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Question 1741Question

Match each quadratic equation with its correct set of real solutions.

Click a left item, then click its matching right item

Items

x25x+6=0x^2 - 5x + 6 = 0
x2+5x+6=0x^2 + 5x + 6 = 0
x2x6=0x^2 - x - 6 = 0

Matches

Show answer & explanation

Answer

The equation x25x+6=0x^2 - 5x + 6 = 0 matches with the solutions x=2x = 2 and x=3x = 3. The equation x2+5x+6=0x^2 + 5x + 6 = 0 matches with the solutions x=3x = -3 and x=2x = -2. The equation x2x6=0x^2 - x - 6 = 0 matches with the solutions x=2x = -2 and x=3x = 3.
Each equation is solved by factoring the quadratic trinomial into two binomials, then applying the zero product property to find the values of xx that make each factor zero. Specifically, x25x+6=0x^2 - 5x + 6 = 0 factors into (x2)(x3)=0(x - 2)(x - 3) = 0, yielding solutions x=2x = 2 and x=3x = 3. The equation x2+5x+6=0x^2 + 5x + 6 = 0 factors into (x+2)(x+3)=0(x + 2)(x + 3) = 0, yielding solutions x=2x = -2 and x=3x = -3. Finally, x2x6=0x^2 - x - 6 = 0 factors into (x3)(x+2)=0(x - 3)(x + 2) = 0, yielding solutions x=3x = 3 and x=2x = -2.

Step-by-Step Solution

1
Factor the quadratic equation x25x+6=0x^2 - 5x + 6 = 0.
(x2)(x3)=0(x - 2)(x - 3) = 0
Identify two integers that multiply to 66 and add up to 5-5. These integers are 2-2 and 3-3.
2
Solve for xx by setting each linear factor equal to zero: x2=0x - 2 = 0 and x3=0x - 3 = 0.
x=2x = 2 and x=3x = 3
Applying the zero product property means if the product of two numbers is zero, at least one of them must be zero.
3
Factor the quadratic equation x2+5x+6=0x^2 + 5x + 6 = 0.
(x+2)(x+3)=0(x + 2)(x + 3) = 0
Identify two integers that multiply to 66 and add up to 55. These integers are 22 and 33.
4
Solve for xx by setting each linear factor equal to zero: x+2=0x + 2 = 0 and x+3=0x + 3 = 0.
x=2x = -2 and x=3x = -3
Applying the zero product property gives the solutions as the negations of the terms inside the binomials.
5
Factor the quadratic equation x2x6=0x^2 - x - 6 = 0.
(x3)(x+2)=0(x - 3)(x + 2) = 0
Identify two integers that multiply to 6-6 and add up to 1-1. These integers are 3-3 and 22.
6
Solve for xx by setting each linear factor equal to zero: x3=0x - 3 = 0 and x+2=0x + 2 = 0.
x=3x = 3 and x=2x = -2
Setting the linear factors to zero yields the roots of the equation.

Key Concept

Solving Quadratic Equations by Factoring
Estimated Time:1m 30s
Question 1742Question

For all real values of aa and bb, what is the simplified form of the expression 3a3b2ab2(a3b5ab2)3a^3b - 2ab^2 - (a^3b - 5ab^2)?

Show answer & explanation

Answer: 2a3b+3ab22a^3b + 3ab^2

Answer

2a3b+3ab22a^3b + 3ab^2
The correct expression is found by distributing the negative sign across the parentheses to change the signs of the terms inside, yielding a3b+5ab2-a^3b + 5ab^2. Combining the like terms 3a3b3a^3b and a3b-a^3b gives 2a3b2a^3b, and combining 2ab2-2ab^2 and +5ab2+5ab^2 gives 3ab23ab^2.

Step-by-Step Solution

1
Distribute the negative sign to each term inside the parenthetical expression: (a3b5ab2)-(a^3b - 5ab^2).
a3b+5ab2-a^3b + 5ab^2
To remove the parentheses, we multiply each term inside by 1-1.
2
Rewrite the full expression with the parentheses removed: 3a3b2ab2a3b+5ab23a^3b - 2ab^2 - a^3b + 5ab^2.
3a3b2ab2a3b+5ab23a^3b - 2ab^2 - a^3b + 5ab^2
This sets up the expression for combining like terms.
3
Group and combine the like terms: (3a3ba3b)(3a^3b - a^3b) and (2ab2+5ab2)(-2ab^2 + 5ab^2).
2a3b+3ab22a^3b + 3ab^2
Only terms with the exact same variable parts and exponents can be combined by adding or subtracting their coefficients.

Key Concept

Simplifying algebraic expressions by distributing negative signs and combining like terms.
Estimated Time:45s
Question 1743Question

For what real value of xx is the rational expression 2x+1x7\frac{2x + 1}{x - 7} undefined? Fill in the blank with the correct number.

Fill in the blanks below

The rational expression 2x+1x7\frac{2x + 1}{x - 7} is undefined when $x = .
Show answer & explanation

Answer

The rational expression is undefined when the variable x equals 7.
A rational expression is undefined when its denominator is equal to 00. For the expression 2x+1x7\frac{2x + 1}{x - 7}, setting the denominator x7=0x - 7 = 0 and solving for xx yields x=7x = 7.

Step-by-Step Solution

1
Set the denominator of the rational expression equal to 0.
x7=0x - 7 = 0
A rational expression is undefined when its denominator is equal to 0 because division by zero is undefined in the real number system.
2
Solve the linear equation for xx by adding 7 to both sides.
x=7x = 7
Isolating the variable xx gives the value that makes the denominator zero.

