solve by fully factoring
x^3-7x-6=0
How would I fully factor this?
To fully factor the given equation, we need to find the roots of the equation and then express it as a product of linear factors.
Here's how you can do it step by step:
Step 1: Begin by trying to find one of the roots of the equation by performing a trial and error. Start with some small values, like -1, -2, 1, or 2, which are factors of the constant term (in this case, -6).
Let's try x = -1:
Plugging -1 into the equation, we get:
(-1)^3 - 7(-1) - 6 = -1 + 7 - 6 = 0
So, x = -1 is one of the roots.
Step 2: Now that we've found one root, we can factor the equation using synthetic division or long division. Divide the polynomial by the linear factor (x + 1).
(x^3 - 7x - 6) ÷ (x + 1)
-1 | 1 0 -7 -6
| -1 1 6
______________
1 -1 -6 0
The result of the division is x^2 - x - 6.
Step 3: Now, factorize the obtained quadratic expression x^2 - x - 6.
(x^2 - x - 6) can be factored as (x - 3)(x + 2).
So, the fully factored form of the equation x^3 - 7x - 6 = 0 is:
(x + 1)(x - 3)(x + 2) = 0.
Therefore, the roots of the equation are x = -1, x = 3, and x = -2.