PSAT 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 PSAT 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. $-3$
2. $3$
3. $5$
4. $13$ (correct answer)

Explanation: We need to find $k$ such that $4(x-2)+k = 4x+5$ for all values of $x$. First, expand the left side: $4(x-2)+k = 4x-8+k$. For this to equal $4x+5$, we need $4x-8+k = 4x+5$. The $4x$ terms already match on both sides, so we need the constant terms to be equal: $-8+k = 5$. Solving for $k$: $k = 5+8 = 13$. A common error is forgetting to distribute the 4 to the -2, which would give $4x-2+k$ instead of $4x-8+k$. Always verify by substituting back: $4(x-2)+13 = 4x-8+13 = 4x+5$ ✓.

Question 2

1. $(9a-25)(a+1)$
2. $(3a-5)(3a+5)$ (correct answer)
3. $(9a-5)(a+5)$
4. $(3a-5)^2$

Explanation: We need to factor $9a^2-25$, which is a difference of squares. The pattern for difference of squares is $A^2-B^2 = (A-B)(A+B)$. Here, $9a^2 = (3a)^2$ and $25 = 5^2$, so we have $(3a)^2-5^2$. Applying the formula: $(3a)^2-5^2 = (3a-5)(3a+5)$. The fully factored form is $(3a-5)(3a+5)$. A common mistake is not recognizing that $9a^2 = (3a)^2$ and trying to factor as $(9a-5)(a+5)$, which doesn't work. Always verify by expanding: $(3a-5)(3a+5) = 9a^2+15a-15a-25 = 9a^2-25$ ✓.

Question 3

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

Explanation: We need to find $k$ such that $2(x-3)+5x=7x+k$ for all values of $x$. First, distribute on the left side: $2(x-3) = 2x - 6$. The left side becomes $2x - 6 + 5x = 7x - 6$. For the equation $7x - 6 = 7x + k$ to be true for all $x$, the coefficients of $x$ must match (they do: both are 7) and the constant terms must match. Therefore, $-6 = k$, so $k = -6$. A common mistake is to write $k = 6$ by forgetting the negative sign.

Question 4

1. $-12$
2. $-4$ (correct answer)
3. $4$
4. $12$

Explanation: We need to expand $(x-6)(x+2)$ and identify the coefficient $b$ in the standard form $x^2+bx+c$. Using FOIL: First terms give $x cdot x = x^2$, Outer terms give $x cdot 2 = 2x$, Inner terms give $-6 cdot x = -6x$, and Last terms give $-6 cdot 2 = -12$. Combining these: $x^2 + 2x - 6x - 12 = x^2 - 4x - 12$. The coefficient of $x$ is $b = -4$. A common mistake is adding the inner and outer products incorrectly, getting $2x + 6x = 8x$ instead of $2x - 6x = -4x$.

Question 5

1. $6$ (correct answer)
2. $12$
3. $18$
4. $-6$

Explanation: We need to find $m$ such that $x^2+12x+36 = (x+m)^2$. Expanding the right side: $(x+m)^2 = x^2+2mx+m^2$. Comparing coefficients with $x^2+12x+36$, we need $2m = 12$ (coefficient of $x$) and $m^2 = 36$ (constant term). From $2m = 12$, we get $m = 6$. Let's verify: if $m = 6$, then $m^2 = 36$ ✓, confirming our answer. Therefore, $x^2+12x+36 = (x+6)^2$ and $m = 6$. This is a perfect square trinomial, and recognizing the pattern $a^2+2ab+b^2 = (a+b)^2$ helps identify $m$ quickly.

Question 6

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

Explanation: The question asks to simplify (dfrac{3x^2 - 12x}{3x}) for (x eq 0) by canceling common factors. First, factor the numerator: (3x^2 - 12x = 3x(x - 4)). Then the expression is (dfrac{3x(x - 4)}{3x}), and cancel 3x top and bottom, leaving (x - 4). A common error is canceling incorrectly, like dividing only one term, resulting in something like x - 12. Another mistake could be factoring out 3 instead of 3x, leading to x-12 or 3x-12. Always factor the numerator completely and cancel entire common factors. In tests, verify by plugging in a number like x=1 into original and simplified expressions.

Question 7

1. $5x^2-2x-4$
2. $7x^2-2x+4$
3. $5x^2-2x+4$ (correct answer)
4. $7x^2+2x-4$

Explanation: To simplify $2x(3x-1)-(x-2)(x+2)$, first expand each product. For $2x(3x-1)$: $2x(3x) - 2x(1) = 6x^2 - 2x$. For $(x-2)(x+2)$, recognize this as a difference of squares: $(x-2)(x+2) = x^2 - 4$. Now combine: $6x^2 - 2x - (x^2 - 4) = 6x^2 - 2x - x^2 + 4 = 5x^2 - 2x + 4$. The key insight is recognizing $(x-2)(x+2) = x^2 - 4$ has no middle term, and remembering to distribute the negative sign to get $+4$.

