# Thread: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

1. ## Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

So I am working my way through an algebra practice book, and I am stumped on this question:

4x^4 + (3x^2)(y^2) + y^4

Now the answer is listed as (2x^2 + xy + y^2)(2x^2 - xy + y^2)

What steps does one take to obtain this answer?

Thanks

2. ## Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

We have two fourth powers and one sqare beneath monomyals Let us try enshorted multiplyig formulas
$(2x^{2}+x^{2})^{2}=4x^{4}+4x^{2}y^{2}+y^{4}=4x^{4} +3x^{2}y^{2}+x^{2}y^{2}$
if we odd bothsidely x^{2}y^{2} we have
$(2x^{2}+x^{2})^{2}-x^{2}y^{2}$=our formula

3. ## Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

Originally Posted by Cartesius24
We have two fourth powers and one sqare beneath monomyals Let us try enshorted multiplyig formulas
$(2x^{2}+x^{2})^{2}=4x^{4}+4x^{2}y^{2}+y^{4}=4x^{4} +3x^{2}y^{2}+x^{2}y^{2}$
if we odd bothsidely x^{2}y^{2} we have
$(2x^{2}+x^{2})^{2}-x^{2}y^{2}$=our formula
Thanks, but I am still confused. What steps do you use to derive that answer? Also, you wrote initially $(2x^2 + x^2)^2$, did you mean $(2x^2 + y^2)^2$?

Another example is the following problem:

$x^8 - 12x^4 + 16$

Do I just add and subtract $x^4$ from this? I find that the two factors of 16 are -4 and -4, so if I add and subtract $4x^4$ from the problem, I get $(x^8 - 8x^2 + 16) - 4x^2$. This I can then factor into the answer.

I used some guesswork to get this though, by knowing that $(x^4 - 4)^2$ is $x^8 - 8x^4 + 16$. So I assumed then that since $-12x^4$ + $4x^4$ is $-8x^4$, then I then assumed I subtract $4x^4$ from the other side.

Are problems like this solved by some guessing/trial-and-error, or is there a specific step I am missing?

4. ## Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

We can also factor like this

$x^8-12x^4+16=\left(x^2+1-\sqrt{5}\right) \left(x^2-1-\sqrt{5}\right) \left(x^2-1+\sqrt{5}\right) \left(x^2+1+\sqrt{5}\right)$

or like this

$x^8-12x^4+16=\left(x^4-6-\sqrt{20}\right)\left(x^4-6+\sqrt{20}\right)$

or as a product of linear factors

5. ## Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

Originally Posted by Idea
We can also factor like this

$x^8-12x^4+16=\left(x^2+1-\sqrt{5}\right) \left(x^2-1-\sqrt{5}\right) \left(x^2-1+\sqrt{5}\right) \left(x^2+1+\sqrt{5}\right)$

or like this

$x^8-12x^4+16=\left(x^4-6-\sqrt{20}\right)\left(x^4-6+\sqrt{20}\right)$

or as a product of linear factors
Where do the $\sqrt{5}$ and the $\sqrt{20}$ come from?

6. ## Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

$u^2-12u+16=\left(u-6+\sqrt{20}\right)\left(u-6-\sqrt{20}\right)$

Then let $u=x^4$ to get

$x^8-12x^4+16=\left(x^4-6+\sqrt{20}\right)\left(x^4-6-\sqrt{20}\right)$

Next,

$\sqrt{6+\sqrt{20}}=1+\sqrt{5}$

7. ## Re: Solving Fourth Degree Trinomial With Two Variables to Fourth Degree

Originally Posted by Idea
$u^2-12u+16=\left(u-6+\sqrt{20}\right)\left(u-6-\sqrt{20}\right)$
Then let $u=x^4$ to get
$x^8-12x^4+16=\left(x^4-6+\sqrt{20}\right)\left(x^4-6-\sqrt{20}\right)$
$\sqrt{6+\sqrt{20}}=1+\sqrt{5}$