# All the real zeros of the given polynomial are integers.

dedica66em 2021-12-19 Answered
All the real zeros of the given polynomial are integers. Find the zeros, and write the polynomial in factored form.
$P\left(x\right)={x}^{3}+12{x}^{2}+48x+64$
You can still ask an expert for help

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it

## Expert Answer

Cheryl King
Answered 2021-12-20 Author has 36 answers
Step 1
Definition used -
Real zeros of a cubic polynomial -
We can factor a given cubic polynomial using the grouping method. Then we will equate all factors to 0 and solve for x. Those values of x are real zeros.
Step 2
Given -
$P\left(x\right)={x}^{3}+12{x}^{2}+48x+64$
${x}^{3}+12{x}^{2}+48x+64=0$
${x}^{3}+4{x}^{2}+8{x}^{2}+32x+16x+64=0$
${x}^{2}\left(x+4\right)+8x\left(x+4\right)+16\left(x+4\right)=0$
$\left(x+4\right)\left({x}^{2}+8x+16\right)=0$
(x+4)(x+4)(x+4)=0
x+4=0, x+4=0, x+4=0
x=-4
Step 3
Polynomial in factored form can be written as-
$P\left(x\right)={\left(x+4\right)}^{3}$

We have step-by-step solutions for your answer!

jgardner33v4
Answered 2021-12-21 Author has 35 answers
$P\left(x\right)={x}^{3}+12{x}^{2}+48x+64$
$=\left(x+4\right)\left({x}^{2}+8x+16\right)$
$=\left(x+4\right){\left(x+4\right)}^{2}$
$={\left(x+4\right)}^{3}$
Real zeroes (x=-4 is root with multiplicity 3)
x=-4, -4, -4
$P\left(x\right)={\left(x+4\right)}^{3}=\left(x+4\right)\left(x+4\right)\left(x+4\right)$

We have step-by-step solutions for your answer!

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it