Question

factor the trinomial
3u^2+1u-24

Answer 1

First find the roots:
\[u_{1,2}=-3; 8/3\]
Then make a factor (x-x1)(x-x2)
\[(x+3)(x-8/3)\]

Answer 2

Do you know how to find the roots of equation \[3u^2+u-24=0\]?

Answer 3

no

Answer 4

\[ax^2+bx+c=0\]
\[x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

For your case:
\[3u^2+u-24=0\]
\[u_{1,2}=\frac{-1 \pm \sqrt{1^2-4 \times 3 \times (-24)}}{2\times3}=\frac{-1\pm \sqrt{289}}{6}\]
\[u_{1,2}=\frac{-18}{6}, \frac{16}{6}\]

Answer 5

Suggest factoring: (u+3)(3u-8) set to 0 solve getting
u=-3, and u=8/3

Answer 6

same results as id21, but a different method.