Answered

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Which second degree polynomial function has a leading coefficient of –1 and root 4 with multiplicity 2? f(x) = –x2 – 8x – 16 f(x) = –x2 8x – 16 f(x) = –x2 – 8x 16 f(x) = –x2 8x 16

Sagot :

Using the Factor Theorem, the second degree polynomial function that has a leading coefficient of –1 and root 4 with multiplicity 2 is given by:

f(x) = -x² + 8x - 16

What is the Factor Theorem?

The Factor Theorem states that a polynomial function with roots [tex]x_1, x_2, \codts, x_n[/tex] is given by:

[tex]f(x) = a(x - x_1)(x - x_2) \cdots (x - x_n)[/tex]

In which a is the leading coefficient.

In this problem, we have that:

  • The leading coefficient is of a = -1.
  • There is a root of 4 with multiplicity 2, hence [tex]x_1 = x_2 = 4[/tex].

Thus, the polynomial is given by:

f(x) = -(x - 4)(x - 4)

f(x) = -(x² - 8x + 16)

f(x) = -x² + 8x - 16

More can be learned about the Factor Theorem at https://brainly.com/question/24380382

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