1. **Problem:** Test whether $(x - 2)$ is a factor of $f(x) = 2x^3 + 2x^2 - 17x + 10$ and factorize $f(x)$ if it is. 2. **Step 1: Use the Factor Theorem.** The Factor Theorem states that $(x - a)$ is a factor of $f(x)$ if and only if $f(a) = 0$. 3. **Step 2: Evaluate $f(2)$.** $$f(2) = 2(2)^3 + 2(2)^2 - 17(2) + 10 = 2(8) + 2(4) - 34 + 10 = 16 + 8 - 34 + 10 = 0$$ Since $f(2) = 0$, $(x - 2)$ is a factor. 4. **Step 3: Perform polynomial division to factorize $f(x)$ by $(x - 2)$.** Divide $f(x)$ by $(x - 2)$: Using synthetic division: - Coefficients: 2, 2, -17, 10 - Bring down 2 - Multiply 2 by 2 = 4; add to 2 = 6 - Multiply 6 by 2 = 12; add to -17 = -5 - Multiply -5 by 2 = -10; add to 10 = 0 (remainder) So quotient is $2x^2 + 6x - 5$. 5. **Step 4: Factorize the quadratic $2x^2 + 6x - 5$.** Find two numbers that multiply to $2 imes (-5) = -10$ and add to $6$. These numbers are $10$ and $-1$. Rewrite: $$2x^2 + 10x - x - 5 = 0$$ Group: $$(2x^2 + 10x) - (x + 5) = 0$$ Factor: $$2x(x + 5) - 1(x + 5) = (2x - 1)(x + 5)$$ 6. **Step 5: Write the full factorization.** $$f(x) = (x - 2)(2x - 1)(x + 5)$$ **Final answer:** $$\boxed{f(x) = (x - 2)(2x - 1)(x + 5)}$$