How do you use the remainder theorem to see if the $n + 2$ $n+2$ is a factor of $n^{4} + 10 n^{3} + 21 n^{2} + 6 n - 8$ $n^4+10n^3+21n^2+6n-8$ ?

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Answer 1 · 2017-01-14T20:18:35

$n + 2$ $n+2$ is a factor of the given polynomial

The remainder theorem states that the bynomial $(x - a)$ $(x-a)$ is a factor of a polynomial P(x) if P(a)=0.

Let it be

$P (n) = n^{4} + 10 n^{3} + 21 n^{2} + 6 n - 8$ $P(n)=n^4+10n^3+21n^2+6n-8$ ,

then

$P (- 2) = {(- 2)}^{4} + 10 \cdot {(- 2)}^{3} + 21 \cdot {(- 2)}^{2} + 6 \cdot (- 2) - 8$ $P(-2)=(-2)^4+10*(-2)^3+21*(-2)^2+6*(-2)-8$

$= 16 - 80 + 84 - 12 - 8 = 0$ $=16-80+84-12-8=0$

Then $n + 2$ $n+2$ is a factor of the given polynomial

How do you use the remainder theorem to see if the n+2n+2 is a factor of n4+10n3+21n2+6n−8n^4+10n^3+21n^2+6n-8?