# Math

how would you factor 4x-x^2-3/ x-1...so that x-1 would cancel out?

could you explain step by step..plz

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1. (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

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2. 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!

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3. Yes, Reiny answer's is correct.

I went back to redo the question and got the same answer as Reiny.

Sorry for the mistake.

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