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Elementary Algebra

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Question 1Question →

The quadratic expression x2−5x−14x^2 - 5x - 14 can be factored completely into (x+a)(x+b)(x + a)(x + b), where aa and bb are integers and a>ba > b. What is the value of aa?

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Answer: 2

Answer

2
To factor x2−5x−14x^2 - 5x - 14, we look for two integers that multiply to −14-14 and add to −5-5. These integers are −7-7 and 22. Thus, the factored form is (x−7)(x+2)(x - 7)(x + 2), which corresponds to (x+a)(x+b)(x + a)(x + b) where the two constant values are −7-7 and 22. Given the condition a>ba > b, the larger value must be assigned to aa. Since 2>−72 > -7, we find a=2a = 2.

Step-by-Step Solution

1
Find two integers that multiply to the constant term −14-14 and add to the linear coefficient −5-5.
The two integers are −7-7 and 22.
Since (−7)×2=−14(-7) \times 2 = -14 and −7+2=−5-7 + 2 = -5, these integers satisfy the requirements for factoring the quadratic trinomial.
2
Write the quadratic expression in its factored form (x+p)(x+q)(x + p)(x + q).
(x−7)(x+2)(x - 7)(x + 2)
The quadratic expression x2+Bx+Cx^2 + Bx + C factors into (x+p)(x+q)(x + p)(x + q) where pp and qq are the found integers.
3
Compare the factored form to the template (x+a)(x+b)(x + a)(x + b) under the condition a>ba > b.
The set of constants is {−7,2}\{-7, 2\}. Since 2>−72 > -7, we assign a=2a = 2 and b=−7b = -7.
This satisfies the requirement that the integer aa is strictly greater than the integer bb.

Key Concept

Factoring quadratic trinomials with a leading coefficient of 1
Estimated Time:45s
Question 2Question →

Match each of the following quadratic equations to its correct set of real solutions.

Click a left item, then click its matching right item

Items

2x(x+1)=12−3x2x(x + 1) = 12 - 3x
3x(x−1)=2(x+4)3x(x - 1) = 2(x + 4)
4x(x−2)=54x(x - 2) = 5

Matches

Show answer & explanation

Answer

The equation 2x(x+1)=12−3x2x(x + 1) = 12 - 3x matches with the solution set {−4,32}\{-4, \frac{3}{2}\}; the equation 3x(x−1)=2(x+4)3x(x - 1) = 2(x + 4) matches with the solution set {−1,83}\{-1, \frac{8}{3}\}; and the equation 4x(x−2)=54x(x - 2) = 5 matches with the solution set {−12,52}\{-\frac{1}{2}, \frac{5}{2}\}.
Each equation is correctly solved by first distributing, moving all terms to one side to set the equation to zero, factoring the resulting trinomial over the integers, and then applying the zero product property to find the corresponding solution set.

Step-by-Step Solution

1
Rearrange the first equation 2x(x+1)=12−3x2x(x + 1) = 12 - 3x into standard form ax2+bx+c=0ax^2 + bx + c = 0.
2x2+5x−12=02x^2 + 5x - 12 = 0
Distributing the 2x2x gives 2x2+2x=12−3x2x^2 + 2x = 12 - 3x. Adding 3x3x and subtracting 1212 from both sides moves all terms to one side.
2
Factor the rearranged first equation 2x2+5x−12=02x^2 + 5x - 12 = 0 and solve for xx.
x=−4x = -4 or x=32x = \frac{3}{2}
Finding two integers that multiply to −24-24 and add to 55 gives 88 and −3-3. Splitting the middle term and factoring by grouping yields (2x−3)(x+4)=0(2x - 3)(x + 4) = 0. Setting each factor to zero gives the solutions.
3
Rearrange the second equation 3x(x−1)=2(x+4)3x(x - 1) = 2(x + 4) into standard form ax2+bx+c=0ax^2 + bx + c = 0.
3x2−5x−8=03x^2 - 5x - 8 = 0
Distributing on both sides gives 3x2−3x=2x+83x^2 - 3x = 2x + 8. Subtracting 2x2x and 88 from both sides sets the quadratic expression to zero.
4
Factor the rearranged second equation 3x2−5x−8=03x^2 - 5x - 8 = 0 and solve for xx.
x=−1x = -1 or x=83x = \frac{8}{3}
Finding two integers that multiply to −24-24 and add to −5-5 gives −8-8 and 33. Grouping terms gives (3x−8)(x+1)=0(3x - 8)(x + 1) = 0. Setting the factors to zero gives the solutions.
5
Rearrange the third equation 4x(x−2)=54x(x - 2) = 5 into standard form ax2+bx+c=0ax^2 + bx + c = 0.
4x2−8x−5=04x^2 - 8x - 5 = 0
Distributing the 4x4x yields 4x2−8x=54x^2 - 8x = 5. Subtracting 55 from both sides sets the equation to zero.
6
Factor the rearranged third equation 4x2−8x−5=04x^2 - 8x - 5 = 0 and solve for xx.
x=−12x = -\frac{1}{2} or x=52x = \frac{5}{2}
Finding two integers that multiply to −20-20 and add to −8-8 gives −10-10 and 22. Grouping terms gives (2x+1)(2x−5)=0(2x + 1)(2x - 5) = 0. Solving each linear factor for xx provides the solutions.

