Question

factor t^3+3t^2-12t+4

Answer 1

hint:
we can factor out t^2 between the first and second term, so we get:
\[\begin{gathered}
{t^3} + 3{t^2} - 12t + 4 = \hfill \\
\hfill \\
= {t^2}\left( {t + 3} \right) - 12t + 4 \hfill \\
\end{gathered} \]

Answer 2

no, I'm sorry, we have to apply another procedure

Answer 3

okay which one? I was trying to factor by grouping too, but it didn't work for me either

Answer 4

we can see that such polynomial is divisible by 2, since
\[P\left( 2 \right) = 0\]
where
\[P\left( t \right) = {t^3} + 3{t^2} - 12t + 4\]

Answer 5

so I apply the Ruffini's method:

Answer 6

oops...

Answer 7

wait where did the 2 come from? how did you get it? how do you know that p(2)=0?

Answer 8

if I replace t=2 into the polynomial you provided, I get:
\[\begin{gathered}
P\left( 2 \right) = {2^3} + 3 \cdot {2^2} - 12 \cdot 2 + 4 = \hfill \\
\hfill \\
= 8 + 12 - 24 + 4 = 0 \hfill \\
\end{gathered} \]

Answer 9

so, according to the Ruffini's theorem, such polynomial, is divisible by 2, and we can write this:
\[\Large {t^3} + 3{t^2} - 12t + 4 = \left( {t - 2} \right)\left( {{t^2} + 5t - 2} \right)\]

Answer 10

oooh thank you! I will learn about Ruffini's theorem

Answer 11

next we have to factor this trinomial:
\[{{t^2} + 5t - 2}\]

Answer 12

such trinomial, can be factored using this formula:
\[\Large {t^2} + 5t - 2 = \left( {t - {t_1}} \right)\left( {t - {t_2}} \right)\]
where \t_1,\;t_2\) are the zeros of such trinomal:
\[\Large {t^2} + 5t - 2 = 0\]

Answer 13

oops.. I meant \(t_1,\;t_2\)

Answer 14

please try to solve this quadratic equation:
\[\Large {t^2} + 5t - 2 = 0\]

Answer 15

that's okay, it is factored enough for the limit i solved. Thank you

Answer 16

:)