How do you divide $4 v^{3} + 6 v^{2} - 8 v - 12$ $4v^3+6v^2-8v-12$ by 2v-3 and is it a factor of the polynomial?

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How do you divide $4 v^{3} + 6 v^{2} - 8 v - 12$ $4v^3+6v^2-8v-12$ by 2v-3 and is it a factor of the polynomial?

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Answer 1 · 2017-02-11T11:03:32

$not a factor of the polynomial$ $" not a factor of the polynomial"$

Explanation:

Using some $algebraic manipulation$ $color(blue)"algebraic manipulation"$ by substituting the divisor into the numerator as a factor.

$\frac{2 v^{2} (2 v - 3) + (6 v^{2} + 6 v^{2}) - 8 v - 12}{2 v - 3}$ $(color(red)(2v^2)(2v-3)+(color(blue)(6v^2)+6v^2)-8v-12)/(2v-3)$

$= \frac{2 v^{2} (2 v - 3) + 6 v (2 v - 3) + (+ 18 v - 8 v) - 12}{2 v - 3}$ $=(color(red)(2v^2)(2v-3)color(red)(+6v)(2v-3)+(color(blue)(+18v)-8v)-12)/(2v-3)$

$= \frac{2 v^{2} (2 v - 3) + 6 v (2 v - 3) + 5 (2 v - 3) + (+ 15 - 12)}{2 v - 3}$ $=(color(red)(2v^2)(2v-3)color(red)(+6v)(2v-3)color(red)(+5)(2v-3)+(color(blue)(+15)-12))/(2v-3)$

$\Rightarrow 4 v^{3} + 6 v^{2} - 8 v - 12 = 2 v^{2} + 6 v + 5 + \frac{3}{2 v - 3}$ $rArr4v^3+6v^2-8v-12=2v^2+6v+5+3/(2v-3)$

Since the remainder of the division is not zero.

$Then 2 v - 3 is not a factor of the polynomial$ $"Then "2v-3" is not a factor of the polynomial"$

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How do you divide $4 v^{3} + 6 v^{2} - 8 v - 12$ $4v^3+6v^2-8v-12$ by 2v-3 and is it a factor of the polynomial?

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Explanation:

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How do you divide $4 v^{3} + 6 v^{2} - 8 v - 12$ $4v^3+6v^2-8v-12$ by 2v-3 and is it a factor of the polynomial?