Equivalent Expressions - PSAT Math

1. Which expression is equivalent to $\ln\bigl(e^{3x}\cdot e^{-x^{2}}\bigr)$ ?
   Answer: $3x-x^{2}$ When you encounter logarithms and exponentials together, remember that these functions are inverses of each other, which creates powerful simplification opportunities. To solve this problem, you need two key logarithm properties: the product rule $\ln(ab) = \ln(a) + \ln(b)$ and the fundamental relationship $\ln(e^x) = x$. Start by applying the product rule to the expression inside the logarithm: $\ln\bigl(e^{3x}\cdot e^{-x^{2}}\bigr) = \ln(e^{3x}) + \ln(e^{-x^{2}})$. Now use the fact that logarithm and exponential are inverse functions: $\ln(e^{3x}) = 3x$ and $\ln(e^{-x^{2}}) = -x^{2}$. Therefore, the expression equals $3x + (-x^{2}) = 3x - x^{2}$, which is choice A. Choice B gives $3x + x^{2}$, which incorrectly treats the exponent $-x^{2}$ as positive $x^{2}$. This is a sign error trap. Choice C shows $\ln(e^{3x})-\ln(e^{x^{2}})$, which represents an intermediate step but with two errors: it uses subtraction instead of addition (wrong logarithm property), and it changes $-x^{2}$ to $x^{2}$ in the second term. Choice D gives $e^{3x-x^{2}}$, which removes the outer logarithm entirely. This confuses the original expression with its exponential form. Study tip: When simplifying $\ln(e^{\text{anything}})$, the logarithm and exponential cancel out, leaving just the exponent. Always double-check your signs when dealing with negative exponents.
2. Which expression is equivalent to $2x(3x-1)-(x-2)(x+2)$? Fully expand each product, watch for the difference of squares in $(x-2)(x+2)$, and then combine like terms.
   Answer: $5x^2-2x+4$ 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 $+4$.
3. For $x\neq1$, which expression is equivalent to $\dfrac{(x^{2}-1)^{2}}{x-1}$ ?
   Answer: $(x-1)(x+1)^{2}$ When you see a rational expression with a quadratic in the numerator that shares a factor with the denominator, your goal is to factor and simplify by canceling common terms. Start by factoring the numerator $(x^2-1)^2$. Since $x^2-1$ is a difference of squares, it factors as $(x-1)(x+1)$. Therefore: $(x^2-1)^2 = [(x-1)(x+1)]^2 = (x-1)^2(x+1)^2$ Now substitute this back into the original expression: $\frac{(x^2-1)^2}{x-1} = \frac{(x-1)^2(x+1)^2}{x-1}$ Since $x \neq 1$, we can cancel one factor of $(x-1)$ from numerator and denominator: $\frac{(x-1)^2(x+1)^2}{x-1} = (x-1)(x+1)^2$ Looking at the wrong answers: Choice B gives $(x-1)^2(x+1)$, which keeps too much of the $(x-1)$ factor and has the wrong power on $(x+1)$. Choice C gives $(x+1)^2$, which eliminates the $(x-1)$ factor entirely—this would only be correct if we had $(x-1)^2$ in the denominator instead of just $(x-1)$. Choice D gives $\frac{(x+1)^2}{x-1}$, which incorrectly leaves a factor of $(x-1)$ in the denominator rather than canceling it. Study tip: When simplifying rational expressions, always factor completely first, then look for common factors to cancel. Remember that you can only cancel factors, not terms that are added or subtracted.
4. Simplify $2(x-3)^2-(x^2-6x+9)$. A common mistake is to forget that $(x-3)^2$ expands to $x^2-6x+9$ before combining like terms.
   Answer: $x^2-6x+9$ We need to simplify $2(x-3)^2-(x^2-6x+9)$. First, expand $(x-3)^2 = x^2-6x+9$. So we have $2(x^2-6x+9)-(x^2-6x+9)$. Distributing the 2 gives: $2x^2-12x+18-(x^2-6x+9)$. Now distribute the negative sign: $2x^2-12x+18-x^2+6x-9$. Combine like terms: $(2x^2-x^2)+(-12x+6x)+(18-9) = x^2-6x+9$. The key insight is recognizing that this equals $(x-3)^2$, and a common error is forgetting to distribute the negative sign to all terms in the second parentheses.
5. The expression $9y^2-25$ can be rewritten using a special factoring pattern. Which option gives a factored form that is equivalent to $9y^2-25$?
