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Nady [450]
3 years ago
9

Which of the following binomials is a factor of x^3+4x^2+x-6

Mathematics
1 answer:
larisa [96]3 years ago
5 0
The answer could be 3, 2, or 1
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The length of a rectangular pool is 20 feet and the width is 15 feet. The
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Evaluate the function for the given values to determine if the value is a root. p(−2) = p(2) = The value is a root of p(x).
bija089 [108]

<em>Note: Since you missed to mention the the expression of the function </em>p(x)<em> . After a little research, I was able to find the complete question. So, I am assuming the expression as </em>p(x)=x^4-9x^2-4x+12<em> and will solve the question based on this assumption expression of  </em>p(x)<em>, which anyways would solve your query.</em>

Answer:

As

p\left(-2\right)=0

Therefore, x=-2 is a root of the polynomial <em> </em>p(x)=x^4-9x^2-4x+12

As

p\left(2\right)=-16

Therefore, x=2 is not a root of the polynomial <em> </em>p(x)=x^4-9x^2-4x+12

Step-by-step explanation:

As we know that for any polynomial let say<em> </em>p(x)<em>, </em>c is the root of the polynomial if p(c)=0.

In order to find which of the given values will be a root of the polynomial, p(x)=x^4-9x^2-4x+12<em>, </em>we must have to evaluate <em> </em>p(x)<em> </em>for each of these values to determine if the output of the function gets zero.

So,

Solving for p\left(-2\right)

<em> </em>p(x)=x^4-9x^2-4x+12

p\left(-2\right)=\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12

\mathrm{Simplify\:}\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12:\quad 0

\left(-2\right)^4-9\left(-2\right)^2-4\left(-2\right)+12

\mathrm{Apply\:rule}\:-\left(-a\right)=a

=\left(-2\right)^4-9\left(-2\right)^2+4\cdot \:2+12

\mathrm{Apply\:exponent\:rule}:\quad \left(-a\right)^n=a^n,\:\mathrm{if\:}n\mathrm{\:is\:even}

=2^4-2^2\cdot \:9+8+12

=2^4+20-2^2\cdot \:9

=16+20-36

=0

Thus,

p\left(-2\right)=0

Therefore, x=-2 is a root of the polynomial <em> </em>p(x)=x^4-9x^2-4x+12<em>.</em>

Now, solving for p\left(2\right)

<em> </em>p(x)=x^4-9x^2-4x+12

p\left(2\right)=\left(2\right)^4-9\left(2\right)^2-4\left(2\right)+12

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

p\left(2\right)=2^4-9\cdot \:2^2-4\cdot \:2+12

p\left(2\right)=2^4-2^2\cdot \:9-8+12

p\left(2\right)=2^4+4-2^2\cdot \:9

p\left(2\right)=16+4-36

p\left(2\right)=-16

Thus,

p\left(2\right)=-16

Therefore, x=2 is not a root of the polynomial <em> </em>p(x)=x^4-9x^2-4x+12<em>.</em>

Keywords: polynomial, root

Learn more about polynomial and root from brainly.com/question/8777476

#learnwithBrainly

7 0
2 years ago
Read 2 more answers
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