Key Concept

Solving quadratic equations by rearranging them into standard form, factoring by grouping, and applying the zero product property.
Question 3Question →

The polynomial x2+8x+15x^2 + 8x + 15 can be factored into the form (x+a)(x+b)(x + a)(x + b), where aa and bb are integers such that a<ba < b. What is the value of 2a+b2a + b?

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Answer: 11

Answer

The value of 2a+b2a + b is 11.
Factoring the trinomial x2+8x+15x^2 + 8x + 15 gives (x+3)(x+5)(x + 3)(x + 5). Since a<ba < b, we must have a=3a = 3 and b=5b = 5. Thus, 2a+b=2(3)+5=112a + b = 2(3) + 5 = 11.

Step-by-Step Solution

1
Factor the quadratic expression x2+8x+15x^2 + 8x + 15.
(x+3)(x+5)(x + 3)(x + 5)
To factor the trinomial, we find two integers that multiply to the constant term 15 and add to the linear coefficient 8. The integers 3 and 5 satisfy these requirements.
2
Assign the values to aa and bb under the condition a<ba < b.
a=3a = 3 and b=5b = 5
Comparing (x+3)(x+5)(x + 3)(x + 5) to (x+a)(x+b)(x + a)(x + b) gives the values 3 and 5. The condition a<ba < b dictates that the smaller value 3 goes to aa and the larger value 5 goes to bb.
3
Calculate the value of 2a+b2a + b.
11
Substitute a=3a = 3 and b=5b = 5 into the expression: 2(3)+5=6+5=112(3) + 5 = 6 + 5 = 11.

Key Concept

Factoring quadratic trinomials
Question 4Question →

If 3x(x−3)=2(x+3)−123x(x - 3) = 2(x + 3) - 12, what is the product of the two solutions to this equation?

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Answer: 2

Answer

The product of the two solutions to the equation is 2.
The correct answer is 2. Expanding both sides of the equation 3x(x−3)=2(x+3)−123x(x - 3) = 2(x + 3) - 12 gives 3x2−9x=2x+6−123x^2 - 9x = 2x + 6 - 12, which simplifies to 3x2−9x=2x−63x^2 - 9x = 2x - 6. Subtracting 2x−62x - 6 from both sides results in the standard form quadratic equation 3x2−11x+6=03x^2 - 11x + 6 = 0. Factoring this expression gives (3x−2)(x−3)=0(3x - 2)(x - 3) = 0. Setting each factor to zero yields the solutions x=23x = \frac{2}{3} and x=3x = 3. Multiplying these solutions gives 23×3=2\frac{2}{3} \times 3 = 2. Alternatively, by Vieta's formulas, the product of the roots of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 is ca\frac{c}{a}, which directly gives 63=2\frac{6}{3} = 2.

Step-by-Step Solution

1
Expand the expressions on both sides of the equation.
3x2−9x=2x+6−123x^2 - 9x = 2x + 6 - 12, which simplifies to 3x2−9x=2x−63x^2 - 9x = 2x - 6
To prepare the equation for rearrangement into the standard quadratic form.
2
Move all terms to the left side of the equation to set it equal to zero.
3x2−11x+6=03x^2 - 11x + 6 = 0
To write the quadratic equation in the standard form ax2+bx+c=0ax^2 + bx + c = 0.
3
Factor the quadratic trinomial by grouping.
(3x−2)(x−3)=0(3x - 2)(x - 3) = 0
To break down the quadratic equation into linear factors that can be solved individually.
4
Set each linear factor to zero and solve for xx.
x=23x = \frac{2}{3} and x=3x = 3
According to the zero product property, if a product of factors is zero, at least one factor must be zero.
5
Calculate the product of the two solutions.
23×3=2\frac{2}{3} \times 3 = 2
To find the product of the solutions as requested by the question.

Key Concept

Solving quadratic equations by factoring after expanding and rearranging terms
Question 5Question →

A charity walkathon organizer pledges to donate a base amount of $150\$150 plus $2.50\$2.50 for every kilometer completed by each participant. On the day of the event, a participant completes a distance that is 33 kilometers less than twice their training average distance. If the organizer's donation for this participant is $212.50\$212.50, what is the participant's training average distance, in kilometers?

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Answer: 14

Answer

The participant's training average distance is 1414 kilometers.
The correct answer is 1414. To find this value, we write the donation relationship as 150+2.50d=212.50150 + 2.50d = 212.50, where dd is the distance completed. Solving for dd gives d=25d = 25 kilometers. Next, we translate the phrase '3 kilometers less than twice their training average distance' into the expression 2A−32A - 3, where AA is the training average. Equating this to the completed distance gives 2A−3=252A - 3 = 25. Solving for AA yields A=14A = 14.

Step-by-Step Solution

1
Set up an equation for the total donation as a function of the distance completed, dd, in kilometers.
150+2.50d=212.50150 + 2.50d = 212.50
The total donation is the sum of the flat base donation of $150\$150 and the rate of $2.50\$2.50 per kilometer completed.
2
Solve the equation for the completed distance, dd.
d=25d = 25
Subtracting 150150 from both sides of the equation yields 2.50d=62.502.50d = 62.50. Dividing both sides by 2.502.50 yields d=25d = 25.
3
Translate the relationship between the completed distance, dd, and the training average distance, AA, into an equation.
2A−3=252A - 3 = 25
The phrase '3 kilometers less than twice their training average distance' translates mathematically to 2A−32A - 3. Since the completed distance is 2525 kilometers, we set the expression equal to 2525.
4
Solve the equation for the training average distance, AA.
A=14A = 14
Adding 33 to both sides of the equation yields 2A=282A = 28. Dividing both sides by 22 yields A=14A = 14.

