Question

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

Answer 1

`" not a factor of the polynomial"`

Explanation:

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

`(color(red)(2v^2)(2v-3)+(color(blue)(6v^2)+6v^2)-8v-12)/(2v-3)`

`=(color(red)(2v^2)(2v-3)color(red)(+6v)(2v-3)+(color(blue)(+18v)-8v)-12)/(2v-3)`

`=(color(red)(2v^2)(2v-3)color(red)(+6v)(2v-3)color(red)(+5)(2v-3)+(color(blue)(+15)-12))/(2v-3)`

`rArr4v^3+6v^2-8v-12=2v^2+6v+5+3/(2v-3)`

Since the remainder of the division is not zero.

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