Answer
The rational zeros are $x=-3$ or $x=\frac{1}{2}$ or $x=2$.
The polynomial in factored form is $(2 x - 1) (x - 2)^2 (x + 3) (x^2 + 1)$.
Work Step by Step
Alternate form:
$2 x^6 - 3 x^5 - 13 x^4 + 29 x^3 - 27 x^2 + 32 x - 12 = (x - 2)^2 (2 x^4 + 5 x^3 - x^2 + 5 x - 3)$
Solve for x over the real numbers:
$(x - 2)^2 (2 x^4 + 5 x^3 - x^2 + 5 x - 3) = 0$
Split into two equations:
$(x - 2)^2 = 0$ or $2 x^4 + 5 x^3 - x^2 + 5 x - 3 = 0$
Take the square root of both sides:
$x - 2 = 0$ or $2 x^4 + 5 x^3 - x^2 + 5 x - 3 = 0$
Add 2 to both sides:
$x = 2$ or $2 x^4 + 5 x^3 - x^2 + 5 x - 3 = 0$
The left hand side factors into a product with three terms:
$x = 2$ or $(2 x - 1) (x^2 + 1) (x + 3) = 0$
Split into three equations:
$x = 2$ or $x + 3 = 0$ or $2 x - 1 = 0$ or $x^2 + 1 = 0$
Subtract 3 from both sides:
$x = 2$ or $x = -3$ or $2 x - 1 = 0$ or $x^2 + 1 = 0$
Add 1 to both sides:
$x = 2$ or $x = -3$ or $2 x = 1$ or $x^2 + 1 = 0$
Divide both sides by 2:
$x = 2$ or $x = -3$ or $x = 1/2$ or $x^2 + 1 = 0$
Subtract 1 from both sides:
$x = 2$ or $x = -3$ or $x = 1/2$ or $x^2 = -1$
$x^2 = -1$ has no solution since for all $x$ on the real line, $x^2 \geq 0$ and $-1<0$:
Answer:
The rational zeros are $x=-3$ or $x=\frac{1}{2}$ or $x=2$.
The polynomial in factored form is $(2 x - 1) (x - 2)^2 (x + 3) (x^2 + 1)$.