   Answer: $(3y-5)(3y+5)$ 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.
6. Factor the expression $4x^2-9$ completely over the integers. Use the correct pattern and signs so that multiplying the factors returns the original expression.
   Answer: $(2x-3)(2x+3)$ To factor $4x^2 - 9$, recognize this as a difference of squares since $4x^2 = (2x)^2$ and $9 = 3^2$. Using the pattern $a^2 - b^2 = (a-b)(a+b)$, we get $4x^2 - 9 = (2x)^2 - 3^2 = (2x-3)(2x+3)$. To verify, expand: $(2x-3)(2x+3) = 4x^2 + 6x - 6x - 9 = 4x^2 - 9$ ✓. A common mistake is factoring as $(4x-3)(x+3)$, but this expands to $4x^2 + 12x - 3x - 9 = 4x^2 + 9x - 9$, which has an unwanted middle term.
7. Simplify the expression $\dfrac{6x^2y}{3xy^2}$, assuming $x\ne 0$ and $y\ne 0$. Which expression is equivalent?
   Answer: $\dfrac{2x}{y}$ The question requires simplifying ($\frac{6x^2 y}{3x y^2}$) to find the equivalent expression, assuming x and y nonzero. Simplify coefficients: ($\frac{6}{3} = 2$); for x terms, ($\frac{x^2}{x} = x$); for y terms, ($\frac{y}{y^2} = \frac{1}{y}$). Combine: (2 cdot x cdot $\frac{1}{y}$ = $\frac{2x}{y}$). A key error is inverting the fraction, leading to ($\frac{2y}{x}$) or ($\frac{3x}{2y}$). Another mistake is squaring terms incorrectly to get ($\frac{2x^2}{y^2}$). Cancel common factors carefully before combining, and check by multiplying back.
8. The expression $x^2-16$ can be rewritten using a special factoring pattern. Which expression is equivalent to $x^2-16$ when it is fully factored over the integers?
   Answer: $(x-4)(x+4)$ We need to factor $x^2 - 16$ using the difference of squares pattern. Recognize that $16 = 4^2$, so we have $x^2 - 4^2$. The difference of squares pattern states that $a^2 - b^2 = (a-b)(a+b)$. Applying this: $x^2 - 16 = x^2 - 4^2 = (x-4)(x+4)$. A common mistake is trying to factor this as $(x-8)(x+2)$, but expanding that gives $x^2 - 6x - 16$, not $x^2 - 16$. Remember that difference of squares has no middle term when expanded.
9. Which of the following is equivalent to $\sqrt{50}-2\sqrt{8}+\sqrt{2}$ ?
   Answer: $2\sqrt{2}$ When you encounter radical expressions like this, you need to simplify each term by factoring out perfect squares from under the radicals, then combine like terms. Start by simplifying each radical separately. For $\sqrt{50}$, factor 50 as $25 \times 2$, so $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$. For $\sqrt{8}$, factor 8 as $4 \times 2$, so $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$. The term $\sqrt{2}$ is already simplified. Now substitute these simplified forms: $5\sqrt{2} - 2(2\sqrt{2}) + \sqrt{2} = 5\sqrt{2} - 4\sqrt{2} + \sqrt{2}$. Since all terms have the same radical part ($\sqrt{2}$), you can combine the coefficients: $5 - 4 + 1 = 2$. Therefore, the expression equals $2\sqrt{2}$, which is choice A. Choice B ($\sqrt{2}$) would result from incorrectly combining coefficients as $5 - 4 - 1 = 0$ and then adding 1, or from other arithmetic errors. Choice C ($4\sqrt{2}$) might come from forgetting the coefficient 2 in front of $\sqrt{8}$, making the calculation $5 - 2 + 1 = 4$. Choice D ($10\sqrt{2}$) could result from adding all coefficients without considering the subtraction: $5 + 4 + 1 = 10$. Study tip: Always factor out perfect squares first, then treat radical expressions like algebraic terms with the same variable—combine coefficients while keeping the radical part unchanged.
10. The polynomial $9x^2-24x+16$ is a perfect square trinomial. Which expression is an equivalent factored form? Watch for sign and the correct middle term when squaring a binomial.
