Question

How do I do this?

Answer 1

you make a bow first, then you stand on one leg

Answer 2

well... hmmm ok... so
first off, you'd want to expand the binomial
keep in mind that \(\bf x^2+3)^2 \implies (x^2+3)(x^2+3)\) and you can use FOIL to expand that
then simplify, add like-terms

now, for "solution", it really means, zeros or x-intercepts
so, you move everything to one-side, either left or right
and = 0 and solve for "x"

Answer 3

When fully expanded, it leaves, x^3 + 6x - 10 = 0. How would you solve this?

Answer 4

well... not quite... gimme a second

Answer 5

Hmm should continue from where jddoe0001 stopped or wait for him to get back?

Answer 6

I*

Answer 7

Please continue

Answer 8

\(\bf {\color{brown}{ (x^2-3)^2 }}+21=10x^2+30
\\ \quad \\
{\color{brown}{ x^4-6x^2+9 }}+21=10x^2+30
\\ \quad \\
x^4-6x^2\cancel{+9+21}-10x^2\cancel{-30}=0
\\ \quad \\
x^4-16x^2=0\impliedby \textit{see the difference of squares?}\)

Answer 9

Ahhhh, thanks. I had accidentally wrote 10x^2 on the right side as 10x.

Answer 10

Sorry jdoe0001 its \[(x^{2}+3)^{2}\]

Answer 11

hmm yeah... shoot, lemme fix that is + =(

Answer 12

Oh yeah

Answer 13

\(\bf {\color{brown}{ (x^2+3)^2 }}+21=10x^2+30
\\ \quad \\
{\color{brown}{ x^4+6x^2+9 }}+21=10x^2+30
\\ \quad \\
x^4+6x^2\cancel{+9+21}-10x^2\cancel{-30}=0
\\ \quad \\
x^4-4x^2=0\impliedby \textit{see the difference of squares?}\)

Answer 14

Ok, thanks!

Answer 15

An alternate way can be to factor out the x^2. Which would give you \[x^{2}(x^{2}-4) = 0\]

Answer 16

hmmmm \(\bf x^4-4x^2=0\impliedby (x^2)^2-2^2x^2=0\implies (x^2)^2-(2x)^2=0\)

Answer 17

Then you can use the zero product rule.

Answer 18

Thanks again