Key Concept

Translating word problems into multi-step linear equations and solving for the unknown variable.
Estimated Time:1m 30s
Question 6Question →

If x=−2x = -2, y=5y = 5, and z=−12z = -\frac{1}{2}, what is the value of the algebraic expression x3y+z−2y−2x\frac{x^3 y + z^{-2}}{y - 2x}?

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Answer: -4

Answer

The value of the expression is -4.
Substituting x=−2x = -2, y=5y = 5, and z=−12z = -\frac{1}{2} into the expression yields x3y=(−2)3(5)=−40x^3 y = (-2)^3(5) = -40 and z−2=(−12)−2=4z^{-2} = \left(-\frac{1}{2}\right)^{-2} = 4, making the numerator −40+4=−36-40 + 4 = -36. Evaluating the denominator gives y−2x=5−2(−2)=9y - 2x = 5 - 2(-2) = 9. Dividing −36-36 by 99 results in −4-4.

Step-by-Step Solution

1
Evaluate the terms in the numerator individually.
x3y=(−2)3(5)=−40x^3 y = (-2)^3(5) = -40 and z−2=(−12)−2=4z^{-2} = \left(-\frac{1}{2}\right)^{-2} = 4.
Negative bases raised to odd powers retain a negative sign, while negative exponents represent the reciprocal raised to a positive power.
2
Calculate the total numerator value.
−40+4=−36-40 + 4 = -36.
Summing the two evaluated terms gives the complete numerator.
3
Evaluate the denominator expression.
y−2x=5−2(−2)=5+4=9y - 2x = 5 - 2(-2) = 5 + 4 = 9.
Subtracting a negative quantity is equivalent to adding its positive counterpart.
4
Divide the numerator by the denominator.
−369=−4.\frac{-36}{9} = -4.
Dividing a negative integer by a positive integer yields a negative quotient.

Key Concept

Evaluating Algebraic Expressions with Negative Integers and Negative Exponents
Question 7Question →

For all non-zero real numbers aa and bb, the expression (a2bk)3(a^2 b^k)^3 is equivalent to a6b15a^6 b^{15}. What is the value of the integer kk?

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Answer: 5

Answer

The value of the integer kk is 5.
To find the value of kk, we simplify the expression (a2bk)3(a^2 b^k)^3 using exponent rules. According to the power of a product property, (xy)z=xzyz(xy)^z = x^z y^z, so (a2bk)3=(a2)3(bk)3(a^2 b^k)^3 = (a^2)^3 (b^k)^3. Next, applying the power of a power property, (xy)z=xyz(x^y)^z = x^{yz}, we get a2⋅3bk⋅3=a6b3ka^{2 \cdot 3} b^{k \cdot 3} = a^6 b^{3k}. Since this expression is equivalent to a6b15a^6 b^{15}, we set the exponents of bb equal to each other: 3k=153k = 15. Dividing both sides by 3 yields k=5k = 5.

Step-by-Step Solution

1
Apply the power of a product rule to the expression (a2bk)3(a^2 b^k)^3.
(a2)3⋅(bk)3(a^2)^3 \cdot (b^k)^3
The power of a product rule states that (xy)z=xzyz(xy)^z = x^z y^z.
2
Apply the power of a power rule to simplify the exponents.
a6b3ka^6 b^{3k}
The power of a power rule states that (xy)z=xy⋅z(x^y)^z = x^{y \cdot z}, so (a2)3=a2⋅3=a6(a^2)^3 = a^{2 \cdot 3} = a^6 and (bk)3=b3k(b^k)^3 = b^{3k}.
3
Set the exponent of bb in a6b3ka^6 b^{3k} equal to the exponent of bb in the equivalent expression a6b15a^6 b^{15}.
3k=153k = 15
Since the expressions are equivalent for all non-zero real numbers, the exponents of like bases must be equal.
4
Solve the linear equation for kk.
k=5k = 5
Dividing both sides of 3k=153k = 15 by 3 isolates the variable kk.

Key Concept

Properties of exponents, specifically the power of a product rule (xy)z=xzyz(xy)^z = x^z y^z and the power of a power rule (xy)z=xy⋅z(x^y)^z = x^{y \cdot z}.
Estimated Time:45s
Question 8Question →

What is the value of xx that satisfies the equation 34(x−8)=12x+2\frac{3}{4}(x - 8) = \frac{1}{2}x + 2?

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Answer: 32

Answer

The correct value of xx is 3232.
Distributing 34\frac{3}{4} on the left side yields 34x−6\frac{3}{4}x - 6. Subtracting 12x\frac{1}{2}x (which is 24x\frac{2}{4}x) from both sides gives 14x−6=2\frac{1}{4}x - 6 = 2. Adding 6 to both sides gives 14x=8\frac{1}{4}x = 8, and multiplying by 4 yields the correct value of 3232.

Step-by-Step Solution

1
Distribute the fraction 34\frac{3}{4} to both terms inside the parentheses.
34x−6=12x+2\frac{3}{4}x - 6 = \frac{1}{2}x + 2
To simplify the equation by removing the parentheses.
2
Subtract 12x\frac{1}{2}x from both sides of the equation.
14x−6=2\frac{1}{4}x - 6 = 2
To collect all terms with the variable xx on one side of the equation.
3
Add 6 to both sides of the equation.
14x=8\frac{1}{4}x = 8
To isolate the term containing the variable.
4
Multiply both sides of the equation by 4.
x=32x = 32
To solve for xx by eliminating the coefficient of 14\frac{1}{4}.

