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
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
We have two fourth powers and one sqare beneath monomyals Let us try enshorted multiplyig formulas
$\displaystyle (2x^{2}+x^{2})^{2}=4x^{4}+4x^{2}y^{2}+y^{4}=4x^{4} +3x^{2}y^{2}+x^{2}y^{2}$
if we odd bothsidely x^{2}y^{2} we have
$\displaystyle (2x^{2}+x^{2})^{2}-x^{2}y^{2}$=our formula
Thanks, but I am still confused. What steps do you use to derive that answer? Also, you wrote initially $\displaystyle (2x^2 + x^2)^2$, did you mean $\displaystyle (2x^2 + y^2)^2$?
Another example is the following problem:
$\displaystyle x^8 - 12x^4 + 16$
Do I just add and subtract $\displaystyle x^4$ from this? I find that the two factors of 16 are -4 and -4, so if I add and subtract $\displaystyle 4x^4$ from the problem, I get $\displaystyle (x^8 - 8x^2 + 16) - 4x^2$. This I can then factor into the answer.
I used some guesswork to get this though, by knowing that $\displaystyle (x^4 - 4)^2$ is $\displaystyle x^8 - 8x^4 + 16$. So I assumed then that since $\displaystyle -12x^4$ + $\displaystyle 4x^4$ is $\displaystyle -8x^4$, then I then assumed I subtract $\displaystyle 4x^4$ from the other side.
Are problems like this solved by some guessing/trial-and-error, or is there a specific step I am missing?
We can also factor like this
$\displaystyle x^8-12x^4+16=\left(x^2+1-\sqrt{5}\right) \left(x^2-1-\sqrt{5}\right) \left(x^2-1+\sqrt{5}\right) \left(x^2+1+\sqrt{5}\right)$
or like this
$\displaystyle x^8-12x^4+16=\left(x^4-6-\sqrt{20}\right)\left(x^4-6+\sqrt{20}\right)$
or as a product of linear factors
First, factor the quadratic polynomial
$\displaystyle u^2-12u+16=\left(u-6+\sqrt{20}\right)\left(u-6-\sqrt{20}\right)$
Then let $\displaystyle u=x^4$ to get
$\displaystyle x^8-12x^4+16=\left(x^4-6+\sqrt{20}\right)\left(x^4-6-\sqrt{20}\right)$
Next,
$\displaystyle \sqrt{6+\sqrt{20}}=1+\sqrt{5}$