Question 8

1. $4x-19$ (correct answer)
2. $4x-15$
3. $8x-19$
4. $4x+19$

Explanation: The task is to simplify $3(2x-5)-4(x+1)+2x$ by distributing and combining like terms. First, distribute the 3: $3(2x-5) = 6x - 15$. Next, distribute the -4: $-4(x+1) = -4x - 4$. Now combine all terms: $6x - 15 - 4x - 4 + 2x = (6x - 4x + 2x) + (-15 - 4) = 4x - 19$. A common error is distributing -4 as $-4x + 4$ instead of $-4x - 4$, which would incorrectly yield $4x - 11$. When distributing a negative number, both terms inside the parentheses change sign.

Question 9

1. $(9y-25)(y+1)$
2. $(3y-5)(3y+5)$ (correct answer)
3. $(9y-5)(y+5)$
4. $(3y-25)(3y+1)$

Explanation: The question requires factoring ($9y^2$ - 25) to find the equivalent expression. Recognize this as a difference of squares: ( (3y)^2 - 5^2 ). Apply the formula (a^2 - b^2 = (a - b)(a + b)), so it factors to ((3y - 5)(3y + 5)). A key error is mismatching the factors, such as using 9y and 1 incorrectly to get ((9y - 25)(y + 1)). Another mistake is confusing it with sum of squares or incorrect coefficients like ((9y - 5)(y + 5)). To check, expand the factored form back to verify it matches the original.

Question 10

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

Explanation: To expand $(2x-3)(x+5)$, we use FOIL method systematically. 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 all terms: $2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15$. The most common error is getting the sign wrong on the last term, writing $+15$ instead of $-15$ because students forget that $(-3)(+5) = -15$.

Question 11

1. $1$
2. $2$
3. $4$ (correct answer)
4. $8$

Explanation: We need to solve $2(3x+k)-5x = x+8$ for $k$. First, distribute on the left side: $6x + 2k - 5x = x + 8$. Combine like terms on the left: $(6x-5x) + 2k = x + 2k = x + 8$. Since the coefficients of $x$ on both sides are already equal (both are 1), we need the constant terms to match: $2k = 8$. Dividing both sides by 2 gives $k = 4$. The key insight is recognizing that once the $x$ terms match, the constant terms must also be equal.

Question 12

1. $2x-8$ (correct answer)
2. $2x+8$
3. $2x^2-8x$
4. $\dfrac{2x-8}{x}$

Explanation: We need to simplify $\frac{6x^2-24x}{3x}$ for $x \neq 0$. First, factor the numerator: $6x^2-24x = 6x(x-4)$. So we have $\frac{6x(x-4)}{3x}$. Since $x \neq 0$, we can cancel the common factor $x$: $\frac{6x(x-4)}{3x} = \frac{6(x-4)}{3}$. Now simplify the coefficient: $\frac{6}{3} = 2$, giving us $2(x-4) = 2x-8$. The simplified expression is $2x-8$. A common error is dividing each term separately without factoring first, which can lead to mistakes with the algebra. Always factor before simplifying rational expressions.

Question 13

1. $(x+5)^2$ (correct answer)
2. $(x-5)^2$
3. $(x+25)(x+1)$
4. $(x+5)(x+25)$

Explanation: We need to recognize that $x^2 + 10x + 25$ is a perfect square trinomial and write it as a squared binomial. A perfect square trinomial has the form $a^2 + 2ab + b^2 = (a+b)^2$. Here, the first term is $x^2 = (x)^2$ and the last term is $25 = 5^2$. The middle term should be $2 cdot x cdot 5 = 10x$, which matches. Therefore, $x^2 + 10x + 25 = (x+5)^2$. Students often confuse the sign and write $(x-5)^2$, but that expands to $x^2 - 10x + 25$.

Question 14

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

Explanation: The question asks for the expanded equivalent of ((2m - 1)^2). Expand using the square formula: ((2m)^2 - 2 cdot 2m cdot 1 + (-1)^2 = 4m^2 - 4m + 1). Alternatively, FOIL: First (2m cdot 2m = $4m^2$), Outer (2m cdot (-1) = -2m), Inner (-1 cdot 2m = -2m), Last (-1 cdot -1 = 1), then combine ($4m^2$ - 2m - 2m + 1 = $4m^2$ - 4m + 1). A common error is forgetting the middle term, resulting in ($4m^2$ - 1). Another mistake is halving coefficients incorrectly, like ($4m^2$ - 2m + 1). Always expand fully and combine like terms when squaring binomials.

