1
$\begingroup$

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:

enter image description here

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

$\endgroup$
2
  • 1
    $\begingroup$ 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 $10$. $\endgroup$
    – Brian Tung
    Mar 5 at 21:22
  • $\begingroup$ Oh, ok that makes much more sense, thank you so much! $\endgroup$ Mar 5 at 21:24
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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:

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

$\endgroup$
1
$\begingroup$

"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:

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.