How would I factor: 3sin(t)^2 + 6sin(t) - 9?

Question

Thank you!

Answer 1

3(sin²t + 2sin(t)-3)
3(sin(t)+3)(sin(t)-1)

Just treat the sin(t) like an x

Answer 2

3Sin^2(t) + 6Sin(t) - 9
3(Sin^2(t) + 2Sin(t) - 3)
3(Sin(t) + 3)(Sin(t) - 1) - fully factorized

Answer 3

to help, you can let x= sin(t)
so you have,
3x^2 + 6x -9
you can factor by using the AC method, multiply a by c and you get -27. you have to figure out what two numbers multiply to-27 and add up to 6, so you figure out that they are 9 and -3.
so then you rewrite the coefficients of x using these values:
3x^2 + 9X-3X -9
then you factor by grouping and get:
3x(x+3) -3(x+3)
(3x-3) (x+3)
so those are your factors and you can substitute sin back in and get:
(3sint(t)-3) (sint(t) +3)

Answer 4

The exact same way that you factor a 2nd degree polyomial. Treat sin(t) like it is x. Then 3x^2+6x-9 is your expression. Factor this then replace x with sin(t).

Answer 5

substitute sin(t) with x:
3x^2+6x-9
then factor as usual:
(3x-3)(x+3)
then put back in the sin(t):
(3sin t -3)(sint t +3)

Answer 6

3sin²(t) + 6sin(t) - 9

3(sin²(t) + 2sin(t) - 3)

3(sin(t) + 3)(sin(t) - 1)