Having trouble w/ a few Factoring Polynomials and Radical Expression problems

**Connor** · January 12th 2011, 02:19 PM

I'm having a tough time understanding these:

X^6 - 1

And if you can, please give me a little explanation on how you got the answer.

**dwsmith** · January 12th 2011, 02:21 PM

$48=3*16$

$75=3*25$

$\displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq rt{3}}{5\sqrt{3}}$

Does this help for the first one?

**dwsmith** · January 12th 2011, 02:22 PM

$x^6-1=0\Rightarrow x^6=1 \ \ \ x=\pm 1$

Do you want all the solutions or just the two real solutions?

**dwsmith** · January 12th 2011, 02:24 PM

$(4\sqrt{2}-3\sqrt{5})2\sqrt{25}=4\sqrt{2}*2\sqrt{25}-3\sqrt{5}*2\sqrt{25}=\cdots$

**pickslides** · January 12th 2011, 02:27 PM

Originally Posted by Connor

I'm having a tough time understanding these:

X^6 - 1

Do you need to factor this guy?

$\displaystyle x^6-1 = (x^3)^2-1^2$

Now apply the difference of two squares. $\displaystyle a^2-b^2 = (a-b)(a+b)$

**Connor** · January 12th 2011, 02:30 PM

Originally Posted by dwsmith

$48=3*16$

$75=3*25$

$\displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq rt{3}}{5\sqrt{3}}$

Does this help for the first one?

No not really. You can't have a radical as a denominator and I have to simplify it more.

Originally Posted by dwsmith

$x^6-1=0\Rightarrow x^6=1 \ \ \ x=\pm 1$

Do you want all the solutions or just the two real solutions?

X^6 - 1 isn't equal to anything in this problem. I'm supposed to factor a polynomial here.

**Connor** · January 12th 2011, 02:34 PM

Originally Posted by pickslides

Do you need to factor this guy?

$\displaystyle x^6-1 = (x^3)^2-1^2$

Now apply the difference of two squares. $\displaystyle a^2-b^2 = (a-b)(a+b)$

Are you sure? I don't have to use this formula?:
$(a \pm b)(a^2 \pm ab + b^2)$

**dwsmith** · January 12th 2011, 02:34 PM

Originally Posted by Connor

No not really. You can't have a radical as a denominator and I have to simplify it more.

You need to factor.

$\displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq rt{3}}{5\sqrt{3}}\Rightarrow \left(\frac{\sqrt{3}}{\sqrt{3}}\right)\left(\frac{ 5\sqrt{16}+\sqrt{25}}{5}\right)$

Then you can further simplify.

Refer to pickslides post then.

**dwsmith** · January 12th 2011, 02:35 PM

Originally Posted by Connor

Are you sure? I don't have to use this formula?:
$(a \pm b)(a^2 \pm ab + b^2)$

That is for cubics.

**e^(i*pi)** · January 12th 2011, 02:36 PM

Once you've used the difference of two squares (as Pickslides suggests) then you can use the sum/difference of two cubes

$(a \pm b)^3 = (a \pm b)(a^2 \mp ab + b^2)$

Since x^6 and 1 are both square numbers and cube numbers you can start with either formula

**Connor** · January 12th 2011, 02:41 PM

Originally Posted by dwsmith

You need to factor.

$\displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq rt{3}}{5\sqrt{3}}\Rightarrow \left(\frac{\sqrt{3}}{\sqrt{3}}\right)\left(\frac{ 5\sqrt{16}+\sqrt{25}}{5}\right)$

Then you can further simplify.

Refer to pickslides post then.

Three of my friends + my mom got an answer of 5 for this problem. I still don't understand how they got that...

**dwsmith** · January 12th 2011, 02:44 PM

Originally Posted by Connor

Three of my friends + my mom got an answer of 5 for this problem. I still don't understand how they got that...

$\displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq rt{3}}{5\sqrt{3}}\Rightarrow \left(\frac{\sqrt{3}}{\sqrt{3}}\right)\left(\frac{ 5\sqrt{16}+\sqrt{25}}{5}\right)\Rightarrow\frac{5\ sqrt{16}+5}{5}\Rightarrow\left(\frac{5}{5}\right)( \sqrt{16}+1)=4+1=5$

It should probably really be $\pm 5$ but you may only need the positive solution.

**Connor** · January 12th 2011, 02:46 PM

Originally Posted by dwsmith

$\displaystyle\frac{5\sqrt{16}\sqrt{3}+\sqrt{25}\sq rt{3}}{5\sqrt{3}}\Rightarrow \left(\frac{\sqrt{3}}{\sqrt{3}}\right)\left(\frac{ 5\sqrt{16}+\sqrt{25}}{5}\right)\Rightarrow\frac{5\ sqrt{16}+5}{5}\Rightarrow\left(\frac{5}{5}\right)( \sqrt{16}+1)=4+1=5$

Well I certainly feel retarded for not being able to figure that out.

**dwsmith** · January 12th 2011, 02:48 PM

Do the last problem in the same method. I have already completed one step for you, see post 4.

**Connor** · January 12th 2011, 02:55 PM

Originally Posted by dwsmith

Do the last problem in the same method. I have already completed one step for you, see post 4.

Ok. And for the first problem. Is this the answer?
$(x + 1)(x - 1)(x^2 - x + 1)(x^2 + x + 1)$

And for the last one is this the answer?
$40\sqrt{2} - 30\sqrt{5}$

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