Question

Solve each polynomial equation by factoring
(Question in comments, medal to whoever helps!)

Answer 1

\[x ^{4}-12x ^{2}-36=0\]

Please explain how to get here, I don't just want the answer.
What I got (I don't know if I'm correct)
\[(x ^{4}-6x ^{2})(-6x ^{2}+36)\]
which reduced to
\[x ^{2}(x ^{2}-6)-6(x ^{2}-6)\]
so does that mean the answer is
\[x = \pm \sqrt{6}\]

Answer 2

Thanks!

Answer 3

@theloaad the first step - multiplies x^4 with x^2. this would get x^6 ?

Answer 4

\[(x ^{4}-6x ^{2})(-6x ^{2}+36)\]multiplying everything with each other:\[-6x^6+36x^4+36x^4-121x^2\]

Answer 5

That was just separating it to factor..

Answer 6

I see, I am not good with this myself, trying to learn it :)

Answer 7

Oh okay haha awesome, yeah I split the 12x^2 into a like factor for 36 so I could factor :)

Answer 8

\[x ^{2}(x ^{2}-6)-6(x ^{2}-6)\]what is the next step towards x=sqrt(6) ?

Answer 9

since the two equations in quotes are the same the can be combined (and you square it), then you take the two numbers you factored out of the equation (\[x ^{2}-6\] since it just so happens thats the same equation it ends up joining the others to make \[(x^{2}-6)^{3}\] and then you just solve for x! (since you're finding a solution you can inore the square)

Answer 10

thanks for all your help!!

if I "take out" the (x^2-6) from the terms, I get
\[(x^2-6)(x ^{2}-6)=0\]
They are connected by addition, so I would not combine the terms, rather I would make parentheses?
\[a(x^2-6)-b(x^2-6)=(x^2-6)(a-b)\]