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Quizzes > High School Quizzes > Electives

Algebra 1 Unit 7 Practice Quiz: Polynomials & Factoring

Conquer factoring polynomials and master Unit 7 Algebra

Difficulty: Moderate
Grade: Grade 9
Study OutcomesCheat Sheet
Paper art promoting Factor Frenzy, a factoring quiz for middle school students.

This Algebra 1 Unit 7 polynomials and factoring quiz helps you practice the skills you need for class and tests. Answer 20 quick questions on adding, multiplying, factoring, and zeros, then see what to review to fix gaps before the unit exam.

What is the factored form of 3x + 6?
(x + 3)2
3x(x + 2)
3(x + 2)
x(3 + 6)
The greatest common factor of 3x + 6 is 3. Factoring out 3 yields 3(x + 2), which is the correct factored form.
Factor the expression x² - 9.
(x + 9)(x - 1)
(x - 9)(x + 1)
(x - 3)²
(x - 3)(x + 3)
x² - 9 is a difference of squares since 9 is 3². It factors as (x - 3)(x + 3) because the product gives back the original expression.
What is the factored form of x² + 5x + 6?
(x + 3)²
(x + 1)(x + 6)
(x + 6)(x - 1)
(x + 2)(x + 3)
The factors of 6 that add up to 5 are 2 and 3. Therefore, x² + 5x + 6 factors to (x + 2)(x + 3).
Factor the expression 4x² + 8x.
4x(x + 2)
4x²(1 + 2)
x(4x + 8)
2x(2x + 8)
Both terms have a common factor of 4x. Factoring this common term yields 4x(x + 2), which is the simplified form.
Factor the expression 25 - y².
(5 - y)²
(5 + y)(5 - y)
(25 - y)(1 + y)
(5 - y²)
25 - y² is a difference of squares since 25 is 5². The expression factors as (5 + y)(5 - y), which is its complete factored form.
Factor the quadratic x² + 7x + 12.
(x + 4)²
(x + 3)(x + 4)
(x + 2)(x + 6)
(x + 1)(x + 12)
The factors of 12 that add up to 7 are 3 and 4. Thus, x² + 7x + 12 factors to (x + 3)(x + 4).
Factor the quadratic 6x² + 11x + 3.
(3x + 3)(2x + 1)
(2x - 3)(3x - 1)
(6x + 3)(x + 1)
(2x + 3)(3x + 1)
Using the AC method, the product of 6 and 3 is 18, and the numbers 2 and 9 can be adjusted to yield the proper factors. The correct factorization is (2x + 3)(3x + 1), which expands to 6x² + 11x + 3.
Factor the expression 2x² - 8.
(2x - 8)(x + 1)
2x(x - 4)
2(x + 2)²
2(x - 2)(x + 2)
First, factor out the common factor 2 to get 2(x² - 4). Then, recognize that x² - 4 is a difference of squares, which factors into (x - 2)(x + 2).
Factor the expression x³ + x² - x - 1 by grouping.
x(x² - 1)
(x + 1)²(x - 1)
(x + 1)(x² - 1)
(x - 1)(x² + 1)
Group the polynomial as (x³ + x²) and (-x - 1). Factor out the common terms to obtain x²(x + 1) - 1(x + 1). Factoring (x + 1) out leads to (x + 1)(x² - 1), which further factors as (x + 1)²(x - 1).
Factor the expression 9x² - 49.
(3x - 7)(3x + 7)
(9x + 7)(x - 7)
(3x - 49)(x + 1)
(9x - 7)(x + 7)
9x² - 49 is a difference of squares because 9x² is (3x)² and 49 is 7². Thus, it factors neatly into (3x - 7)(3x + 7).
Factor the quadratic 2x² + 5x - 3.
(2x + 1)(x - 3)
(2x - 1)(x + 3)
(x - 1)(2x + 3)
(2x + 3)(x - 1)
The quadratic can be factored using the AC method where the numbers multiply to -6 and add to 5. The correct grouping leads to the factors (2x - 1) and (x + 3).
Factor the perfect square trinomial x² - 16x + 64.
(x - 4)(x - 12)
(x - 16)(x + 4)
(x + 8)²
(x - 8)²
Notice that 64 is 8² and the middle term, -16x, is twice -8x. This confirms the expression is a perfect square trinomial, factoring to (x - 8)².
Factor the expression 4x² - 12x + 9.
(2x - 3)²
(2x - 3)(2x + 3)
(4x - 3)(x - 3)
(2x + 3)²
Recognize that (2x - 3)² expands to 4x² - 12x + 9. This confirms that the expression factors perfectly as (2x - 3)².
Factor the expression 2x³ + x² - 8x - 4 completely.
(2x + 1)(x - 2)(x + 2)
(2x + 1)(x - 2)²
(x + 1)(2x - 4)(x + 2)
(2x - 1)(x - 2)(x + 2)
By grouping the terms into (2x³ + x²) and (-8x - 4), you can factor out x² and -4 respectively. This leads to a common binomial (2x + 1) and further factors as (2x + 1)(x² - 4), which is a difference of squares factoring to (x - 2)(x + 2).
Factor the expression x² - 2xy + y².
x(x - 2y)
(x + y)²
(x - y)²
(x - y)(x + y)
The expression matches the pattern of a perfect square trinomial. Recognizing that (x - y)² expands to x² - 2xy + y² confirms the factorization.
Factor the polynomial 3x³ + 6x² - 3x - 6 completely.
(3x + 2)(x - 1)(x + 1)
3(x + 2)(x² - 1)
(x + 2)(3x² - 3x - 6)
3(x + 2)(x - 1)(x + 1)
Begin by grouping the terms: (3x³ + 6x²) and (-3x - 6). Factoring out 3x² from the first and -3 from the second yields 3x²(x + 2) - 3(x + 2). Factoring out (x + 2) gives 3(x + 2)(x² - 1), and further factoring the difference of squares results in 3(x + 2)(x - 1)(x + 1).
Factor the expression 4x❴ - 16x² completely.
4x(x - 2)(x + 2)
4x²(x - 2)(x + 2)
2x²(x - 2)(x + 2)
(2x - 4)(2x + 4)x²
First, factor out the common factor 4x² to obtain 4x²(x² - 4). Recognize that x² - 4 is a difference of squares, which factors as (x - 2)(x + 2). Thus, the complete factorization is 4x²(x - 2)(x + 2).
Factor the difference of cubes 8x³ - 27.
(2x + 3)(4x² - 6x + 9)
(4x - 9)(2x² + 3x + 1)
(2x - 3)(4x² - 6x + 9)
(2x - 3)(4x² + 6x + 9)
Recognize that 8x³ is (2x)³ and 27 is 3³. Applying the difference of cubes formula a³ - b³ = (a - b)(a² + ab + b²) with a = 2x and b = 3 gives (2x - 3)(4x² + 6x + 9).
Factor the polynomial x❴ - 16 completely over the reals.
(x - 2)(x + 2)(x² + 4)
(x - 4)(x + 4)(x² + 4)
(x - 2)(x + 2)(x² - 4)
(x² - 4)(x² + 4)
Rewrite x❴ - 16 as (x²)² - 4², which is a difference of squares. This factors to (x² - 4)(x² + 4), and further factoring x² - 4 into (x - 2)(x + 2) gives the complete factorization as (x - 2)(x + 2)(x² + 4).
Factor the expression 6x²y + 9xy² - 12xy - 18y² completely.
3y(2x + 3y)(x - 2)
(2x + 3y)(3xy - 6y)
(2x + 3y)(x - 2)(3y + 1)
3xy(2x + 3y)(x - 2)
Group the expression as (6x²y + 9xy²) and (-12xy - 18y²). Factor out 3xy from the first group and -6y from the second group to obtain 3xy(2x + 3y) - 6y(2x + 3y). Factoring out the common binomial (2x + 3y) gives 3y(2x + 3y)(x - 2).
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Study Outcomes

