# Use the remainder theorem to determine if the given number

Use the remainder theorem to determine if the given number c is a zero of the polynomial.
${x}^{4}+9{x}^{3}+22{x}^{2}+19x+45;c=-3$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Piosellisf
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
${\left(-3\right)}^{4}+9{\left(-3\right)}^{3}+22{\left(-3\right)}^{2}+19\left(-3\right)+45$
=81-243+198-57+45
$=24\ne 0$
So c=-3 is not a zero.
###### Did you like this example?
Annie Gonzalez
${x}^{4}+9{x}^{3}+22{x}^{2}+19x+45,c=-3$
By remainder theorem-
Replace x=-3,
So,
${\left(-3\right)}^{4}+9{\left(-3\right)}^{3}+22{\left(-3\right)}^{2}+19\left(-3\right)+45$
=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.