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.
I'm guessing that "shorten" means that some simplification/cancelling occurs in the rational function.
This occurs when there is a common factor of the numerator polynomial and the denominator polynomial.
The only factor of the denominator polynomial is evidently $$x+3.$$ So we require that $x+3$ is a factor of $2x^2+bx-30$, so that it will simplify.
By the polynomial remainder theorem, $x+3$ is a factor of a polynomial $P(x)$ iff $P(-3)=0$, hence $x+3$ will be a factor of $2x^2+bx-30$ iff $$2(-3)^2+b(-3)-30=0$$ $$18-3b-30=0$$ $$3b=-12$$ $$b=-4$$
Hence, $f(x)$ "shortens" when $b=-4$. We can use polynomial division or factoring to show that $2x^2-4x-30=(2x-10)(x+3)$, hence $$f(x)=\frac{(2x-10)(x+3)}{x+3}$$ $$f(x)=2x-10~\text{ }\forall x\neq -3.$$