Question 15

1. $(9y-25)(y+1)$
2. $(3y-5)(3y+5)$ (correct answer)
3. $(9y-5)(y+5)$
4. $(3y-5)^2$

Explanation: The expression $9y^2-25$ is a difference of squares since $9y^2 = (3y)^2$ and $25 = 5^2$. The difference of squares pattern states that $a^2-b^2 = (a-b)(a+b)$. Here, $a = 3y$ and $b = 5$, so $9y^2-25 = (3y)^2-5^2 = (3y-5)(3y+5)$. A common mistake is trying to factor out a common factor when none exists, or not recognizing that $9y^2$ is a perfect square. When you see a binomial with subtraction and both terms are perfect squares, always check for the difference of squares pattern.

Question 16

1. $(x-4)(2x+2)$ (correct answer)
2. $(x-4)(2x+4)$
3. $2x^2-5x-12$
4. $2x^2-9x-12$

Explanation: We need to simplify $(2x+3)(x-4) - (x-4)$. Notice that $(x-4)$ is a common factor. We can rewrite this as $(2x+3)(x-4) - 1(x-4) = (x-4)[(2x+3)-1] = (x-4)(2x+3-1) = (x-4)(2x+2)$. The expression simplifies to $(x-4)(2x+2)$. Alternatively, we could expand everything first: $(2x+3)(x-4) = 2x^2-8x+3x-12 = 2x^2-5x-12$, then subtract $(x-4)$ to get $2x^2-5x-12-x+4 = 2x^2-6x-8$, which factors as $2(x^2-3x-4) = 2(x-4)(x+1) = (x-4)(2x+2)$. The factored form $(x-4)(2x+2)$ is simpler than the expanded form.

Question 17

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

Explanation: We need to factor $5x^2-20x$ by finding the greatest common factor (GCF). Looking at both terms, $5x^2 = 5 cdot x cdot x$ and $20x = 5 cdot 4 cdot x$, so the GCF is $5x$. Factoring out $5x$: $5x^2-20x = 5x(x) - 5x(4) = 5x(x-4)$. The expression in factored form is $5x(x-4)$. A common error is factoring out only $5$ or only $x$ instead of the full GCF $5x$, which would give incomplete factorization. Always check your factoring by distributing back to verify you get the original expression.

Question 18

1. $2x-12$ (correct answer)
2. $x-12$
3. $2x-8$
4. $10x-12$

Explanation: We need to simplify $3(2x-5)-4(x+1)+7$ by distributing and combining like terms. First, distribute: $3(2x-5) = 6x-15$ and $-4(x+1) = -4x-4$. Now we have $6x-15-4x-4+7$. Combine the $x$ terms: $6x-4x = 2x$. Combine the constants: $-15-4+7 = -12$. Therefore, the expression simplifies to $2x-12$. A common error is forgetting to distribute the negative sign with $-4$, which would incorrectly give $-4x+4$ instead of $-4x-4$.

Question 19

1. $(x+5)^2$ (correct answer)
2. $(x-5)^2$
3. $(x+25)^2$
4. $(x+5)(x+25)$

Explanation: The expression $x^2+10x+25$ is a perfect square trinomial of the form $a^2+2ab+b^2 = (a+b)^2$. We need to identify what values of $a$ and $b$ give us this trinomial. Since the first term is $x^2$, we have $a = x$. The last term is $25 = 5^2$, so $b = 5$. To verify: the middle term should be $2ab = 2(x)(5) = 10x$, which matches. Therefore, $x^2+10x+25 = (x+5)^2$. A common mistake is confusing the sign and writing $(x-5)^2$, which would expand to $x^2-10x+25$.

Question 20

1. $x^2+4x-21$ (correct answer)
2. $x^2-4x-21$
3. $x^2+10x-21$
4. $x^2+4x+21$

Explanation: The question requires expanding ((x + 7)(x - 3)) using FOIL to find the equivalent expression. Apply FOIL: First terms (x cdot x = $x^2$), Outer terms (x cdot (-3) = -3x), Inner terms (7 cdot x = 7x), Last terms (7 cdot (-3) = -21). Combine like terms: (x^2 + (-3x + 7x) - 21 = x^2 + 4x - 21). A key error is mishandling signs in the outer and inner products, such as treating both as positive to get ($x^2$ + 10x - 21). Another common mistake is adding instead of subtracting the last term, leading to ($x^2$ + 4x + 21). When expanding binomials, always double-check the signs and combine like terms carefully.