SAT Math Quiz: Equivalent Expressions

What this quiz covers

This quiz focuses on Equivalent Expressions, giving you a quick way to practice the rules, question types, and explanations that matter most for SAT Math.

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. $2x^2+13x-15$
2. $2x^2+7x-15$ (correct answer)
3. $2x^2-7x-15$
4. $2x^2+7x+15$

Explanation: To expand $(2x-3)(x+5)$, we use FOIL or distribution method. First terms: $2x cdot x = 2x^2$; Outer terms: $2x cdot 5 = 10x$; Inner terms: $-3 cdot x = -3x$; Last terms: $-3 cdot 5 = -15$. Combining these gives $2x^2+10x-3x-15 = 2x^2+7x-15$. A common sign error occurs when multiplying the inner terms, forgetting that $-3 imes x = -3x$, which would incorrectly lead to $2x^2+13x-15$. When using FOIL, carefully track the sign of each term throughout the multiplication.

Question 2

1. $x-4$
2. $x+4$ (correct answer)
3. $\dfrac{x^2-16}{x-4}$
4. $\dfrac{x-4}{x+4}$

Explanation: To simplify $\frac{x^2-16}{x-4}$ for $x \neq 4$, first factor the numerator. Recognize that $x^2 - 16 = x^2 - 4^2$, which is a difference of squares: $x^2 - 16 = (x-4)(x+4)$. Now the fraction becomes $\frac{(x-4)(x+4)}{x-4} = x+4$ after canceling the common factor $(x-4)$. A common error is trying to cancel before factoring, or not recognizing that 16 is a perfect square. Always factor completely before attempting to simplify rational expressions.

Question 3

1. $2x-13$ (correct answer)
2. $2x-3$
3. $6x-13$
4. $6x-3$

Explanation: To simplify $4(x-2)-[3x-(x-5)]$, we must work from the inside out. First, simplify inside the brackets: $3x-(x-5) = 3x-x+5 = 2x+5$. Now we have $4(x-2)-[2x+5]$. Distributing the 4 gives $4x-8$, and subtracting the bracketed expression gives $4x-8-2x-5 = 2x-13$. The key error to avoid is forgetting to distribute the negative sign when removing the brackets, which would incorrectly yield $4x-8-2x+5 = 2x-3$. When subtracting a bracketed expression, change the sign of every term inside.

Question 4

1. $\dfrac{x - 3}{x + 3}$
2. $\dfrac{x + 3}{x + 3}$
3. $\dfrac{x + 3}{x - 3}$ (correct answer)
4. $\dfrac{x - 3}{x - 3}$

Explanation: Factor to get $\dfrac{(x - 3)(x + 3)}{(x - 3)^2} = \dfrac{x + 3}{x - 3}$ for $x \ne 3$. The other choices come from inverting or canceling incorrectly.

Question 5

1. $9y - 1$
2. $7y + 11$
3. $8y - 8$
4. $7y - 11$ (correct answer)

Explanation: Compute $8y - 6 - y - 5 = 7y - 11$. The distractors come from not distributing the negative or miscombining constants and coefficients.

Question 6

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

Explanation: Factor the common $(x - 3)^2$: $(x - 3)^2[(x - 3) - 1] = (x - 3)^2(x - 4)$. The other options reflect incorrect factoring or exponents.

Question 7

1. $6x^2-18x-5$
2. $6x^2-18x+5$
3. $6x^2-8x-5$
4. $6x^2+2x-5$ (correct answer)

Explanation: To simplify $2x(3x-4)-5(1-2x)$, distribute both parts carefully. First: $2x(3x-4) = 6x^2 - 8x$. Second: $-5(1-2x) = -5 + 10x$ (note the positive $10x$). Combining: $6x^2 - 8x - 5 + 10x = 6x^2 + 2x - 5$. The critical step is correctly distributing $-5$ to get $+10x$, since $-5 imes (-2x) = +10x$. Many students incorrectly get $-10x$, forgetting that negative times negative equals positive.

Question 8

1. $5x+4$ (correct answer)
2. $5x-4$
3. $x+4$
4. $5x+16$

Explanation: Distribute and combine like terms: $3x-6+2x+10=5x+4$. The other choices result from sign mistakes, failing to distribute, or incorrect constant addition.

Question 9

1. $(x - 5)(x - 4)$ (correct answer)
2. $(x + 5)(x - 4)$
3. $(x - 5)(x + 4)$
4. $(x - 10)(x + 2)$

Explanation: Find two numbers that multiply to 20 and add to -9: -5 and -4, giving $(x - 5)(x - 4)$. The other options either give the wrong middle term or the wrong product.

Question 10

1. $-9$
2. $9$ (correct answer)
3. $11$
4. $13$

Explanation: The question asks for the value of (k) such that (4(2x - 1) - k) is equivalent to (8x - 13) for all (x), by matching coefficients after simplifying. Start by distributing the 4: (4 cdot 2x = 8x) and (4 cdot (-1) = -4), so the left side is (8x - 4 - k). For equivalence, the x-coefficients already match (both 8), so set the constants equal: (-4 - k = -13). Solve for k: (-k = -13 + 4 = -9), so (k = 9). A common error is mishandling signs when isolating k, such as adding 4 to both sides incorrectly to get k = 13. Another mistake is not distributing first, leading to mismatched terms like comparing 2x to 8x. When finding missing constants in equivalent expressions, expand fully and equate like terms to solve systematically.

