Factor Completely:
3t^3 + 5t^2 - 12t
first take out the t
so you have:
t(3t^2 + 5t - 12)
you should be able to factor it easily from there-- what numbers multiply to get (12 x 3) 36 and add to get 5?
4 and 9.
so, you will take 9 and 4 and replace it with the 5, for -9t + 4t = =5t.
so, you have:
t(3t^2 + 5t - 12)
t(3t^2 - 9t + 4t - 12)
t(3t(t - 3) 4(t - 3)) =
t(3t + 4)(t - 3)
To factor the expression 3t^3 + 5t^2 - 12t completely, you can follow these steps:
1. Start by taking out the common factor, which is "t." That gives you t(3t^2 + 5t - 12).
2. Now, you need to factor the quadratic expression 3t^2 + 5t - 12. To do this, you need to find two numbers that multiply to give you -12 (the product of the coefficients of the t^2 and the constant term) and add up to give you 5 (the coefficient of the linear term).
3. In this case, the numbers that satisfy this requirement are 9 and -4. So, you can rewrite the quadratic expression as 3t^2 - 9t + 4t - 12.
4. Next, group the terms: t(3t^2 - 9t + 4t - 12).
5. Now, factor by grouping. Take out the greatest common factor from the first two terms (3t) and the second two terms (4): t(3t(t - 3) + 4(t - 3)).
6. Notice that you have a common factor (t - 3) in both groups. You can factor it out: t(3t + 4)(t - 3).
So, the completely factored form of the expression 3t^3 + 5t^2 - 12t is t(3t + 4)(t - 3).