how would you factor 4x-x^2-3/ x-1...so that x-1 would cancel out?
could you explain step by step..plz
(4x-x^2-3)/(x-1)
= -(x^2 - 4x + 3)/(x-1)
= -(x-1)(x-3)/(x-1)
= -(x-3), x not equal to 1
Rearrange the numerator to become
(-x^2 + 4x - 3)/(x - 1)
Factor the numerator.
The numerator becomes
-x^2 + 3x + x - 3.
We divide the numerator's polynomial into two groups:
(-x^2 + 3x) is group a.
(x - 3) is group b.
We factor group a. We cannot factor group b because it is already in lowest term.
So, (-x^2 + 3x) becomes
-x(x - 3)(x - 3)
We now this fraction:
[-x(x-3)(x-3)]/(x - 1) = final answer.
You can also write the final answer this way:
[-x(x - 3)^2]/(x - 1)
Done!
Yes, Reiny answer's is correct.
I went back to redo the question and got the same answer as Reiny.
Sorry for the mistake.
Certainly! To factor the expression (4x - x^2 - 3)/(x - 1) such that (x - 1) cancels out, we can follow these steps:
Step 1: Simplify the expression:
First, simplify the numerator by factoring out a common factor, which is -1:
= (-1)(x^2 - 4x + 3)/(x - 1)
Step 2: Factor the quadratic expression:
The quadratic expression x^2 - 4x + 3 can be factored as follows:
= (-1)(x - 1)(x - 3)/(x - 1)
Step 3: Cancel out the common factor:
Now, we can cancel out the common factor of (x - 1):
= (-1)(x - 3)
Therefore, the expression (4x - x^2 - 3)/(x - 1) can be factored as (-1)(x - 3), after canceling out (x - 1).