Question

(27x^4-18x^3+9x^2) divided by 3x

Answer 1

Answer:

$9x^3-6x^2+3x$

Step-by-step explanation:

We have the following:

$\frac{27x^4-18x^3+9x^2}{3x}$

which can be separeted like this:

$\frac{27x^4}{3x} -\frac{18x^3}{3x} +\frac{9x^2}{3x}$

and using the rule:

$\frac{x^m}{x^n}=x^{m-n}$

in other words we substract the exponents at the moment we make the division, and also dividing the coefficients  ($\frac{27}{3} =9,\frac{18}{3} =6, \frac{9}{3} =3$) we get the following:

$9x^{4-1}-6x^{3-1}+3x^{2-1}\\=9x^3-6x^2+3x$

the answer is: $9x^3-6x^2+3x$

Answer 2

9x^3-6x^2+3x

Basically with this you just wanna simplify each term one-by-one.

For instance, look at the first term, 27x^4

Imagine 27x^4/3x

3 goes into 27 9 times, and x goes into x^4 4 times (cuz x to the power of 4 is just x times x times x times x)

So you’re left with 9x^3