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sweet-ann [11.9K]
3 years ago
6

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 th

e zeros of g(x)? (show your work!)
Enter your answers in the blanks:


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


Zeros: ______ and ______
Mathematics
1 answer:
mariarad [96]3 years ago
8 0

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.

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