  1. Apply various factoring techniques to simplify polynomial expressions.
  2. Analyze polynomial structures to identify common factors and special products.
  3. Evaluate factorizations by expanding expressions to verify their correctness.
  4. Interpret and solve algebraic equations using appropriate factoring methods.
  5. Synthesize multiple strategies to factor complex polynomial expressions confidently.

Algebra Unit 7 Test: Polynomials & Factoring Cheat Sheet

  1. Understand the Greatest Common Factor (GCF) - Kick off your factoring journey by hunting for the largest factor shared by every term - like finding the coolest common denominator in a group. For example, 6x² + 9x becomes 3x(2x + 3), clearing the path for smoother steps.
  2. Master Factoring Trinomials - Trinomials are like puzzle pieces: find two numbers that multiply to the constant term and add to the linear coefficient. So x² + 5x + 6 splits into (x + 2)(x + 3), and you'll feel like a math wizard in no time.
  3. Apply the Difference of Squares Formula - Spot a² − b² patterns to instantly split expressions into (a − b)(a + b). For instance, x² − 16 transforms into (x − 4)(x + 4), turning any thorny problem into a friendly hug.
  4. Utilize the Sum and Difference of Cubes - Cubes beckon for special treatment - use a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²). Watch x³ + 8 flip to (x + 2)(x² − 2x + 4) faster than you can say "factor fiesta!"
  5. Practice Factoring by Grouping - Four-term polynomials? Split them into pairs and factor each, like forming dynamic duos. For x³ + 3x² + x + 3, grouping yields (x + 3)(x² + 1) and a high-five from algebra itself.
  6. Recognize Perfect Square Trinomials - Perfect square trinomials follow a² + 2ab + b² = (a + b)² - easy to spot once you know the secret handshake. So x² + 6x + 9 neatly becomes (x + 3)², adding instant style points.
  7. Understand the Factor Theorem - The Factor Theorem says that if f(c) = 0, then (x − c) is a factor. Use this to connect roots and factors, making you a detective on the hunt for x‑intercepts.
  8. Apply the Remainder Theorem - The Remainder Theorem is your shortcut to checking factors: f(x) divided by (x − c) leaves a remainder of f(c). If that remainder is zero, you've struck factoring gold.
  9. Use the Rational Root Theorem - The Rational Root Theorem is like a talent scout for possible roots ±(factors of constant)/(factors of leading coefficient). For P(x) = 2x³ − 3x² − 8x + 3, test ±1, ±3, ±1/2, ±3/2 to find winning roots.
  10. Practice with Various Factoring Techniques - Finally, mix and match these methods in rapid-fire practice sessions to cement your skills. Regular drill adds muscle memory, turning you into a factoring champion and readying you for any polynomial challenge.
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