Question

Solve for \(x\).

Answer 1

\(\large \color{black}{\begin{align} (x^2+3x+1)(x^2+3x-3)\geq 5\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

.

Answer 3

One dumb method is to expand everything out and factor it again absorbing that 5 on right hand side

Answer 4

And that gives
$$
(x-1) (x+1) (x+2) (x+4)-5
$$
Then find for what values of x is this greater or equal to 0

Answer 5

Pretty sure you mean

$$
(x-1) (x+1) (x+2) (x+4)\ge 0
$$
Then find for what values of x is this greater or equal to 0

Answer 6

.

Answer 7

how did u factor that

\(\large \color{black}{\begin{align} (x-1) (x+1) (x+2) (x+4)\ge 0\hspace{.33em}\\~\\
\end{align}}\)

Answer 8

x(x+3) +1
x(x+3) -3