Key Concept

Solving linear equations with variables on both sides and fractional coefficients

Alternative Method

Multiply the entire equation by the least common denominator of the fractions, which is 4, to clear the fractions before solving: 4⋅[34(x−8)]=4⋅[12x+2]  ⟹  3(x−8)=2x+84 \cdot [\frac{3}{4}(x - 8)] = 4 \cdot [\frac{1}{2}x + 2] \implies 3(x - 8) = 2x + 8. Then distribute: 3x−24=2x+83x - 24 = 2x + 8. Subtract 2x2x from both sides: x−24=8x - 24 = 8. Add 24 to both sides: x=32x = 32.
Estimated Time:45s
Question 9Question →

If the expression (p3q−2)−1(p−1q2)3\frac{(p^3 q^{-2})^{-1}}{(p^{-1} q^2)^3} is simplified to the form pxqyp^x q^y for all non-zero real numbers pp and qq, what is the value of 2x−y2x - y?

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Answer: 4

Answer

The correct answer is 4.
Applying the power of a power rule to the numerator yields (p3q−2)−1=p−3q2(p^3 q^{-2})^{-1} = p^{-3} q^2. Applying the same rule to the denominator yields (p−1q2)3=p−3q6(p^{-1} q^2)^3 = p^{-3} q^6. Dividing the terms by subtracting exponents gives p−3−(−3)q2−6=p0q−4p^{-3 - (-3)} q^{2-6} = p^0 q^{-4}. Thus, x=0x = 0 and y=−4y = -4. Evaluating 2x−y2x - y gives 2(0)−(−4)=42(0) - (-4) = 4.

Step-by-Step Solution

1
Simplify the numerator using the power of a power property, which states that (am)n=amn(a^m)^n = a^{mn}.
(p3q−2)−1=p3⋅(−1)q−2⋅(−1)=p−3q2(p^3 q^{-2})^{-1} = p^{3 \cdot (-1)} q^{-2 \cdot (-1)} = p^{-3} q^2
This distributes the exponent of −1-1 to both factors inside the parentheses in the numerator.
2
Simplify the denominator using the power of a power property, which states that (am)n=amn(a^m)^n = a^{mn}.
(p−1q2)3=p−1⋅3q2⋅3=p−3q6(p^{-1} q^2)^3 = p^{-1 \cdot 3} q^{2 \cdot 3} = p^{-3} q^6
This distributes the exponent of 33 to both factors inside the parentheses in the denominator.
3
Combine the simplified numerator and denominator using the quotient property of exponents, which states that aman=am−n\frac{a^m}{a^n} = a^{m-n}.
p−3q2p−3q6=p−3−(−3)q2−6=p0q−4\frac{p^{-3} q^2}{p^{-3} q^6} = p^{-3 - (-3)} q^{2 - 6} = p^0 q^{-4}
This simplifies division by subtracting the exponent in the denominator from the exponent in the numerator for each base.
4
Identify the values of xx and yy from the simplified form p0q−4p^0 q^{-4}, and evaluate the final expression 2x−y2x - y.
x=0x = 0 and y=−4y = -4, so 2(0)−(−4)=42(0) - (-4) = 4
This substitutes the values of the exponents into the target algebraic expression to find the final numerical answer.

Key Concept

Properties of exponents including power of a power and quotient properties

Alternative Method

Alternatively, you can write the terms with positive exponents first: (p3q−2)−1(p−1q2)3=(p−1q2)3(p3q−2)1=p−3q6p3q−2=p−3−3q6−(−2)=p−6q8\frac{(p^3 q^{-2})^{-1}}{(p^{-1} q^2)^3} = \frac{(p^{-1} q^2)^3}{(p^3 q^{-2})^1} = \frac{p^{-3} q^6}{p^3 q^{-2}} = p^{-3-3} q^{6-(-2)} = p^{-6} q^8, but this expression is equivalent to the original expression only if inverted properly. Direct distribution is less prone to inversion errors.
Estimated Time:1m 30s
Question 10Question →

Which of the following is the completely factored form of x2−64x^2 - 64?

Show answer & explanation

Answer: (x−8)(x+8)(x - 8)(x + 8)

Answer

(x−8)(x+8)(x - 8)(x + 8)
The polynomial x2−64x^2 - 64 is a difference of two squares because x2x^2 is the square of xx and 6464 is the square of 88. Applying the difference of squares formula, a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b), with a=xa = x and b=8b = 8 gives the factored form (x−8)(x+8)(x - 8)(x + 8).

Step-by-Step Solution

1
Identify the structure of the polynomial.
The expression x2−64x^2 - 64 is a difference of two squares since x2x^2 is (x)2(x)^2 and 6464 is (8)2(8)^2.
Recognizing the difference of squares allows the use of the algebraic identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b).
2
Apply the difference of squares identity with a=xa = x and b=8b = 8.
(x−8)(x+8)(x - 8)(x + 8)
Substituting xx and 88 into the formula yields the factored form.

Key Concept

Factoring the difference of squares using the identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)
Question 11Question →

Which of the following is equivalent to the expression 4x2(x−2y)−3x(x2−3xy)−(x3+xy2)4x^2(x - 2y) - 3x(x^2 - 3xy) - (x^3 + xy^2) for all real values of xx and yy?

