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Which second degree polynomial function has a leading coefficient of –1 and root 4 with multiplicity 2?

2 Answers

0 votes

The correct answer is D, or
f(x) = -x^2 + 8x - 16

Just got it right on Edge 2020, hope this helps!! :)

User Peduarte
by
8.2k points
1 vote

Answer:

-
(x-4)^(2)

or -
x^(2)+8x-16

Explanation:

Given a second degree polynomial has a root 4 with multiplicity 2

That means, 4 is a repeating root of the polynomial.

Any second degree polynomial has at most 2 real roots.

⇒Both roots of the polynomial are 4.

and its expression can be written as
(x-4)^(2) where c is a real number

⇒c·(
x^(2)-8x+16)

⇒c
x^(2)-8cx+16c

Also, the leading coefficient is given as -1

So, c = -1

and the expression becomes (-1)
x^(2)-8·(-1)·x+16·(-1)

-
x^(2)+8x-16

User Mojgan
by
7.7k points

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