Question 11

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

Explanation: To factor $2x^2-8x$, we first identify the greatest common factor (GCF). Both terms have a factor of 2 and a factor of $x$, so the GCF is $2x$. Factoring out $2x$: $2x^2÷2x=x$ and $-8x÷2x=-4$. Therefore, $2x^2-8x=2x(x-4)$. A common mistake is to factor out only part of the GCF, such as just $2$ or just $x$, giving $2(x^2-4x)$ or $x(2x-8)$. While these are correct, they're not completely factored. Always factor out the complete GCF in one step.

Question 12

1. $4x^2-9$
2. $4x^2-12x+9$ (correct answer)
3. $4x^2-6x+9$
4. $2x^2-12x+9$

Explanation: We need to expand $(2x-3)^2$, which means $(2x-3)(2x-3)$. Using the pattern $(a-b)^2=a^2-2ab+b^2$, we get $(2x)^2-2(2x)(3)+3^2=4x^2-12x+9$. Alternatively, using FOIL: First gives $4x^2$, Outer gives $-6x$, Inner gives $-6x$, and Last gives $9$, so we have $4x^2-6x-6x+9=4x^2-12x+9$. The most common error is getting the wrong middle term by forgetting to double the product of the two terms, writing $-6x$ instead of $-12x$. Remember that $(a-b)^2$ always has a negative middle term equal to $-2ab$.

Question 13

1. $4x+1$
2. $4x-1$
3. $8x-29$
4. $4x-29$ (correct answer)

Explanation: This problem asks us to simplify the expression $3(2x-5)-2(x+7)$ by distributing and combining like terms. First, distribute the 3 to get $6x-15$, then distribute the -2 to get $-2x-14$ (note that the negative sign must be distributed to both terms). This gives us $6x-15-2x-14$. Combining like terms: $6x-2x = 4x$ and $-15-14 = -29$, resulting in $4x-29$. A common error is forgetting to distribute the negative sign to the +7, which would incorrectly yield $4x-1$. When subtracting a grouped expression, always distribute the negative to every term inside the parentheses.

Question 14

1. $3y(4y-9)$ (correct answer)
2. $y(12y-27)$
3. $3(4y^2-9y)$
4. $9y(3y-4)$

Explanation: To factor $12y^2-27y$ completely, we need to find the greatest common factor (GCF). Both terms contain $y$, and the coefficients 12 and 27 share a GCF of 3. So the GCF is $3y$. Factoring out $3y$: $12y^2 - 27y = 3y(4y - 9)$. Check by distributing: $3y(4y - 9) = 12y^2 - 27y$ ✓. A common error is factoring out only $y$ or only 3, missing the complete GCF. Always check your factoring by multiplying back out.

Question 15

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

Explanation: To factor $6x^2+11x+3$, we need two binomials whose product gives this trinomial. The first terms must multiply to $6x^2$ (options: $6x cdot x$ or $3x cdot 2x$) and the last terms must multiply to $3$ (options: $3 cdot 1$). Testing $(3x+1)(2x+3)$: First terms give $6x^2$, outer gives $9x$, inner gives $2x$, last gives $3$. Middle term: $9x+2x = 11x$ ✓. Therefore, $6x^2+11x+3 = (3x+1)(2x+3)$. A common error is pairing factors incorrectly, like $(6x+1)(x+3)$ which gives a middle term of $19x$. When factoring trinomials with leading coefficient ≠ 1, systematically test factor pairs.

Question 16

1. $6ab(2a-3b)$ (correct answer)
2. $3ab(4a-6b)$
3. $6ab(2a+3b)$
4. $ab(12a-18b)$

Explanation: To factor $12a^2b-18ab^2$ completely, we first identify the greatest common factor (GCF) of both terms. The GCF of the coefficients 12 and 18 is 6, and both terms contain at least one $a$ and one $b$, so the GCF is $6ab$. Factoring out $6ab$: $12a^2b÷6ab=2a$ and $18ab^2÷6ab=3b$. Therefore, $12a^2b-18ab^2=6ab(2a-3b)$. A common mistake is to factor out only part of the GCF, such as $3ab$ or just $ab$, which leaves the expression not completely factored. Always factor out the complete GCF to ensure the remaining terms have no common factors.

Question 17

1. $5xy(3x-2y)$ (correct answer)
2. $5x(3xy-2y^2)$
3. $10xy(3x-y)$
4. $xy(15x-10y)$

Explanation: To factor $15x^2y-10xy^2$ completely, first find the GCF of both terms. The GCF of coefficients 15 and 10 is 5; the GCF of $x^2y$ and $xy^2$ is $xy$. So the GCF is $5xy$. Factoring out $5xy$: $15x^2y-10xy^2 = 5xy(3x-2y)$. We can verify: $5xy cdot 3x = 15x^2y$ and $5xy cdot (-2y) = -10xy^2$. A common error is factoring out only part of the GCF, like just $5x$ or $xy$, leaving common factors in the parentheses. Always factor out the complete GCF to get the simplest form.

Question 18

1. $(x+5)(x-2)$
2. $(x+1)(x+10)$
3. $(x+4)(x+3)$
4. $(x+5)(x+2)$ (correct answer)

Explanation: $x^2 + 7x + 10$ factors to $(x+5)(x+2)$. The other factorizations expand to different quadratics.

Question 19

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

Explanation: Factor the numerator as $(x-3)(x+3)$ and cancel $x-3$ to get $x+3$. The other choices come from subtracting incorrectly or not factoring the difference of squares.

Question 20

1. $2x^2-7x-15$
2. $2x^2+13x-15$
3. $2x^2+7x-15$ (correct answer)
4. $2x^2+7x+15$

Explanation: FOIL gives $2x^2+10x-3x-15=2x^2+7x-15$. The distractors have a sign error in the middle term, an incorrect sum of the linear terms, or the wrong sign on the constant.