Show answer & explanation

Answer: x2y−xy2x^2y - xy^2

Answer

The correct simplified expression is x2y−xy2x^2y - xy^2.
Distributing the terms yields 4x3−8x2y−3x3+9x2y−x3−xy24x^3 - 8x^2y - 3x^3 + 9x^2y - x^3 - xy^2. Grouping and combining like terms results in (4−3−1)x3+(−8+9)x2y−xy2=x2y−xy2(4-3-1)x^3 + (-8+9)x^2y - xy^2 = x^2y - xy^2.

Step-by-Step Solution

1
Distribute the term 4x24x^2 to the terms inside the first set of parentheses: 4x2(x−2y)4x^2(x - 2y).
4x3−8x2y4x^3 - 8x^2y
Multiplying a monomial by a binomial requires distributing the multiplier to each term inside.
2
Distribute the term −3x-3x to the terms inside the second set of parentheses: −3x(x2−3xy)-3x(x^2 - 3xy).
−3x3+9x2y-3x^3 + 9x^2y
Carefully apply the distributive property, remembering that multiplying a negative term by a negative term yields a positive term: −3x⋅(−3xy)=9x2y-3x \cdot (-3xy) = 9x^2y.
3
Distribute the negative sign (or −1-1) to the terms inside the third set of parentheses: −(x3+xy2)-(x^3 + xy^2).
−x3−xy2-x^3 - xy^2
Distributing the negative sign changes the signs of all terms inside the parentheses.
4
Combine all parts and group the like terms: x3x^3, x2yx^2y, and xy2xy^2.
x2y−xy2x^2y - xy^2
Add the coefficients of the terms with the same variable parts: (4−3−1)x3+(−8+9)x2y−xy2=0x3+x2y−xy2=x2y−xy2(4 - 3 - 1)x^3 + (-8 + 9)x^2y - xy^2 = 0x^3 + x^2y - xy^2 = x^2y - xy^2.

Key Concept

Simplifying algebraic expressions by distributing and combining like terms.
Estimated Time:1m 0s
Question 12Question →

What is the positive difference between the two real solutions to the quadratic equation 2x(3x−5)=3(x−2)2x(3x - 5) = 3(x - 2)?

Show answer & explanation

Answer: 56\frac{5}{6}

Answer

The positive difference between the two solutions is 56\frac{5}{6}.
The correct positive difference is 56\frac{5}{6} because rearranging the equation 2x(3x−5)=3(x−2)2x(3x - 5) = 3(x - 2) yields 6x2−13x+6=06x^2 - 13x + 6 = 0. Factoring this equation gives (2x−3)(3x−2)=0(2x - 3)(3x - 2) = 0, which has the solutions x=32x = \frac{3}{2} and x=23x = \frac{2}{3}. Subtracting these values gives 32−23=9−46=56\frac{3}{2} - \frac{2}{3} = \frac{9 - 4}{6} = \frac{5}{6}.

Step-by-Step Solution

1
Expand both sides of the equation.
6x2−10x=3x−66x^2 - 10x = 3x - 6
To begin solving, distribute the term 2x2x on the left side and the constant 33 on the right side.
2
Rearrange the terms to set the quadratic equation to zero.
6x2−13x+6=06x^2 - 13x + 6 = 0
Subtract 3x3x and add 66 to both sides of the equation to write it in standard form ax2+bx+c=0ax^2 + bx + c = 0.
3
Factor the quadratic equation over the integers.
(2x−3)(3x−2)=0(2x - 3)(3x - 2) = 0
Find two binomials whose product is 6x2−13x+66x^2 - 13x + 6. We search for two numbers that multiply to 3636 (from 6×66 \times 6) and sum to −13-13, which are −9-9 and −4-4, allowing factoring by grouping.
4
Solve for the roots of the equation.
x=32x = \frac{3}{2} or x=23x = \frac{2}{3}
Set each linear factor equal to zero using the Zero Product Property: 2x−3=0  ⟹  x=322x - 3 = 0 \implies x = \frac{3}{2} and 3x−2=0  ⟹  x=233x - 2 = 0 \implies x = \frac{2}{3}.
5
Calculate the positive difference between the two solutions.
32−23=96−46=56\frac{3}{2} - \frac{2}{3} = \frac{9}{6} - \frac{4}{6} = \frac{5}{6}
Subtract the smaller solution from the larger solution to find the positive difference.

Key Concept

Solving quadratic equations by rearranging and factoring over integers.
Question 13Question →

For all real numbers xx and yy, which of the following is equivalent to the expression (x−2y)2−(x2−5xy)(x - 2y)^2 - (x^2 - 5xy)?

Show answer & explanation

Answer: 4y2+xy4y^2 + xy

Answer

The expression is equivalent to 4y2+xy4y^2 + xy.
Expanding (x−2y)2(x - 2y)^2 yields x2−4xy+4y2x^2 - 4xy + 4y^2. Distributing the negative sign to −(x2−5xy)-(x^2 - 5xy) gives −x2+5xy-x^2 + 5xy. Combining the results yields (x2−x2)+(−4xy+5xy)+4y2=xy+4y2(x^2 - x^2) + (-4xy + 5xy) + 4y^2 = xy + 4y^2, which is equivalent to 4y2+xy4y^2 + xy.

