Question

1. a polynomial function has single roots at $x = -3$ and $x = 5$, this function also passes through the point $(2, 90).$

Explanation:

Step1: Form polynomial from roots

If a polynomial has roots at $x=r_1$ and $x=r_2$, it can be written as $f(x)=a(x-r_1)(x-r_2)$. Here, $r_1=-3$, $r_2=5$, so:
$f(x)=a(x+3)(x-5)$

Step2: Substitute given point to find $a$

Substitute $x=2$, $f(2)=90$ into the equation:
$90=a(2+3)(2-5)$
Simplify the right-hand side:
$90=a(5)(-3)$
$90=-15a$
Solve for $a$:
$a=\frac{90}{-15}=-6$

Step3: Expand to get full polynomial

Substitute $a=-6$ back into the factored form and expand:
$f(x)=-6(x+3)(x-5)$
First multiply $(x+3)(x-5)=x^2-5x+3x-15=x^2-2x-15$
Then multiply by $-6$:
$f(x)=-6(x^2-2x-15)=-6x^2+12x+90$

Answer:

The polynomial function is $\boldsymbol{f(x)=-6x^2+12x+90}$ (or in factored form $\boldsymbol{f(x)=-6(x+3)(x-5)}$)