    Answer: $(3x-4)^2$ The question involves recognizing ($9x^2$ - 24x + 16) as a perfect square trinomial and finding its factored form. A perfect square is ($(ax pm b)^2$ = $a^2$ $x^2$ pm 2ab x + $b^2$); here, ($a^2$ = 9) so a=3, ($b^2$=16) so b=4, and middle term -24x matches -2*3*4=-24, so it's $(3x - 4)^2$. Verify by expanding: $(3x)^2$=$9x^2$, 2*3x*(-4)=-24x, $(-4)^2$=16. An error might be using +4 instead of -4, leading to a difference of squares like (3x+4)(3x-4). Another mistake is choosing $(9x-4)^2$, which gives $81x^2$ -72x +16. Check the middle term matches 2ab with correct sign. Identify the square root of the leading and constant terms to match quickly.
11. The expression $(x+7)(x-3)$ is expanded using FOIL. Which expression is equivalent to $(x+7)(x-3)$?
    Answer: $x^2+4x-21$ 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.
12. If $2(3x+k)-5x= x+8$, what is the value of $k$? Distribute first, then combine like terms so both sides have matching coefficients of $x$.
    Answer: $4$ 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.
13. Expand and simplify the expression $(2x-3)(x+5)$. Use FOIL carefully, paying attention to the sign of $-3$ and combining like terms into standard form $ax^2+bx+c$.
    Answer: $2x^2+7x-15$ 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 $(-3)(+5) = -15$.
14. A student rewrites $4x^2-16x$ by factoring out the greatest common factor. Which expression is equivalent to $4x^2-16x$?
    Answer: $4x(x-4)$ The question requires factoring out the greatest common factor from $4x^2 - 16x$ to find the equivalent expression. Identify the GCF as 4x, since 4 divides both coefficients and x divides both terms. Factor: $4x \cdot x - 4x \cdot 4 = 4x(x - 4)$. A key error is factoring incorrectly, such as $4(x^2 - 4)$ which misses the x in the second term. Another mistake is using an incorrect GCF like x, leading to $x(4x - 16x) = x(-12x)$, not equivalent. To verify, expand the factored form back to the original polynomial.
15. Which expression is equivalent to $3(2x-5)-4(x+1)+2x$? Be careful to distribute the $3$ and the $-4$ correctly, and then combine like terms to write the result as a simplified linear expression in $x$.
    Answer: $4x-19$ 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.
16. Which expression is equivalent to $3(2x-5)-4(x+1)+7$? Be careful to distribute the negatives and then combine like terms so the final result is written in the form $ax+b$.
    Answer: $2x-12$ 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 $-4x-4$.
17. Which expression is equivalent to $(2x+3)(x-4) - (x-4)$? Factor or expand as needed, but choose the simplest equivalent expression.
    Answer: $(x-4)(2x+2)$ 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.
18. Which expression is equivalent to $(4x+1)^2$? Use the perfect square trinomial pattern and ensure the middle term is correct.
    Answer: $16x^2+8x+1$ This question asks for the expanded form of ($(4x + 1)^2$) using the perfect square trinomial pattern. The pattern is ($(a + b)^2$ = $a^2$ + 2ab + $b^2$), where (a = 4x) and (b = 1), so ( $(4x)^2$ = $16x^2$ ), (2 cdot 4x cdot 1 = 8x), and ($1^2$ = 1), giving ($16x^2$ + 8x + 1). You can also FOIL: (4x cdot 4x = $16x^2$), (4x cdot 1 = 4x), (1 cdot 4x = 4x), (1 cdot 1 = 1), then combine (4x + 4x = 8x). A key error is halving the middle term, leading to 4x instead of 8x. Another mistake might be using $8x^2$ if confusing the coefficient. Recognize perfect squares to avoid full expansion and ensure the middle term is twice the product.
19. Which expression is equivalent to $(2m-1)^2$ after expanding?
    Answer: $4m^2-4m+1$ 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.
20. Simplify the expression $\dfrac{6x^2-12x}{3x}$. Assume $x\ne 0$. A tempting error is to cancel only one term in the numerator instead of factoring first.
    Answer: $2x-4$ To simplify $\frac{6x^2-12x}{3x}$ where $x \neq 0$, first factor the numerator. The numerator $6x^2 - 12x = 6x(x - 2)$. Now we have $\frac{6x(x-2)}{3x}$. Cancel the common factor of $3x$: $\frac{6x(x-2)}{3x} = \frac{6x}{3x} \cdot (x-2) = 2(x-2) = 2x - 4$. A common error is to cancel only one term, like writing $\frac{6x^2}{3x} - 12x = 2x - 12x$, which is incorrect. Always factor before canceling in rational expressions.

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