Question

How do you factor `3t^3-21t^2-12t`?

Answer 1

`3t(t-1/2*(7-sqrt(65)))(t-1/2*(7+sqrt(65)))`

Explanation:

We can write
`3t(t^2-7t-4)`

to get the other factors we must solve the quadrat6ic

`t_{1,2}=7/2pmsqrt(49/4+16/4)`
this is

`t_{1,2}=7/2pmsqrt(65)/2`

Answer 2

`3t^3-21t^2-12t = 3t(t-7/2-sqrt(65)/2)(t-7/2+sqrt(65)/2)`

Explanation:

Given:

`3t^3-21t^2-12t`

Note that all of the terms are divisible by `3t`. So we can separate that out as a monomial factor:

`3t^3-21t^2-12t = 3t(t^2-7t-4)`

The remaining quadratic is in the form:

`at^2+bt+c`

with `a=1`, `b=-7` and `c=-4`

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (color(blue)(-7))^2-4(color(blue)(1))(color(blue)(-4)) = 49+16 = 65`

Since `Delta > 0` this quadratic does have real zeros and hence linear factors with real coefficients, but since `Delta` is not a perfect square those zeros and coefficients are irrational.

We can use the quadratic formula to find them:

`t = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(t) = (-b+-sqrt(Delta))/(2a)`

`color(white)(t) = (7+-sqrt(65))/2`

Hence:

`t^2-7t-4 = (t-7/2-sqrt(65)/2)(t-7/2+sqrt(65)/2)`

Putting it all together:

`3t^3-21t^2-12t = 3t(t-7/2-sqrt(65)/2)(t-7/2+sqrt(65)/2)`