Answer:
[tex]\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}[/tex]
So the quotient is [tex]1x^2+5x+-1[/tex] and the remainder is [tex]-6[/tex].
Step-by-step explanation:
We could do this by synthetic division since the denominator is a linear factor in the form [tex]x-c[/tex].
Since we are dividing by [tex]x+2=x-(-2)[/tex], this is our setup for the synthetic division:
-2 | 1 7 9 -8
| -2 -10 2
______________
1 5 -1 -6
So the quotient is [tex]1x^2+5x+-1[/tex] and the remainder is [tex]-6[/tex].
So [tex]\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}[/tex].
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, [tex]x^2+5x-1[/tex]. and the same remainder, [tex]-6[/tex].
Now let's check our result that:
[tex]\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}[/tex].
So I'm going to rewrite the right hand side as a single fraction:
[tex]\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)}{x+2}+\frac{-6}{x+2}[/tex].
[tex]\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}[/tex]
Now let's focus on multiplying [tex](x+2)(x^2+5x-1)[/tex].
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 ( ).
[tex]x(x^2)=x^3[/tex]
[tex]x(5x)=5x^2[/tex]
[tex]x(-1)=-x[/tex]
[tex]2(x^2)=2x^2[/tex]
[tex]2(5x)=10x[/tex]
[tex]2(-1)=-2[/tex]
---------------------------Combine like terms:
[tex]x^3+(5x^2+2x^2)+(-x+10x)+-2[/tex]
[tex]x^3+7x^2+9x-2[/tex]
So let's go back where we were in our check of [tex]\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}[/tex]:
[tex]\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}[/tex]
[tex]\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-2-6}{x+2}[/tex]
[tex]\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-8}{x+2}[/tex]
We have the exact same thing on both sides so we did good.
