Welcome to Your Algebra 2 Trial

Problem: Factor completely: x² − 5x + 6

Step 1: Find two integers that multiply to +6 (constant term) and add to −5 (middle coefficient).

Step 2: The pair −2 and −3 works: (−2)(−3) = 6 and (−2) + (−3) = −5.

Step 3: Write the factors with those numbers: (x − 2)(x − 3).

Step 4 (check): Multiply—(x − 2)(x − 3) expands to x² − 5x + 6.

Final answer: (x − 2)(x − 3)

Problem: Factor completely: 2x² + 7x + 3

Step 1: Identify a, b, and c.
a = 2, b = 7, c = 3

Step 2: Multiply a · c.
2 · 3 = 6

Step 3: Find two numbers that multiply to 6 and add to 7.
6 and 1

Step 4: Split the middle term.
2x² + 6x + x + 3

Step 5: Group the terms.
(2x² + 6x) + (x + 3)

Step 6: Factor each group.
2x(x + 3) + 1(x + 3)

Step 7: Factor out the common binomial.
(2x + 1)(x + 3)

Final answer: (2x + 1)(x + 3)

Problem: For y = (x − 3)² + 1, name the vertex and whether the parabola opens up or down.

Step 1: Compare to vertex form y = a(x − h)² + k.

Step 2: Read h and k from (x − 3)² + 1. Here h = 3 and k = 1, so the vertex is (3, 1).

Step 3: Read a. There is no number in front of the square, so a = 1.

Step 4: Since a > 0, the parabola opens up.

Final answer: Vertex (3, 1); opens up (a = 1 > 0).

Problem: Compute the discriminant of 2x² − 5x − 3 = 0.

Step 1: Identify a, b, and c in ax² + bx + c = 0.
a = 2, b = −5, c = −3

Step 2: Use D = b² − 4ac.

Step 3: Substitute.
b² = (−5)² = 25

Step 4: Compute −4ac.
−4(2)(−3) = −8(−3) = +24

Step 5: Add.
D = 25 + 24 = 49

Final answer: D = 49

Problem: Describe end behavior of p(x) = −3x⁵ + 2x − 1 as x → ∞ and as x → −∞.

Step 1: Focus on the leading term—the term with the highest degree. Here it is −3x⁵.

Step 2: Note the degree is 5 (odd) and the leading coefficient is −3 (negative).

Step 3: For large |x|, lower-degree terms (like 2x and −1) are dominated by −3x⁵.

Step 4: Odd degree with negative leading coefficient: as x → ∞, p(x) → −∞; as x → −∞, p(x) → ∞.

Final answer: As x → ∞, p(x) → −∞; as x → −∞, p(x) → ∞.

Problem: Is (x − 2) a factor of q(x) = x³ − 4x² + x + 6? Explain how you know.

Step 1: Use the Factor Theorem: (x − 2) is a factor if and only if q(2) = 0.

Step 2: Substitute x = 2 into q(x).
q(2) = (2)³ − 4(2)² + (2) + 6

Step 3: Compute each term.
8 − 16 + 2 + 6

Step 4: Add.
8 − 16 = −8; −8 + 2 = −6; −6 + 6 = 0

Step 5: Since q(2) = 0, (x − 2) is a factor.

Final answer: q(2) = 0, so yes—(x − 2) is a factor.

Problem: For f(x) = (x² − 9)/(x² − 4), identify vertical asymptotes (after factoring).

Step 1: Factor the numerator: x² − 9 = (x − 3)(x + 3) (difference of squares).

Step 2: Factor the denominator: x² − 4 = (x − 2)(x + 2).

Step 3: Write the simplified form (no common factors cancel between top and bottom here).
f(x) = (x − 3)(x + 3) / ((x − 2)(x + 2))

Step 4: Vertical asymptotes occur where the simplified denominator is 0 and not canceled by the numerator.

Step 5: Set (x − 2)(x + 2) = 0 → x = 2 or x = −2.

Final answer: Vertical asymptotes at x = 2 and x = −2.

Problem: For g(x) = (x² − 1)/(x − 1), describe any hole and the simplified form.

Step 1: Factor the numerator: x² − 1 = (x − 1)(x + 1).

Step 2: Rewrite g(x).
g(x) = (x − 1)(x + 1) / (x − 1)

Step 3: Cancel the common factor (x − 1), for x ≠ 1.
g(x) = x + 1, with x ≠ 1

Step 4: A canceled factor (x − 1) means a hole at the value that made that factor zero: x = 1.

Final answer: Hole at x = 1; simplified g(x) = x + 1 (with x ≠ 1).