Question

1. a polynomial function has single roots at $x=-4$ and $x=6$, and a triple root at $x=1$. this function also passes through the point $(-2, 192)$.

Explanation:

Step1: Write general polynomial form

If a polynomial has a root $x=r$, then $(x-r)$ is a factor. For a triple root at $x=1$, the factor is $(x-1)^3$. So the general form is:
$P(x) = a(x+4)(x-6)(x-1)^3$

Step2: Substitute given point to find $a$

Substitute $x=-2$, $P(-2)=192$ into the equation:
$192 = a(-2+4)(-2-6)(-2-1)^3$
Simplify the right-hand side:
$192 = a(2)(-8)(-27)$
$192 = a(432)$
Solve for $a$:
$a = \frac{192}{432} = \frac{2}{3}$

Step3: Write final polynomial

Substitute $a=\frac{2}{3}$ back into the general form:
$P(x) = \frac{2}{3}(x+4)(x-6)(x-1)^3$

Answer:

$P(x) = \frac{2}{3}(x+4)(x-6)(x-1)^3$
(Expanded form optional: $P(x)=\frac{2}{3}x^5 - \frac{8}{3}x^4 - \frac{34}{3}x^3 + \frac{116}{3}x^2 - 20x - 48$)