To factor x² - 6x + 8, we need two numbers that multiply to 8 and add to -6. The numbers are -4 and -2 since (-4)(-2) = 8 and -4 + (-2) = -6. Therefore, x² - 6x + 8 = (x - 4)(x - 2). Choice C represents a perfect square that would give x² - 6x + 9.

This is a polynomial substitution question testing careful evaluation with negative inputs. Choice A (−21) is correct — substitute x = −2 term by term: 2(−2)³ = 2(−8) = −16. −(−2)² = −(4) = −4. 3(−2) = −6. +5 = 5. Sum: −16 − 4 − 6 + 5 = −21. Choice B (−13) results from computing 2(−2)³ as −8 rather than −16 — forgetting the coefficient of 2: (−2)³ = −8, but 2 × (−8) = −16. Choice C (−5) comes from multiple sign errors throughout, often including treating (−2)³ as +8 for some terms. Choice D (11) likely results from treating (−2)³ as +8 (a sign error on the cube of a negative): 2(8) − 4 − 6 + 5 = 16 − 5 = 11. Pro tip: Evaluate each term separately and write out each result before adding. Compute 2(−2)³ as: step 1: (−2)³ = −8, step 2: 2 × (−8) = −16. Rushing through in one mental step is where sign errors sneak in.

This is a polynomial factors and zeros question testing synthetic division and factoring. Choice B (5 and −2) is correct — since (x − 1) is a factor, divide p(x) by (x − 1) using synthetic division: coefficients 1, −4, −7, 10 divided by root 1: bring down 1; 1×1 − 4 = −3; −3×1 − 7 = −10; −10×1 + 10 = 0 ✓. Quotient: x² − 3x − 10 = (x − 5)(x + 2). Zeros: x = 5 and x = −2. Choice A (−5 and 2) flips both signs of the correct zeros. Choice C (5 and 2) gets x = 5 correct but uses x = 2 instead of x = −2, likely from factoring x² − 3x − 10 as (x − 5)(x − 2) (wrong sign on the second factor). Choice D (−5 and −2) negates both correct zeros. Pro tip: After dividing out the known factor, you're left with a simpler polynomial to factor. Here: (x − 1)(x² − 3x − 10) = 0. Factor x² − 3x − 10: find two numbers that multiply to −10 and sum to −3: −5 and +2. So (x − 5)(x + 2) = 0 → x = 5 or x = −2.

To find $f(3)$ for $f(x) = x^2 - 4x + 4$, we substitute $x = 3$ into the function. We get $f(3) = (3)^2 - 4(3) + 4 = 9 - 12 + 4 = 1$. Notice that this function can also be written as $(x - 2)^2$, so $f(3) = (3 - 2)^2 = 1^2 = 1$.

To multiply $(x - 1)(x + 4)$ using FOIL: First terms: $x \cdot x = x^2$; Outer terms: $x \cdot 4 = 4x$; Inner terms: $(-1) \cdot x = -x$; Last terms: $(-1) \cdot 4 = -4$. Combining these gives $x^2 + 4x - x - 4 = x^2 + 3x - 4$.

To factor $x^2 - 9$, we recognize this as a difference of squares pattern $a^2 - b^2 = (a + b)(a - b)$. Here we have $x^2 - 3^2$, so the factorization is $(x + 3)(x - 3)$. We can verify: $(x + 3)(x - 3) = x^2 - 3x + 3x - 9 = x^2 - 9$.

To solve x² - 4x + 4 = 0, we recognize this as a perfect square trinomial. The equation factors as (x - 2)² = 0, which means x - 2 = 0, so x = 2. Since this is a repeated root, both solutions are x = 2. The solutions are 2 and 2.

To find $f(0)$ for $f(x) = 3x^2 - 4x + 7$, we substitute $x = 0$ into the function. We get $f(0) = 3(0)^2 - 4(0) + 7 = 0 - 0 + 7 = 7$. When evaluating polynomial functions at zero, all terms with $x$ become zero, leaving only the constant term.

The expression x² - 6x + 9 is a perfect square trinomial of the form a² - 2ab + b² where a = x and b = 3. Since (-6x) = 2(x)(-3) and 9 = (-3)², we get x² - 6x + 9 = (x - 3)². We can verify: (x - 3)² = x² - 6x + 9. Choice A would be a difference of squares.

To factor x² + x - 12, find two numbers that multiply to -12 and add to 1. Since (4)(-3) = -12 and 4 + (-3) = 1, the factorization is (x + 4)(x - 3). Rearranging gives (x - 3)(x + 4). We can verify: (x - 3)(x + 4) = x² + 4x - 3x - 12 = x² + x - 12.