# What are the Zero's of this function?

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• Jun 1st 2011, 12:07 PM
bkbowser
What are the Zero's of this function?
the problem is {x}^{4} - 16

I get;
{x}^{4} - {2}^{4} = {(x+2)}^{2} + {(x-2)}^{2}

for zeros of 2 and -2 each of multiplicity of 2.

However the book states zeros of 2 and -2 each of multiplicity 1. Could someone be so kind as to explain to me what I have done wrong?
• Jun 1st 2011, 12:25 PM
TheEmptySet
Quote:

Originally Posted by bkbowser
the problem is {x}^{4} - 16

I get;
{x}^{4} - {2}^{4} = {(x+2)}^{2} + {(x-2)}^{2}

for zeros of 2 and -2 each of multiplicity of 2.

However the book states zeros of 2 and -2 each of multiplicity 1. Could someone be so kind as to explain to me what I have done wrong?

You can applied the difference of square incorrectly.

$x^4-16=(x^2)^2-4^2=(x^2-4)(x^2+4)$

Now use the difference of squares again!
• Jun 1st 2011, 12:54 PM
bkbowser
Quote:

Originally Posted by TheEmptySet
You can applied the difference of square incorrectly.

$x^4-16=(x^2)^2-4^2=(x^2-4)(x^2+4)$

Now use the difference of squares again!

$(x^2+4)(x-2)(x+2)$ Is this it? It does give zeros of 2 and -2 each of multiplicity 1. And since the question just asked about real zeros the other 2 wouldn't count?
• Jun 1st 2011, 01:19 PM
TheEmptySet
Quote:

Originally Posted by bkbowser
$(x^2+4)(x-2)(x+2)$ Is this it? It does give zeros of 2 and -2 each of multiplicity 1. And since the question just asked about real zeros the other 2 wouldn't count?

yes
• Jun 4th 2011, 08:36 AM
HallsofIvy
Another way to solve $x^4- 16= 0$ is to write it as $x^4= 16$. Taking the square root of each side we have $x^2= 4$ and $x^2= -4$. From $x^2= 4$, taking the square root again, x= 2 and x= -2. There are no real number x such that $x^2= -4$.