How would I express $f(x)=2x^3-5x^2+3$ in the form $f(x)=(x-c)q(x)+r$ for $c = -2$?
I have no idea where to even start. Could anyone help me out?
How would I express $f(x)=2x^3-5x^2+3$ in the form $f(x)=(x-c)q(x)+r$ for $c = -2$?
I have no idea where to even start. Could anyone help me out?
Use polynomial long division to divide $2x^3 - 5x^2 + 3$ by $x - (-2) = x + 2$. That is, you want to set up $x + 2 \overline{)2x^3 - 5x^2 + 0x + 3}$.
If you're not sure how to do this, I would recommend watching the process on a YouTube video, such as the one here.
You do it by polynomial long division, similar to arithmetic long division. There is a detailed example on Wikipedia
Another way is this.
First note that $f$ is of degree 3, and hence $q$ must have degree 2. Thus we get the equation $f(x)=(x-(-2))(Ax^2+Bx+C)+r$ but $f(x)=2x^3-5x^2 +3$, hence multiplying out the right hand-side and identifying coefficients we get $A=2,\quad B+2A = -5,\quad C+2B =0,\quad 2C+r=3 $
which is easily solved.