Question

(2X^2+11X+8) DIVIDED BY (X+4)

Answer 1

\[(x+4)\times\text{something} = x^2+11x+8\]
The something has to have an \(x\) in it to get the \(x^2\)

What you can try doing is multiplying out
\((x+4)(Ax+B)+C\)
and then compare coefficients with \(x^2+11x+8\)
For example the coefficient of \(x^2\) is 1, therefore A = 1

Answer 2

how do I find quotient and remainder?

Answer 3

The \(Ax+B\) is the quotient, the \(C\) is the remainder

Answer 4

im confused

Answer 5

I really only need the answer. this is the final question on my test and I don't know how to do it at all

Answer 6

Ok, well, the way I like to think of these problems (when you're dividing polynomials) is not to think of it as a division, but to think:
"What do I need to multiply the thing on the bottom to get the thing on the top?"
The thing on the bottom has \(x^1 = x\) as the highest power of \(x\), but the thing on the top has \(x^2\) as the highest power of \(x\).
This means that the answer needs and \(x\) and the most general solution will be
\[Ax+B+\frac{C}{(x+4)}\]
meaning
\[(x+4)(Ax+B)+C = x^2+11x+8\]
All you need to do is multiply out the left hand side and compare it to the right hand side.
So first multiply this out:
\[(x+4)(Ax+B)+C = Ax^2+(4A+B)x+4B+C\]
Now let's compare coefficients
The coefficient of \(x^2\) is \(A = 1\)
The coefficient of \(x\) is \(4A+B = 11\)
The coefficient of \(1\) is \(4B+C = 8\)

Now you can solve for \(A\), \(B\) and \(C\)