Determine a value for the constant b so that we can shorten the expression for the function $f\left(x\right)=\frac{2x^2+bx-30}{x+3}$. Shorten the expression.
Here is the step by step solution we got:
The zero point for the denominator $x+3$ is $x=-3$
The function $f$ is defined when $x\ne-3$
We can shorten the expression only if the denominator and numerator have a common factor. The numerator has the factor $x+3$ only if $-3$ is the zero point in the numerator.
We determine the constant b:
The numerators value is zero for the variable value $-3$.
$2\cdot\left(-3\right)^2+b\cdot\left(-3\right)-30=0$
$b=-4$
Now I don't understand why x is replaced with $-3$ above. What is the reason behind it, why can't it be for example be $-4$?
edit: need an easy explanation, I'm not too good at math.
By long division, there exist polynomials $q(x)$ and $r(x)$, with $\deg r(x)<\deg(x+3)$ such that $$2x^2+bx-30=q(x)(x+3)+r(x).$$ Once we have $q(x)$ and $r(x)$ we can simplify $f(x)$ as follows: $$f(x)=\frac{2x^2+bx-30}{x+3}=\frac{q(x)(x+3)+r(x)}{x+3}=q(x)+\frac{r(x)}{x+3}.$$ Now note that $r(x)$ is a constant, i.e. a number, because $\deg r(x)<\deg(x+3)$, and so $$r=r(-3)=q(-3)(-3+3)+r(-3)=2(-3)^2+b(-3)-30.$$ In particular we see that $f(x)$ reduces to a polynomial if $r=0$, which is equivalent to $b=-4$.