Question

(x^4-81)/(x^3+3x^2+9x+27) divide

Answer 1

The quotient for the given expression is x-3

Step-by-step explanation:

Given that the fractional expression is $\frac{x^4-81}{x^3+3x^2+9x+27}$

To divide the given expression :

The expression $x^4-81$ can be written as $x^4-81=x^4+0x^3+0x^2+0x-81$ . since for missed terms we can put 0

                                         x-3

                           __________________

$x^3+3x^2+9x+27$  ) $x^4+0x^3+0x^2+0x-81$

                                $x^4+3x^3+9x^2+27x$  

                               (-)__(-)__(-)__(-)_______

                                        $-3x^3-9x^2-27x-81$

                                        $-3x^2-9x^2-27x-81$

                                  ___(+)__(+)___(+)__(+)_______

                                                                       0

                                           _____________________

Therefore the quotient to the given expression is x-3

When we divide the given fractional expression we get the quotient is x-3

Therefore $\frac{x^4-81}{x^3+3x^2+9x+27}=x-3$

Answer 2

Answer:

(x-3) is the answer.☺

hope this help u