College Algebra Quiz: Factoring Trinomials And Special Products

What this quiz covers

This quiz focuses on Factoring Trinomials And Special Products, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $(x^2 - 4)(x^2 - 9)$
2. $(x - 2)(x + 2)(x - 3)(x + 3)$ (correct answer)
3. $(x^2 + 4)(x^2 + 9)$
4. $(x - 6)(x + 6)(x^2 + 1)$

Explanation: First, recognize this as a quadratic in $x^2$. Let $u = x^2$, so we have $u^2 - 13u + 36$. Factoring: $(u - 4)(u - 9) = (x^2 - 4)(x^2 - 9)$. Since both factors are differences of squares, we can factor further: $x^2 - 4 = (x-2)(x+2)$ and $x^2 - 9 = (x-3)(x+3)$. Therefore, the complete factorization is $(x-2)(x+2)(x-3)(x+3)$. Choice A shows incomplete factorization, C has incorrect signs, and D results from incorrect factoring.

Question 2

1. $(2x - 3y)(4x^2 + 6xy + 9y^2)$ (correct answer)
2. $(2x - 3y)(4x^2 - 6xy + 9y^2)$
3. $(2x + 3y)(4x^2 - 6xy + 9y^2)$
4. $(2x - 3y)^2(2x + 3y)$

Explanation: This is a difference of cubes: $8x^3 - 27y^3 = (2x)^3 - (3y)^3$. Using the formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ with $a = 2x$ and $b = 3y$: $(2x - 3y)((2x)^2 + (2x)(3y) + (3y)^2) = (2x - 3y)(4x^2 + 6xy + 9y^2)$. Choice B incorrectly uses the sum of cubes middle term, C has the wrong sign in the first factor, and D represents an incorrect factorization pattern.

Question 3

1. $6x^2y^2$
2. $6xy^2$ (correct answer)
3. $6x^3y^4$
4. $12xy$

Explanation: To find the GCF, we need the greatest common factor of the coefficients and the lowest powers of each variable. Coefficients: $\gcd(12, 18, 24) = 6$. For $x$: the powers are 3, 2, and 1, so the lowest is $x^1 = x$. For $y$: the powers are 2, 3, and 4, so the lowest is $y^2$. Therefore, GCF = $6xy^2$. Choice A uses $x^2$ instead of $x^1$, C uses the highest powers instead of lowest, and D uses the wrong coefficient and $y^1$ instead of $y^2$.

Question 4

1. $k = 12$
2. $k = 15$ (correct answer)
3. $k = 18$
4. $k = 21$

Explanation: Expanding $(2x + a)(x + b) = 2x^2 + 2bx + ax + ab = 2x^2 + (2b + a)x + ab$. Comparing with $2x^2 + kx + 18$, we need $ab = 18$ and $k = 2b + a$. The integer factor pairs of 18 are: (1,18), (2,9), (3,6), (-1,-18), (-2,-9), (-3,-6). Testing each: For $a = 3, b = 6$: $k = 2(6) + 3 = 15$. We can verify: $(2x + 3)(x + 6) = 2x^2 + 12x + 3x + 18 = 2x^2 + 15x + 18$. The other choices correspond to different factor pairs that don't work or computational errors.

Question 5

1. $x^2 - 6x + 5$
2. $x^2 - 2x + 5$
3. $x^2 + 2x + 5$
4. $x^2 + 2x - 5$ (correct answer)

Explanation: This question tests polynomial division and the relationship between factors and products. When you see a problem asking what polynomial times a given factor produces a specific result, you're essentially being asked to perform polynomial division. Since we need a polynomial that, when multiplied by $(x - 4)$, gives $x^3 - 2x^2 - 13x + 20$, we can find it by dividing $x^3 - 2x^2 - 13x + 20$ by $(x - 4)$. Using polynomial long division or synthetic division with $x = 4$: $x^3 - 2x^2 - 13x + 20 = (x - 4)(x^2 + 2x - 5)$ We can verify this by expanding: $(x - 4)(x^2 + 2x - 5) = x^3 + 2x^2 - 5x - 4x^2 - 8x + 20 = x^3 - 2x^2 - 13x + 20$ ✓ Choice A: $x^2 - 6x + 5$ would give $(x - 4)(x^2 - 6x + 5) = x^3 - 10x^2 + 29x - 20$, which has different coefficients and the wrong constant term. Choice B: $x^2 - 2x + 5$ would give $(x - 4)(x^2 - 2x + 5) = x^3 - 6x^2 + 13x - 20$, which has the wrong signs on several terms. Choice C: $x^2 + 2x + 5$ would give $(x - 4)(x^2 + 2x + 5) = x^3 - 2x^2 - 3x - 20$, which has the wrong middle coefficient and constant term. Study tip: When checking polynomial multiplication problems, always verify your answer by expanding the product. Also, remember that if $(x - a)$ is a factor of a polynomial, then $x = a$ should be a root—plugging $x = 4$ into the original polynomial should give zero.

