Urgent! Help! A cubic polynomial has zeros at x=-1, x=1, and x=3. It has a y-intercept of -6. What is the remainder when we divide this polynomial by x^2+1 ??

Oct 29, 2016

I used WolframAlpha to do the division the remainder is $4 x - 12$

Explanation:

Start is a factor, k that allows one to adjust the y intercept:

k

Multiply that by the factor corresponding to the zero, $x = - 1$ -- that is $\left(x + 1\right)$

$k \left(x + 1\right)$

Multiply by the factor corresponding to the zero, $x = - 1$ -- that is $\left(x - 1\right)$

$k \left(x + 1\right) \left(x - 1\right)$

The last factor is the one corresponding to the zero, $x = 3$ -- $\left(x - 3\right)$

$k \left(x + 1\right) \left(x - 1\right) \left(x - 3\right)$

To find the value of k, we set the factors equal to -6 and x within the factors equal to 0

$- 6 = k \left(+ 1\right) \left(- 1\right) \left(- 3\right)$

$k = - 2$

Because the divisor is not of the form $\left(x - a\right)$ but, instead, of the form $\left({x}^{2} - a\right)$, one cannot use the remainder theorem. Therefore, the only way to find the remainder is by using division.

Here is a link to WolframAlpha for the division.

Please notice that the remainder is $4 x - 12$