Step-by-Step Solution

1
Expand the squared binomial (x−2y)2(x - 2y)^2.
x2−4xy+4y2x^2 - 4xy + 4y^2
Apply the binomial squaring formula (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 to expand the expression.
2
Distribute the negative sign across the second parenthetical expression −(x2−5xy)-(x^2 - 5xy).
−x2+5xy-x^2 + 5xy
Distributing the negative sign changes the signs of both terms inside the parentheses.
3
Combine like terms from the expanded parts.
xy+4y2xy + 4y^2
Group and add coefficients of the like terms: (x2−x2)+(−4xy+5xy)+4y2=0+xy+4y2(x^2 - x^2) + (-4xy + 5xy) + 4y^2 = 0 + xy + 4y^2.

Key Concept

Simplifying Expressions and Combining Like Terms
Question 14Question →

Which of the following expressions represents the complete factorization of 2x4−322x^4 - 32?

Show answer & explanation

Answer: 2(x−2)(x+2)(x2+4)2(x - 2)(x + 2)(x^2 + 4)

Answer

2(x−2)(x+2)(x2+4)2(x - 2)(x + 2)(x^2 + 4)
The correct answer is found by first factoring out the greatest common factor of 22, resulting in 2(x4−16)2(x^4 - 16). Next, recognize that x4−16x^4 - 16 is a difference of squares, (x2)2−42(x^2)^2 - 4^2, which factors into (x2−4)(x2+4)(x^2 - 4)(x^2 + 4). Finally, factor the remaining difference of squares, x2−4x^2 - 4, into (x−2)(x+2)(x - 2)(x + 2). The sum of squares, x2+4x^2 + 4, cannot be factored further. Combining these parts gives the complete factorization.

Step-by-Step Solution

1
Factor out the greatest common factor from the polynomial.
2(x4−16)2(x^4 - 16)
Both terms of 2x4−322x^4 - 32 are divisible by 22, so we factor it out to simplify the remaining expression.
2
Factor the difference of squares inside the parentheses.
2(x2−4)(x2+4)2(x^2 - 4)(x^2 + 4)
The expression x4−16x^4 - 16 is a difference of squares because it can be written as (x2)2−42(x^2)^2 - 4^2.
3
Factor the remaining difference of squares.
2(x−2)(x+2)(x2+4)2(x - 2)(x + 2)(x^2 + 4)
The binomial x2−4x^2 - 4 is also a difference of squares (x2−22x^2 - 2^2), which factors into (x−2)(x+2)(x - 2)(x + 2). The sum of squares x2+4x^2 + 4 cannot be factored further using real numbers.

Key Concept

Factoring polynomials completely by extracting the greatest common factor and repeatedly applying the difference of squares formula.
Estimated Time:1m 0s
Question 15Question →

The square of 33 less than a number xx is equal to 1616. What is the greater of the two possible values for xx?

Show answer & explanation

Answer: 7

Answer

7
The word problem translates to the equation (x−3)2=16(x-3)^2 = 16. Expanding the left side and subtracting 16 from both sides gives the standard quadratic equation x2−6x−7=0x^2 - 6x - 7 = 0. This factors into (x−7)(x+1)=0(x-7)(x+1) = 0, yielding two solutions: x=7x = 7 and x=−1x = -1. The greater of these two values is 7.

Step-by-Step Solution

1
Translate the verbal description into an algebraic equation.
(x−3)2=16(x-3)^2 = 16
'3 less than a number xx' is written as x−3x - 3, and its square is set equal to 16.
2
Expand the squared binomial and rearrange the terms into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−6x−7=0x^2 - 6x - 7 = 0
Expanding (x−3)2(x-3)^2 yields x2−6x+9x^2 - 6x + 9. Subtracting 16 from both sides sets the equation to zero.
3
Factor the quadratic equation over the integers.
(x−7)(x+1)=0(x - 7)(x + 1) = 0
We find two integers that multiply to −7-7 and add to −6-6, which are −7-7 and 11.
4
Solve for xx by setting each factor to zero, then select the greater value.
x=7x = 7 (since the solutions are 77 and −1-1)
Setting the factors to zero gives x−7=0  ⟹  x=7x - 7 = 0 \implies x = 7 and x+1=0  ⟹  x=−1x + 1 = 0 \implies x = -1. The larger of these two values is 77.

Key Concept

Solving Quadratic Equations by Factoring
Estimated Time:1m 0s
Question 16Question →

Which of the following is a factor of the polynomial 8x3−12x2−18x+278x^3 - 12x^2 - 18x + 27?

Show answer & explanation

Answer: 2x+32x + 3

Answer

The correct answer is 2x+32x + 3.
The correct answer is the factor 2x+32x + 3. By grouping the first two terms and the last two terms, we get 4x2(2x−3)−9(2x−3)4x^2(2x - 3) - 9(2x - 3). Factoring out the common binomial yields (4x2−9)(2x−3)(4x^2 - 9)(2x - 3). The term 4x2−94x^2 - 9 is a difference of squares, which factors into (2x−3)(2x+3)(2x - 3)(2x + 3). Therefore, the fully factored expression is (2x−3)2(2x+3)(2x - 3)^2(2x + 3), making the linear expression with a positive constant term the correct factor.

