Question

Fill in the missing portions of the function to rewrite g(x) = 3a^2 − 42a + 135 to reveal the zeros of the function. What are the zeros of g(x)? (show your work!)
Enter your answers in the blanks:


g(x) = 3(_____)(_____)


Zeros: ______ and ______

Answer 1

Answer:

g(x) = 3(x-9)(x-5)

Zeros: x = 9 and x = 5.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$g(x) = 3x^{2} - 42a + 135$

So

$a = 3, b = -42, c = 135$

$\bigtriangleup = (-42)^{2} - 4*3*135 = 144$

$x_{1} = \frac{-(-42) + \sqrt{144}}{2*3} = 9$

$x_{2} = \frac{-(42) - \sqrt{144}}{2*3} = 5$

So

g(x) = 3(x-9)(x-5)

Zeros: x = 9 and x = 5.