Question 6

1. $(x^2 + 4)(x^2 + 1)$
2. $(x + 2)^2(x - 2)^2$
3. $(x^2 + 2)(x^2 - 2)$
4. $(x^2 + 2)^2$ (correct answer)

Explanation: When you encounter a polynomial expression like $x^4 + 4x^2 + 4$, look for patterns that suggest perfect square trinomials or other special factoring forms. This expression has terms with even powers, which often signals a substitution approach. Let's substitute $u = x^2$ to simplify: $u^2 + 4u + 4$. Now you can recognize this as a perfect square trinomial of the form $a^2 + 2ab + b^2 = (a + b)^2$, where $a = u$ and $b = 2$. This gives us $(u + 2)^2$. Substituting back: $(x^2 + 2)^2$. To verify, expand $(x^2 + 2)^2 = (x^2)^2 + 2(x^2)(2) + 2^2 = x^4 + 4x^2 + 4$ ✓ Choice A is incorrect because expanding $(x^2 + 4)(x^2 + 1)$ gives $x^4 + x^2 + 4x^2 + 4 = x^4 + 5x^2 + 4$, which has the wrong middle coefficient. Choice B is wrong because $(x + 2)^2(x - 2)^2 = [(x + 2)(x - 2)]^2 = (x^2 - 4)^2 = x^4 - 8x^2 + 16$, giving completely different coefficients. Choice C fails because $(x^2 + 2)(x^2 - 2) = (x^2)^2 - 4 = x^4 - 4$, which is missing the middle term entirely. Study tip: When factoring polynomials with even powers, try substituting $u = x^2$ to reveal simpler patterns. Always verify your factorization by expanding it back to the original expression.

Question 7

1. $(3x + 4)(2x - 1) = 6x^2 + 5x - 4$ (correct answer)
2. $(3x + 4) + (2x - 1) = 6x^2 + 5x - 4$
3. $2(3x + 4) + 2(2x - 1) = 6x^2 + 5x - 4$
4. $(3x + 4)^2 + (2x - 1)^2 = 6x^2 + 5x - 4$

Explanation: The area of a rectangle equals length times width, so we need $(3x + 4)(2x - 1) = 6x^2 + 5x - 4$. We can verify: $(3x + 4)(2x - 1) = 6x^2 - 3x + 8x - 4 = 6x^2 + 5x - 4$ ✓. Choice B represents adding the dimensions (which would give a linear expression, not quadratic), C represents the perimeter formula, and D represents the sum of squares of the dimensions.

Question 8

1. Check if the first terms multiply to give $3x^2$
2. Check if the last terms multiply to give $-6$
3. Expand $(3x + 2)(x - 3)$ completely and compare (correct answer)
4. Check if $x = 3$ makes the original expression equal zero

Explanation: While checking individual terms (choices A and B) can catch some errors, only expanding completely will definitively verify the factorization. Expanding: $(3x + 2)(x - 3) = 3x^2 - 9x + 2x - 6 = 3x^2 - 7x - 6$ ✓. The factorization is actually correct. Choice D only checks one potential zero, but a polynomial can have the right zeros with wrong factorization. Complete expansion is the most reliable verification method, especially when the middle term coefficient needs to be checked.

Question 9

1. $k = 9$ or $k = -9$
2. $k = 3$ only
3. $k = 9$ only (correct answer)
4. $k = 6$ or $k = -6$

Explanation: When you encounter a perfect square trinomial, you're looking at an expression that can be written as $(a + b)^2$ or $(a - b)^2$. These expansions always follow the pattern $a^2 \pm 2ab + b^2$. For $x^2 - 6x + k$ to be a perfect square trinomial, it must match this pattern. Since the first term is $x^2$, we know $a = x$. The middle term $-6x$ tells us about the relationship between $a$ and $b$. Since $-6x = -2 \cdot x \cdot b$, we can solve: $-6 = -2b$, so $b = 3$. This means our perfect square trinomial should be $(x - 3)^2$. Let's expand to verify: $(x - 3)^2 = x^2 - 6x + 9$. Therefore, $k = 9$ is the only possible value. Looking at the wrong answers: Choice A suggests $k = -9$ is also possible, but $x^2 - 6x - 9$ cannot be factored as a perfect square since perfect squares of real expressions always have positive constant terms when the middle term is negative. Choice B claims $k = 3$ only, but $x^2 - 6x + 3$ doesn't factor as a perfect square. Choice D suggests $k = 6$ or $k = -6$, but neither $x^2 - 6x + 6$ nor $x^2 - 6x - 6$ can be written as perfect squares. Remember: for any perfect square trinomial $x^2 + bx + k$, the constant term $k$ must equal $\left(\frac{b}{2}\right)^2$. This gives you a quick way to check your work.