How do you write a polynomial with Zeros: -2, multiplicity 2; 4, multiplicity 1; degree 3?

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How do you write a polynomial with Zeros: -2, multiplicity 2; 4, multiplicity 1; degree 3?

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Answer 1 · 2016-02-29T23:39:41

$p (x) = x^{3} - 12 x - 16$ $p(x)=x^3-12x-16$

Explanation:

For a polynomial, if $x = a$ $x=a$ is a zero of the function, then $(x - a)$ $(x-a)$ is a factor of the function.

We have two unique zeros: $- 2$ $-2$ and $4$ $4$ . However, $- 2$ $-2$ has a multiplicity of $2$ $2$ , which means that the factor that correlates to a zero of $- 2$ $-2$ is represented in the polynomial twice.

Follow the colors to see how the polynomial is constructed:

$zero at - 2, multiplicity 2$ $"zero at "color(red)(-2)", multiplicity "color(blue)2$
$zero at 4, multiplicity 1$ $"zero at "color(green)4", multiplicity "color(purple)1$

$p (x) = {(x - (- 2))}^{2} {(x - 4)}^{1}$ $p(x)=(x-(color(red)(-2)))^color(blue)2(x-color(green)4)^color(purple)1$

Thus,

$p (x) = {(x + 2)}^{2} (x - 4)$ $p(x)=(x+2)^2(x-4)$

Expand:

$p (x) = (x^{2} + 4 x + 4) (x - 4)$ $p(x)=(x^2+4x+4)(x-4)$

$p (x) = x^{3} - 12 x - 16$ $p(x)=x^3-12x-16$

We can graph the function to understand multiplicities and zeros visually:

graph{x^3-12x-16 [-6, 6, -43.83, 14.7]}

The zero at $x = - 2$ $x=-2$ "bounces off" the $x$ $x$ -axis. This behavior occurs when a zero's multiplicity is even.

The zero at $x = 4$ $x=4$ continues through the $x$ $x$ -axis, as is the case with odd multiplicities.

Note that the function does have three zeros, which it is guaranteed by the Fundamental Theorem of Algebra, but one of such zeros is represented twice.

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How do you write a polynomial with Zeros: -2, multiplicity 2; 4, multiplicity 1; degree 3?

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