Answer
The real zeros are $x=-2$ or $x=3$ or $x=1$ or $x=-3$.
The polynomial in factored form is $(x-3)(x-1)(x+2)^2(x+3)$.
Work Step by Step
Alternate form:
$x^5+3x^4-9x^3-31x^2+36 = (x+2)^2(x^3-x^2-9x+9)$
Solve for $x$ over the real numbers:
$(x+2)^2(x^3-x^2-9x+9)=0$
Split into two equations:
$(x+2)^2=0$ or $x^3-x^2-9x+9=0$
Take the square root of both sides:
$x+2=0$ or $x^3-x^2-9x+9=0$
Subtract $2$ from both sides:
$x=-2$ or $x^3-x^2-9x+9=0$
The left hand side factors into a product with three terms:
$x=-2$ or $(x-3)(x-1)(x+3)=0$
Split into three equations:
$x=-2$ or $x-3=0$ or $x-1=0$ or $x+3=0$
Add $3$ to both sides:
$x=-2$ or $x=3$ or $x-1=0$ or $x+3=0$
Add $1$ to both sides:
$x=-2$ or $x-3$ or $x=1$ or $x+3=0$
Answer:
The real zeros are $x=-2$ or $x=3$ or $x=1$ or $x=-3$.
The polynomial in factored form is $(x-3)(x-1)(x+2)^2(x+3)$.
