Answer
The rational zeros are $x=2$ or $x=-1$ or $x=\frac{1}{2}$ or $x=-\frac{1}{3}$.
The polynomial in factored form is $(x-2)(x+1)(2x-1)(3x+1)$.
Work Step by Step
Alternate form:
$6x^4-7x^3-12x^2+3x+2=(x-2)(x+1)(2x-1)(3x+1)$
The left hand side factors into a product with four terms:
$(x-2)(x+1)(2x-1)(3x+1)=0$
Split into four equations:
$x-2=0$ or $x+1=0$ or $2x-1=0$ or $3x+1=0$
Add $2$ to both sides:
$x=2$ or $x+1=0$ or $2x-1=0$ or $3x+1=0$
Subtract $1$ from both sides:
$x=2$ or $x=-1$ or $2x-1=0$ or $3x+1=0$
Add $1$ to both sides:
$x=2$ or $x=-1$ or $2x=1$ or $3x+1=0$
Divide both sides by 2:
$x=2$ or $x=-1$ or $x=\frac{1}{2}$ or $3x+1=0$
Subtract $1$ from both sides:
$x=2$ or $x=-1$ or $x=\frac{1}{2}$ or $3x=-1$
Answer:
The rational zeros are $x=2$ or $x=-1$ or $x=\frac{1}{2}$ or $x=-\frac{1}{3}$.
The polynomial in factored form is $(x-2)(x+1)(2x-1)(3x+1)$.
