# What are the Zero's of this function?

• 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.

\$\displaystyle 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.

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

Now use the difference of squares again!

\$\displaystyle (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
\$\displaystyle (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 \$\displaystyle x^4- 16= 0\$ is to write it as \$\displaystyle x^4= 16\$. Taking the square root of each side we have \$\displaystyle x^2= 4\$ and \$\displaystyle x^2= -4\$. From \$\displaystyle x^2= 4\$, taking the square root again, x= 2 and x= -2. There are no real number x such that \$\displaystyle x^2= -4\$.