I have no real method for this problem other than trail and error and foil.
I have come up with some amny variations until I am on my second sheet of paper!
The closet I have gotten to factoring/solving this problem so far is....
#46.)
-3t^3+3t^2-6t
(-1t^2+1)(3t -6)
Is there a method to this madness@!?
Factor out a negative three t:
-3t(t^2-t+2) The last term will not factor in the real number system, at least here in Texas.
thanks that made me laugh
2x-4/x2*-9 times x+3/x2*-2x
*means squared
Yes, there is a method to factor this expression. The method is called factorization by grouping, which involves grouping the terms in a way that allows you to factor out common factors from each group. Here's how you can factor the expression -3t^3 + 3t^2 - 6t:
Step 1: Look for common factors among the terms.
In this case, we can see that each term has a common factor of -3t: -3t^3, 3t^2, and -6t. We can factor out -3t from each term.
-3t^3 + 3t^2 - 6t
= -3t(t^2 - t + 2)
Step 2: Factor the trinomial (t^2 - t + 2).
Now we need to factor the trinomial t^2 - t + 2. Since this trinomial cannot be factored using simple integer factorization, we can use the quadratic formula to find its roots. The roots will help us determine the factors.
The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / (2a), where ax^2 + bx + c = 0.
In our case, the trinomial t^2 - t + 2 can be written as t^2 - t + 2 = 0. Comparing this to the quadratic formula, we have a = 1, b = -1, and c = 2.
Using the quadratic formula, we can find the roots:
t = (-(-1) ± √((-1)^2 - 4(1)(2))) / (2(1))
= (1 ± √(1 - 8)) / 2
= (1 ± √(-7)) / 2
Since √(-7) is not a real number, the roots are imaginary. This means that t^2 - t + 2 is not factorable using real numbers.
Therefore, the final factored expression is:
-3t(t^2 - t + 2)