# What are all the possible rational zeros for f(x)=2x^3-3x^2-4x+6 and how do you find all zeros?

Jan 22, 2017

Possible Rational Zeros: $\pm 1 , \pm 2 , \pm 3 , \pm 6 , \pm \frac{1}{2} , \pm \frac{3}{2}$
Actual Zeros: $\frac{3}{2} \mathmr{and} \pm \sqrt{2}$.

#### Explanation:

The way to determine all possible roots is through The Rational Roots Theorem, which states:

If P(x) is a polynomial with integer coefficients and if $\frac{p}{q}$ is a zero of $P \left(x\right) \left(P \left(\frac{p}{q}\right) = 0\right)$, then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient.

Basically, find all integer (positive AND negative) factors of both the constant term (6, in this case) and the leading coefficient (2, in this case), and find all possible quotients $\frac{p}{q}$ where p is a factor of the constant, and q is a factor of the leading coefficient.

The integer factors of 6 are: $\pm 1 , \pm 2 , \pm 3 , \pm 6$
The integer factors of 2 are: $\pm 1 , \pm 2$

Therefore, the possible rational roots are:
$\pm 1 , \pm 2 , \pm 3 , \pm 6 , \pm \frac{1}{2} , \pm \frac{3}{2}$

To find which of these are roots in the actual equation, you could use Guess and Check with Synthetic Division, or simply group and factor the polynomial. I'll show the latter for the sake of space preservation:

$2 {x}^{3} - 4 x - 3 {x}^{2} + 6$
$= 2 x \left({x}^{2} - 2\right) - 3 \left({x}^{2} - 2\right)$
$= \left(2 x - 3\right) \left({x}^{2} - 2\right)$
From this we can conclude that one rational root is $\frac{3}{2}$, which is one of the roots we foretold!

However, we are still stuck with $\left({x}^{2} - 2\right)$. Fortunately, we can use a difference of squares formula to factor this into two linear factors:

$\left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right)$

So your complete factorization of this polynomial is:

#f(x) = (2x - 3)(x - sqrt2)(x + sqrt2)

From this, we can determine that the roots of this equation are $\frac{3}{2} \mathmr{and} \pm \sqrt{2}$.