Factor: f(x)=2x^4-5x^3-30x^2+80x-32 a) f(x)=(2x-1)(x-2)(x+4)(x-4) b) f(x)=(2x-1)(x-2)(x-4)^2 c) f(x)=(2x+1)(x-2)(x-4)^2 d) f(x)=(2x+1)(x+2)(x-4)^2

## Step 1: Rational Root Theorem<br />### The Rational Root Theorem helps identify potential rational roots of a polynomial. In this case, potential rational roots are of the form $\pm \frac{p}{q}$, where $p$ is a factor of the constant term (-32) and $q$ is a factor of the leading coefficient (2). Possible rational roots are $\pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm 32, \pm \frac{1}{2}$.<br /><br />## Step 2: Testing Potential Roots<br />### Substitute the potential roots into the polynomial $f(x)$. We find that $f(2) = 2(16) - 5(8) - 30(4) + 80(2) - 32 = 32 - 40 - 120 + 160 - 32 = 0$. Therefore, $x=2$ is a root, and $(x-2)$ is a factor.<br /><br />## Step 3: Polynomial Division<br />### Divide $f(x)$ by $(x-2)$ using synthetic division or polynomial long division. This yields $2x^3 - x^2 - 32x + 16$.<br /><br />## Step 4: Factoring the Cubic<br />### Now we factor the cubic $2x^3 - x^2 - 32x + 16$. Testing potential roots again, we find that $x = \frac{1}{2}$ is a root since $2(\frac{1}{8}) - (\frac{1}{4}) - 32(\frac{1}{2}) + 16 = \frac{1}{4} - \frac{1}{4} - 16 + 16 = 0$. Thus, $(2x - 1)$ is a factor.<br /><br />## Step 5: Further Polynomial Division<br />### Dividing $2x^3 - x^2 - 32x + 16$ by $(2x - 1)$ gives $x^2 - 16$.<br /><br />## Step 6: Difference of Squares<br />### $x^2 - 16$ is a difference of squares, factoring to $(x-4)(x+4)$.<br /><br />## Step 7: Final Factorization<br />### Combining all factors, we have $f(x) = (2x-1)(x-2)(x-4)(x+4)$.