(x) + r = (x+3)(2x^2-x +7) + r$$

And $$f(x) + 10 = (x+3)(2x^2-x +7) +10$$.

And doesn't that mean the remainder of $$f(x)\div (x+3) = 10$$?

After all if we divide $$(x+3)(2x^2 -x +7) + 10$$ we'd get $$\frac {(x+3)(2x^2 - x+ 7)}{x+3} = 2x^2 -x + 7$$ and then .... we'd be left with the remainder of $$10$$.

So if you are told $$f(x)$$ divided by $$d(x)$$ has a quotient of $$q(x)$$ and a remainder of $$r$$ then you can just do $$f(x) = d(x)q(x) + r$$.

And so $$g(x) = (x+3)(2x^2 - x+7) + 10=$$

$$(2x^3+5x^2+4x+21) + 10=$$

$$2x^3 + 5x^2 +4 + 31$$

has to do it and if we did that we'd have: