Question

PLZ HELP
what is the coefficient of x in the division 18x^3+12x^2-3x/6x^2

Answer 1

\[\frac{18x^3+12x^2-3x}{6x^2}=\frac{18x^3}{6x^2}+\frac{12x^2}{6x^2}-\frac{3x}{6x^2}\] is a start, i.e. divide term by term

Answer 2

You mean simplify?

Answer 3

then what?

Answer 4

@satellite73

Answer 5

hello?

Answer 6

it was 2 if anyone was wondering i got it wrong.

Answer 7

@ShadowFang Sorry you couldn't get help, but thanks for the answer :)

Answer 8

I put 2 and got it wrong, the answer is 3

Answer 9

really?!? that is weird... im really sorry

Answer 10

Divide each of the terms in the numerator with \[6x^{2}\]
This gives:
\[\frac{18x ^{3} }{ 6x^{2} }+\frac{12x^{2} }{ 6x^{2} }-\frac{ 3x }{ 6x^{2} }\]
On simplify;
\[3x+2-\frac{ 3 }{ 2x }\]
The co-efficient of x is 3