Question

What is the result when x^3+ 7x^2+ 9x - 8 is divided by x+ 2?

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$

So the quotient is $1x^2+5x+-1$ and the remainder is $-6$.

Step-by-step explanation:

We could do this by synthetic division since the denominator is a linear factor in the form $x-c$.

Since we are dividing by $x+2=x-(-2)$, this is our setup for the synthetic division:

-2 |     1    7    9    -8

   |         -2   -10     2

   ______________

          1   5    -1       -6

So the quotient is $1x^2+5x+-1$ and the remainder is $-6$.

So $\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$.

We can also do long division.

                 x^2+5x-1

               ____________________

        x+2| x^3+7x^2+9x-8

              -(x^3+2x^2)

              -------------------

                       5x^2+9x-8

                    -( 5x^2+10x)

                      --------------------

                                 -x-8

                               -(-x-2)

                                --------------

                                      -6

So we see here we get the same quotient, $x^2+5x-1$. and the same remainder, $-6$.

Now let's check our result that:

$\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$.

So I'm going to rewrite the right hand side as a single fraction:

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)}{x+2}+\frac{-6}{x+2}$.

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}$

Now let's focus on multiplying $(x+2)(x^2+5x-1)$.

We are going to multiply the first term of the first ( ) to every term in the second ( ).

We are also going to multiply the second term of the first ( ) to every term in the second ( ).

$x(x^2)=x^3$

$x(5x)=5x^2$

$x(-1)=-x$

$2(x^2)=2x^2$

$2(5x)=10x$

$2(-1)=-2$

---------------------------Combine like terms:

$x^3+(5x^2+2x^2)+(-x+10x)+-2$

$x^3+7x^2+9x-2$

So let's go back where we were in our check of $\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$:

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}$

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-2-6}{x+2}$

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-8}{x+2}$

We have the exact same thing on both sides so we did good.