Algebra 1 — Semester B Practice

Question 1 of 10

TEKS 7A-7CEasy Word

A parabola opens downward when a is:

Apositive

Bnegative

Czero

Dequal to 1

Explanation

📌 y = ax² + bx + c. If a < 0 → opens DOWN. If a > 0 → opens UP.

Question 2 of 10

TEKS 8A-8BEasy Calc Word

If discriminant < 0, the equation has:

Ano real solutions

Bone real solution

Ctwo real solutions

Dinfinite solutions

Explanation

📌 D < 0 → no real solutions (no x-intercepts). The parabola doesn't cross the x-axis.

Question 3 of 10

TEKS 10A-10EMedium Word Diagram

An open-top box is constructed from a 12-inch square piece of cardboard by cutting an equal x-inch square from each corner and folding up the sides. Which polynomial represents the volume of the box in cubic inches?

A12x − 4x²

Bx(12 − x)²

Cx(12 − 2x)²

D144 − 4x²

Explanation

After cutting an x-inch square from each corner, the base of the box becomes (12 − 2x) by (12 − 2x), and the height is x. So V = x(12 − 2x)(12 − 2x) = x(12 − 2x)². The x(12 − x)² answer forgets that x is removed from BOTH ends of each side. The 12x − 4x² and 144 − 4x² answers drop the height factor.

Question 4 of 10

TEKS 10A-10EMedium Word Diagram

The area of a rectangular yard is given by the polynomial 2x² + 7x + 3 square feet. Which pair of binomials could represent the length and width of the yard?

A(2x + 3) and (x + 1)

B2(x + 1)(x + 3)

C(2x + 1) and (x + 3)

D(2x − 1) and (x − 3)

Explanation

Factor 2x² + 7x + 3. Find two numbers that multiply to (2)(3) = 6 and add to 7: 6 and 1. Split: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Verify by FOIL: 2x·x + 2x·3 + 1·x + 1·3 = 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓. The (2x + 3)(x + 1) pair gives 2x² + 5x + 3 when expanded.

Question 5 of 10

TEKS 1A-1GMedium Calc

Solve: 3x²−12=0

A±4

B2 only

C±2

D4 only

Explanation

📌 3x²=12 → x²=4 → x=±2

Question 6 of 10

TEKS 6A-6CMedium Calc

Factor: x² − 7x + 12

A(x−2)(x−6)

B(x+3)(x+4)

C(x−3)(x−4)

D(x−1)(x−12)

Explanation

📌 Numbers that multiply to 12, add to −7: −3 and −4.
(x−3)(x−4)

Question 7 of 10

TEKS 10A-10EMedium Word Diagram

A rectangular vegetable garden has a length of (x + 7) feet and a width of (x + 3) feet. Which polynomial represents the area of the garden in square feet?

A2x + 10

Bx² + 10x + 21

Cx² + 21

Dx² + 4x + 21

Explanation

Area = length × width = (x + 7)(x + 3). Use FOIL: x·x + x·3 + 7·x + 7·3 = x² + 3x + 7x + 21 = x² + 10x + 21. The x² + 21 answer only multiplies the first and last terms (skips the middle). The 2x + 10 answer is the perimeter formula 2(L + W) wrongly applied. The x² + 4x + 21 answer mixes up the middle coefficient.

Question 8 of 10

TEKS 10A-10EHard

When the product (2x + k)(3x − 6) is written in standard form, the coefficient of the x-term is −18. What is the constant term of the product?

A−2

B12

C−12

D60

Explanation

Expand with k carried along using the distributive property: (2x + k)(3x − 6) = 6x² − 12x + 3kx − 6k = 6x² + (3k − 12)x − 6k. The parameter is hidden inside the x-coefficient, so match that coefficient to the stated result: 3k − 12 = −18, giving 3k = −6 and k = −2. The question asks for a different coefficient, the constant term, which is −6k = −6(−2) = 12. A quick verification: (2x − 2)(3x − 6) = 6x² − 18x + 12, and the x-coefficient is indeed −18. The most tempting wrong choice is −2, which is k itself — that student did the backward work correctly and then stopped before answering the question actually asked. Choosing −12 comes from forgetting that the constant is −6k, not 6k, so the two negatives fail to cancel. Choosing 60 comes from a sign error in the expansion, writing the x-coefficient as 3k + 12 = −18, which gives k = −10 and −6k = 60.

Question 9 of 10

TEKS 11A-11BHard

When the expression (k·xᵐ)³ · x⁻⁴ is written in simplest form, it equals 27x⁵. Here k is a positive constant and m is a positive integer. What is the value of k + m?

A12

B30

Explanation

Use the power-of-a-product rule: (k·xᵐ)³ = k³·x^(3m), so the coefficient is cubed too. Then the product-of-powers rule gives x^(3m)·x⁻⁴ = x^(3m−4). Matching k³x^(3m−4) = 27x⁵ term by term: k³ = 27 gives k = 3, and 3m − 4 = 5 gives 3m = 9, so m = 3. Therefore k + m = 3 + 3 = 6. The most tempting wrong value comes from leaving the coefficient alone — writing (k·xᵐ)³ = k·x^(3m) and concluding k = 27, which yields 27 + 3 = 30; the outer exponent applies to every factor inside the parentheses, not just the variable. Another common slip is forgetting to multiply the exponent by 3, solving m − 4 = 5 for m = 9 and reporting 12, and a third is reading m = 5 straight off the right-hand side for a sum of 8.

Question 10 of 10

TEKS 12A-12EHard

An arithmetic sequence has first term a₁ = 8 and a positive common difference d. The 1st, 5th, and 17th terms of this sequence, taken in that order, form a geometric sequence. What is the common ratio of that geometric sequence?

Explanation

The arithmetic rule aₙ = a₁ + (n − 1)d gives a₅ = 8 + 4d and a₁₇ = 8 + 16d. Three numbers are geometric exactly when the middle one is the geometric mean of the outer two: (a₅)² = a₁·a₁₇, so (8 + 4d)² = 8(8 + 16d). Expanding gives 64 + 64d + 16d² = 64 + 128d, so 16d² = 64d, and 16d(d − 4) = 0 yields d = 0 or d = 4. The stated condition d > 0 decides the case: d = 4. The three terms are then 8, 24, 72, and the common ratio is 24/8 = 3, confirmed by 72/24 = 3. The most tempting wrong choice is 4: that is the common difference d, reported instead of the common ratio the question actually asks for — the right work followed by the wrong quantity. The choice 1 is the ratio of the rejected root: d = 0 gives the constant sequence 8, 8, 8, which does satisfy the geometric-mean equation but violates the requirement that d be positive. The value 9 comes from 72/8, a jump across two ratio steps rather than one.

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