How do you divide $y^{4} - 8 y^{3} + 10 y^{2} + 2 y + 4$ $y^4-8y^3+10y^2+2y+4$ by y-2 and is it a factor of the polynomial?

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Answer 1 · 2018-06-23T20:22:19

Using factor theorem:

$2^{4} - 8 \cdot 2^{3} + 10 \cdot 2^{2} + 2 \cdot 2 + 4$ $2^4-8*2^3+10*2^2+2*2+4$
$= 16 - 64 + 40 + 4 + 4 = 0$ $=16-64+40+4+4 = 0$

Since $f (2)$ $f(2)$ is $0$ $0$ , the linear function $y - 2$ $y-2$ is a factor.

As for the quotient, long division can be employed to give $y^{3} - 6 y^{2} - 2 y - 2$ $y^3-6y^2-2y-2$

How do you divide y^4-8y^3+10y^2+2y+4y4−8y3+10y2+2y+4y^4-8y^3+10y^2+2y+4 by y-2 and is it a factor of the polynomial?