Step-by-Step Solution

1
Group the four terms of the polynomial into two pairs.
(8x3−12x2)−(18x−27)(8x^3 - 12x^2) - (18x - 27)
Grouping terms allows us to find common binomial factors in a cubic polynomial.
2
Factor out the greatest common factor (GCF) from each group.
4x2(2x−3)−9(2x−3)4x^2(2x - 3) - 9(2x - 3)
The GCF of 8x38x^3 and 12x212x^2 is 4x24x^2, and the GCF of −18x-18x and 2727 is −9-9 (factoring out the negative to match the binomials).
3
Factor out the common binomial factor (2x−3)(2x - 3).
(4x2−9)(2x−3)(4x^2 - 9)(2x - 3)
Both groups share the factor (2x−3)(2x - 3), so it can be factored out.
4
Factor the quadratic difference of squares (4x2−9)(4x^2 - 9).
(2x−3)(2x+3)(2x−3)=(2x−3)2(2x+3)(2x - 3)(2x + 3)(2x - 3) = (2x - 3)^2(2x + 3)
The term 4x2−94x^2 - 9 is a difference of squares of the form a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b), where a=2xa = 2x and b=3b = 3.

Key Concept

Factoring a cubic polynomial completely by grouping and then applying the difference of squares formula.
Question 17Question →

The square of 3 less than twice a certain real number is equal to 3 less than 7 times that number. If pp and qq are the two distinct real solutions to this equation with p>qp > q, what is the value of 4p−4q4p - 4q?

Show answer & explanation

Answer: 13

Answer

13
Correctly expanding the squared expression leads to the quadratic equation 4x2−19x+12=04x^2 - 19x + 12 = 0. Factoring this equation yields the solutions 44 and 3/43/4. Since the problem specifies p>qp > q, we have p=4p = 4 and q=3/4q = 3/4. Substituting these values into the expression 4p−4q4p - 4q gives 4(4)−4(3/4)=16−3=134(4) - 4(3/4) = 16 - 3 = 13.

Step-by-Step Solution

1
Translate the verbal description into an algebraic equation.
(2x−3)2=7x−3(2x - 3)^2 = 7x - 3
'Twice a number' is represented as 2x2x, '3 less than twice the number' is 2x−32x - 3, and 'the square of' that is (2x−3)2(2x - 3)^2. This is set equal to '3 less than 7 times that number', which is 7x−37x - 3.
2
Expand the binomial and write the equation in standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4x2−19x+12=04x^2 - 19x + 12 = 0
Expanding the left side yields 4x2−12x+9=7x−34x^2 - 12x + 9 = 7x - 3. Subtracting 7x7x and adding 33 to both sides results in the standard form.
3
Factor the quadratic equation by grouping.
(4x−3)(x−4)=0(4x - 3)(x - 4) = 0
We need two numbers that multiply to 4×12=484 \times 12 = 48 and add to −19-19. These numbers are −16-16 and −3-3. Rewriting the equation as 4x2−16x−3x+12=04x^2 - 16x - 3x + 12 = 0 allows us to factor out 4x(x−4)−3(x−4)=04x(x - 4) - 3(x - 4) = 0, which simplifies to (4x−3)(x−4)=0(4x - 3)(x - 4) = 0.
4
Find the roots of the equation and evaluate the target expression 4p−4q4p - 4q.
p=4p = 4, q=3/4q = 3/4, and the final value is 1313.
Setting each factor to zero gives x=3/4x = 3/4 and x=4x = 4. Since p>qp > q, we assign p=4p = 4 and q=3/4q = 3/4. Evaluating the expression yields 4(4)−4(3/4)=16−3=134(4) - 4(3/4) = 16 - 3 = 13.

Key Concept

Solving Quadratic Equations by Factoring
Estimated Time:2m 30s
Question 18Question →

When the expression 6x−2x(x−3y)−(2x−12y)2−34x(6x−8)6x - 2x(x - 3y) - \left(2x - \frac{1}{2}y\right)^2 - \frac{3}{4}x(6x - 8) is simplified by combining like terms, what is the coefficient of the x2x^2 term?

Show answer & explanation

Answer: −212-\frac{21}{2}

Answer

−212-\frac{21}{2}
Expanding the entire expression yields the x2x^2 terms −2x2-2x^2, −4x2-4x^2, and −92x2-\frac{9}{2}x^2. Summing these coefficients gives −2−4−92=−212-2 - 4 - \frac{9}{2} = -\frac{21}{2}. Therefore, the coefficient of the x2x^2 term is −212-\frac{21}{2}.

Step-by-Step Solution

1
Expand the first parenthetical expression by distributing the term −2x-2x.
−2x(x−3y)=−2x2+6xy-2x(x - 3y) = -2x^2 + 6xy. The expression becomes: 6x−2x2+6xy−(2x−12y)2−34x(6x−8)6x - 2x^2 + 6xy - \left(2x - \frac{1}{2}y\right)^2 - \frac{3}{4}x(6x - 8).
Distribution is required to eliminate parentheses before terms can be combined.
2
Expand the squared binomial (2x−12y)2\left(2x - \frac{1}{2}y\right)^2 using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2, then distribute the negative sign.
(2x−12y)2=4x2−2xy+14y2\left(2x - \frac{1}{2}y\right)^2 = 4x^2 - 2xy + \frac{1}{4}y^2. Distributing the negative gives −4x2+2xy−14y2-4x^2 + 2xy - \frac{1}{4}y^2.
This simplifies the second parenthetical term of the expression.
3
Expand the third parenthetical term by distributing −34x-\frac{3}{4}x.
−34x(6x−8)=−92x2+6x-\frac{3}{4}x(6x - 8) = -\frac{9}{2}x^2 + 6x.
This simplifies the final parenthetical term of the expression.
4
Combine the coefficients of all the x2x^2 terms.
The x2x^2 terms are −2x2-2x^2, −4x2-4x^2, and −92x2-\frac{9}{2}x^2. Combining their coefficients gives: −2−4−92=−6−92=−122−92=−212-2 - 4 - \frac{9}{2} = -6 - \frac{9}{2} = -\frac{12}{2} - \frac{9}{2} = -\frac{21}{2}.
Combining like terms simplifies the expression to find the final coefficient of x2x^2.

