pogonofor9z
2021-12-16
Answered

Use the remainder theorem to determine if the given number c is a zero of the polynomial.

${x}^{4}+9{x}^{3}+22{x}^{2}+19x+45;c=-3$

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Piosellisf

Answered 2021-12-17
Author has **40** answers

Step 1:Concept

According to remainder theorem, if we plug the given number -3 for x, and -f -3 is a zero, then the result should be 0.

Step 2:Solution

Plugging -3 for x

${(-3)}^{4}+9{(-3)}^{3}+22{(-3)}^{2}+19(-3)+45$

=81-243+198-57+45

$=24\ne 0$

So c=-3 is not a zero.

According to remainder theorem, if we plug the given number -3 for x, and -f -3 is a zero, then the result should be 0.

Step 2:Solution

Plugging -3 for x

=81-243+198-57+45

So c=-3 is not a zero.

Annie Gonzalez

Answered 2021-12-18
Author has **41** answers

By remainder theorem-

Replace x=-3,

So,

=81-243+198-57+45

=324-300

=24

By replace x=-3 in equation, we cannot get 0,

So, c=-3 is not a zero of the polynomial.

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