Thread: Finding real and complex zeros of x^5 + 8x^2

1. Finding real and complex zeros of x^5 + 8x^2

Let $P(x) = x^5 + 8x^2$. Find all the zeros of P, real and complex.

I factor to this:
$0=x^2(x^3+8)$, giving me zeros of 0 and -2. ARE there any complex zeros? The reason I ask is because the next part of the question is "Express P as a product of linear and irreducible quadratic factors with real coefficients"

Where am I going wrong?

2. Originally Posted by cdbmath
Let $P(x) = x^5 + 8x^2$. Find all the zeros of P, real and complex.

I factor to this:
$0=x^2(x^3+8)$, giving me zeros of 0 and -2. ARE there any complex zeros? The reason I ask is because the next part of the question is "Express P as a product of linear and irreducible quadratic factors with real coefficients"

Where am I going wrong?
There are complex zeros.

Note that $P(x)=x^2(x^3+8)={\color{blue}x^2(x+2)}{\color{red} (x^2-2x+4)}$

The part highlighted in blue will give you the real solutions to $P(x)=0$ and the part highlighted in red will give you the complex solutions to $P(x)=0$.

Also note that $P(x)=x^2(x+2)(x^2-2x+4)$ is the answer to the second part of the question.

Does this clarify things?

3. Originally Posted by Chris L T521
There are complex zeros.

Note that $P(x)=x^2(x^3+8)={\color{blue}x^2(x+2)}{\color{red} (x^2-2x+4)}$

The part highlighted in blue will give you the real solutions to $P(x)=0$ and the part highlighted in red will give you the complex solutions to $P(x)=0$.

Also note that $P(x)=x^2(x+2)(x^2-2x+4)$ is the answer to the second part of the question.

Does this clarify things?
Thanks! I used long division to get $(x^4-2x^3+4x^2)$ and then took out $x^2$ to get $(x^2-2x+4)$... I was then able to get my complex zeros.