Key Concept

Simplifying algebraic expressions by distributing coefficients and combining like terms.
Estimated Time:1m 30s
Question 19Question →

An algebraic expression of the form x4−10x2y2+9y4−x2+9y2x^4 - 10x^2y^2 + 9y^4 - x^2 + 9y^2 is defined for all real numbers xx and yy. When this expression is factored completely over the integers, which of the following is one of its factors?

Show answer & explanation

Answer: x+3yx + 3y

Answer

The binomial x+3yx + 3y is a factor of the expression.
To factor the polynomial completely, we group the terms as (x4−10x2y2+9y4)−(x2−9y2)(x^4 - 10x^2y^2 + 9y^4) - (x^2 - 9y^2). The first trinomial is in quadratic form and factors to (x2−9y2)(x2−y2)(x^2 - 9y^2)(x^2 - y^2). The expression is then rewritten as (x2−9y2)(x2−y2)−(x2−9y2)(x^2 - 9y^2)(x^2 - y^2) - (x^2 - 9y^2), which has a GCF of (x2−9y2)(x^2 - 9y^2). Factoring out the GCF yields (x2−9y2)(x2−y2−1)(x^2 - 9y^2)(x^2 - y^2 - 1). Finally, the difference of squares (x2−9y2)(x^2 - 9y^2) factors into (x−3y)(x+3y)(x - 3y)(x + 3y). The fully factored expression is (x−3y)(x+3y)(x2−y2−1)(x - 3y)(x + 3y)(x^2 - y^2 - 1), which contains the factor x+3yx + 3y.

Step-by-Step Solution

1
Group the terms of the polynomial into two parts.
(x4−10x2y2+9y4)−(x2−9y2)(x^4 - 10x^2y^2 + 9y^4) - (x^2 - 9y^2)
Grouping terms allows us to find and extract common algebraic factors from distinct parts of the polynomial.
2
Factor the trinomial from the first group as a quadratic form in terms of x2x^2 and y2y^2.
(x2−9y2)(x2−y2)(x^2 - 9y^2)(x^2 - y^2)
The trinomial x4−10x2y2+9y4x^4 - 10x^2y^2 + 9y^4 can be written as (x2)2−10(x2)(y2)+9(y2)2(x^2)^2 - 10(x^2)(y^2) + 9(y^2)^2, which factors as (x2−9y2)(x2−y2)(x^2 - 9y^2)(x^2 - y^2).
3
Substitute this factorization back into the grouped expression and factor out the greatest common factor.
(x2−9y2)(x2−y2−1)(x^2 - 9y^2)(x^2 - y^2 - 1)
Both parts of the grouped expression share (x2−9y2)(x^2 - 9y^2) as a GCF, leaving (x2−y2−1)(x^2 - y^2 - 1) when factored out.
4
Factor the difference of squares term completely over the integers.
(x−3y)(x+3y)(x2−y2−1)(x - 3y)(x + 3y)(x^2 - y^2 - 1)
The term (x2−9y2)(x^2 - 9y^2) is a difference of squares of the form a2−b2a^2 - b^2, which factors into (a−b)(a+b)(a - b)(a + b).

Key Concept

Factoring high-degree polynomials using quadratic form substitution, grouping, and difference of squares.
Question 20Question →

The sum of 99 and the product of a number and 66 less than that number is equal to 33 more than the number. What is the sum of all possible values of this number?

Show answer & explanation

Answer: 7

Answer

7
The correct answer is 77. The verbal description translates to x(x−6)+9=x+3x(x - 6) + 9 = x + 3. Distributing the left side gives x2−6x+9=x+3x^2 - 6x + 9 = x + 3. Subtracting xx and 33 from both sides results in x2−7x+6=0x^2 - 7x + 6 = 0. Factoring the quadratic expression gives (x−6)(x−1)=0(x - 6)(x - 1) = 0, which yields the solutions x=6x = 6 and x=1x = 1. Summing these values gives 6+1=76 + 1 = 7.

Step-by-Step Solution

1
Translate the verbal statement into an algebraic equation.
x(x−6)+9=x+3x(x - 6) + 9 = x + 3, where xx represents the unknown number.
Establishing the relationship between the algebraic expressions defined by the problem.
2
Expand the left side of the equation and combine like terms to set the equation to zero.
x2−6x+9=x+3  ⟹  x2−7x+6=0x^2 - 6x + 9 = x + 3 \implies x^2 - 7x + 6 = 0.
Quadratic equations must be set to zero before they can be solved by factoring.
3
Factor the quadratic equation over the integers.
(x−6)(x−1)=0(x - 6)(x - 1) = 0.
Finding two numbers that multiply to 66 and add up to −7-7 allows us to factor the trinomial.
4
Apply the zero-product property to find the individual roots.
x=6x = 6 and x=1x = 1.
If the product of two factors is zero, at least one of the factors must be zero.
5
Sum the possible values of the number.
6+1=76 + 1 = 7.
The question asks for the sum of all possible values of the number.

Key Concept

Solving Quadratic Equations by Factoring
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