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

Ernest Ryland 2021-12-15 Answered
All the real zeros of the given polynomial are integers. Find the zeros, and write the polynomial in factored form.
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
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jean2098
Step 1
Given: All the real zeros of the given polynomial are integers.
Find the zeros, and write the polynomial in factored form.
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
Step 2
To Find the zeros, and write the polynomial in factored form:
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
$=\left(x-1\right){\left(x+2\right)}^{2}$
=(x-1)(x+2)(x+2)
Therefore, The zeros of P(x) is P(x)=0
Hence, (x-1)(x+2)(x+2)=0
$⇒x=1,-2,-2$

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Lynne Trussell
Consider the polynomial function in the textbook.
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
The objective of the questions is to find the zeros and write the polynomial in factored form.
Consider $P\left(x\right)={x}^{3}+3{x}^{2}-4$,
The leading coefficient is 1 and the constant term is -4, any rational zero must be a divisior of the constant term -4.
So, the possible rational zeros are $±1,±2$ and $±4$
Test each of these possibilities.
Let x=1
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
$P\left(1\right)={\left(1\right)}^{3}+3{\left(1\right)}^{2}-4$
=1+3-4
=0
Now test for x=-1
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
$P\left(-1\right)={\left(-1\right)}^{3}+3\left(-1\right)-4$
=-1+3-4
=-2
Similarly, for x=2
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
$P\left(2\right)={\left(2\right)}^{3}+3{\left(2\right)}^{2}-4$
=8+12-4
=16
Similarly, for x=-2
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
$P\left(-2\right)={\left(-2\right)}^{3}+3{\left(-2\right)}^{2}-4$
=-8+12-4
=0
Similarly, for x=4
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
$P\left(4\right)={4}^{3}+3{\left(4\right)}^{2}-4$
=64+48-4
=108
Similarly, for x=-4
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
$P\left(-4\right)={\left(-4\right)}^{3}+3{\left(-4\right)}^{2}-4$
=-64-48-4
=-116
Hence, real zeros of $P\left(x\right)={x}^{3}+3{x}^{2}-4$ are 1 and -2.
Now find the factored form of $P\left(x\right)={x}^{3}+3{x}^{2}-4$,
$P\left(x\right)={x}^{3}+3{x}^{2}-4$
$P\left(x\right)={\left(x+2\right)}^{2}\left(x-1\right)$
Hence, the factored form of $P\left(x\right)={x}^{3}+3{x}^{2}-4$ is ${\left(x+2\right)}^{2}\left(x-1\right)$.

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