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

##### 1 Answer

Feb 11, 2017

#### 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"#