Ernest Ryland

2021-12-15

All the real zeros of the given polynomial are integers. Find the zeros, and write the polynomial in factored form.

$P\left(x\right)={x}^{3}+3{x}^{2}-4$

jean2098

Beginner2021-12-16Added 38 answers

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:

$P\left(x\right)={x}^{3}+3{x}^{2}-4$

$=(x-1){(x+2)}^{2}$

=(x-1)(x+2)(x+2)

Therefore, The zeros of P(x) is P(x)=0

Hence, (x-1)(x+2)(x+2)=0

$\Rightarrow x=1,-2,-2$

Given: All the real zeros of the given polynomial are integers.

Find the zeros, and write the polynomial in factored form.

Step 2

To Find the zeros, and write the polynomial in factored form:

=(x-1)(x+2)(x+2)

Therefore, The zeros of P(x) is P(x)=0

Hence, (x-1)(x+2)(x+2)=0

Lynne Trussell

Beginner2021-12-17Added 32 answers

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$\pm 1,\pm 2$ and $\pm 4$

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

$P\left(x\right)={x}^{3}+3{x}^{2}-4$

$P(-1)={(-1)}^{3}+3(-1)-4$

=-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

$P\left(x\right)={x}^{3}+3{x}^{2}-4$

$P(-2)={(-2)}^{3}+3{(-2)}^{2}-4$

=-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

$P\left(x\right)={x}^{3}+3{x}^{2}-4$

$P(-4)={(-4)}^{3}+3{(-4)}^{2}-4$

=-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$P\left(x\right)={x}^{3}+3{x}^{2}-4$ ,

$P\left(x\right)={x}^{3}+3{x}^{2}-4$

$P\left(x\right)={(x+2)}^{2}(x-1)$

Hence, the factored form of$P\left(x\right)={x}^{3}+3{x}^{2}-4$ is ${(x+2)}^{2}(x-1)$ .

The objective of the questions is to find the zeros and write the polynomial in factored form.

Consider

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

Test each of these possibilities.

Let x=1

=1+3-4

=0

Now test for x=-1

=-1+3-4

=-2

Similarly, for x=2

=8+12-4

=16

Similarly, for x=-2

=-8+12-4

=0

Similarly, for x=4

=64+48-4

=108

Similarly, for x=-4

=-64-48-4

=-116

Hence, real zeros of

Now find the factored form of

Hence, the factored form of

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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