# Thread: Need help on Pre Calc Problem...Factoring

1. ## Need help on Pre Calc Problem...Factoring

factor the polynomial as a product of linear and a factor of a constant or an irreducible polynimial

6X^3-5X^2+3X-1

Thanks guys

2. Originally Posted by victorfk06
factor the polynomial as a product of linear and a factor of a constant or an irreducible polynimial

6X^3-5X^2+3X-1

Thanks guys
i dont really have a good way of explaining how i factor, its just a run through of factors in my head to see what guesses might make sense...

$(3x^2-x+1)(2x-1)$

3. Hello, victorfk06!

Factor the polynomial: . $6x^3-5x^2+3x-1$
Are you familiar with the Factor Theorem?

Given a polynomial $p(x)$, if $p(a) = 0$, then $(x-a)$ is a factor of $p(x).$

And the Rational Roots Theorem?

A rational root of a polynomial equation is of the form: . $\frac{n}{d}$
. . where $n$ is a factor of the constant term,
. . and $d$ is a factor of the leading coefficient.

We want a number which makes the polynomial equal zero.

There are eight candidates.
. . The numerator is a factor of 1: . $\pm1$
. . The denominator is a factor of 6: . $\pm1,\:\pm2,\:\pm3,\:\pm6$
The possible rational roots are: . $\pm1,\;\pm\tfrac{1}{2},\:\pm\tfrac{1}{3},\:\pm\tfr ac{1}{6}$

And we find that $x = \tfrac{1}{2}$ works: . $6\left(\tfrac{1}{2}\right)^3 - 5\left(\tfrac{1}{2}\right)^2 + 3\left(\tfrac{1}{2}\right) - 1 \;=\;0$

Hence, $\left(x-\tfrac{1}{2}\right)\;\hdots\;\text{ or }(2x-1)$ is a factor.

Using long division, we find that:

. . $6x^3 - 5x^2 + 3x - 1 \;=\;(2x-1)(3x^2-x+1)$ . . . as emathinstruction already pointed out.

We find that the quadratic expression doesn't factor . . . so we're done!