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 ??

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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 ??

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Answer 1 · 2016-10-29T16:56:43

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

Explanation:

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

k

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

$k (x + 1)$ $k(x + 1)$

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

$k (x + 1) (x - 1)$ $k(x + 1)(x - 1)$

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

$k (x + 1) (x - 1) (x - 3)$ $k(x + 1)(x - 1)(x - 3)$

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

$- 6 = k (+ 1) (- 1) (- 3)$ $-6 = k(+1)(-1)(-3)$

$k = - 2$ $k = -2$

Because the divisor is not of the form $(x - a)$ $(x - a)$ but, instead, of the form $(x^{2} - a)$ $(x^2 - a)$ , 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 $4 x - 12$ $4x - 12$

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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 ??

1 Answer

Explanation:

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