Key Concept

Identifying values that make a rational expression undefined
Question 1744Question

Match each algebraic expression on the left with its equivalent simplified form on the right. Assume all variables represent real numbers.

Click a left item, then click its matching right item

Items

2(3a24b)3(a22b)2(3a^2 - 4b) - 3(a^2 - 2b)
a(3ab)2b(ab)a(3a - b) - 2b(a - b)
(a+b)(3a2b)b2(a + b)(3a - 2b) - b^2
4a2(ab)(a+2b)2b24a^2 - (a - b)(a + 2b) - 2b^2

Matches

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Answer

The expression 2(3a24b)3(a22b)2(3a^2 - 4b) - 3(a^2 - 2b) matches 3a22b3a^2 - 2b; a(3ab)2b(ab)a(3a - b) - 2b(a - b) matches 3a23ab+2b23a^2 - 3ab + 2b^2; (a+b)(3a2b)b2(a + b)(3a - 2b) - b^2 matches 3a2+ab3b23a^2 + ab - 3b^2; and 4a2(ab)(a+2b)2b24a^2 - (a - b)(a + 2b) - 2b^2 matches 3a2ab3a^2 - ab.
Each expression is correctly simplified by distributing coefficients, expanding binomial products, and collecting like terms.

Step-by-Step Solution

1
Simplify the expression 2(3a24b)3(a22b)2(3a^2 - 4b) - 3(a^2 - 2b)
3a22b3a^2 - 2b
Distribute the coefficients to remove the parentheses: 6a28b3a2+6b6a^2 - 8b - 3a^2 + 6b. Group the a2a^2 terms and the bb terms, and then combine: (63)a2+(8+6)b=3a22b(6 - 3)a^2 + (-8 + 6)b = 3a^2 - 2b.
2
Simplify the expression a(3ab)2b(ab)a(3a - b) - 2b(a - b)
3a23ab+2b23a^2 - 3ab + 2b^2
Distribute the variables aa and 2b-2b: 3a2ab2ab+2b23a^2 - ab - 2ab + 2b^2. Combine the like terms ab-ab and 2ab-2ab to get 3ab-3ab, resulting in 3a23ab+2b23a^2 - 3ab + 2b^2.
3
Simplify the expression (a+b)(3a2b)b2(a + b)(3a - 2b) - b^2
3a2+ab3b23a^2 + ab - 3b^2
Multiply the binomial factors using the distributive property: (a+b)(3a2b)=3a22ab+3ab2b2=3a2+ab2b2(a + b)(3a - 2b) = 3a^2 - 2ab + 3ab - 2b^2 = 3a^2 + ab - 2b^2. Subtract the remaining b2b^2 term: 3a2+ab2b2b2=3a2+ab3b23a^2 + ab - 2b^2 - b^2 = 3a^2 + ab - 3b^2.
4
Simplify the expression 4a2(ab)(a+2b)2b24a^2 - (a - b)(a + 2b) - 2b^2
3a2ab3a^2 - ab
Expand (ab)(a+2b)=a2+ab2b2(a - b)(a + 2b) = a^2 + ab - 2b^2. Subtract this product from 4a24a^2 by distributing the negative sign: 4a2a2ab+2b24a^2 - a^2 - ab + 2b^2. Finally, subtract the last term 2b22b^2: 3a2ab+2b22b2=3a2ab3a^2 - ab + 2b^2 - 2b^2 = 3a^2 - ab.

Key Concept

Simplifying algebraic expressions by distributing factors and combining like terms
Estimated Time:2m 0s
Question 1745Question

The polynomial x3+5x29x45x^3 + 5x^2 - 9x - 45 can be factored completely into three linear factors of the form (xa)(xb)(xc)(x - a)(x - b)(x - c), where aa, bb, and cc are integers such that a<b<ca < b < c. What is the value of ab+ca - b + c?

Show answer & explanation

Answer: 1

Answer

The correct answer is 1.
Factoring the polynomial by grouping gives (x29)(x+5)(x^2 - 9)(x + 5), which simplifies to (x3)(x+3)(x+5)(x - 3)(x + 3)(x + 5) after factoring the difference of squares. Writing this expression in the form (xa)(xb)(xc)(x - a)(x - b)(x - c) identifies the values as 33, 3-3, and 5-5. Ordering these values to satisfy the inequality a<b<ca < b < c yields a=5a = -5, b=3b = -3, and c=3c = 3. Evaluating ab+ca - b + c gives 5(3)+3=1-5 - (-3) + 3 = 1.

Step-by-Step Solution

1
Group the terms of the polynomial x3+5x29x45x^3 + 5x^2 - 9x - 45.
(x3+5x2)(9x+45)(x^3 + 5x^2) - (9x + 45)
Grouping allows factoring by finding common binomial terms in a cubic polynomial.
2
Factor out the greatest common factor (GCF) from each grouped term.
x2(x+5)9(x+5)x^2(x + 5) - 9(x + 5)
The GCF of the first group is x2x^2 and the GCF of the second group is 99.
3
Factor out the common binomial factor (x+5)(x + 5).
(x29)(x+5)(x^2 - 9)(x + 5)
Both terms share the common factor (x+5)(x + 5).
4
Factor the quadratic term x29x^2 - 9 as a difference of squares.
(x3)(x+3)(x+5)(x - 3)(x + 3)(x + 5)
x29x^2 - 9 is a difference of squares, which factors into (x3)(x+3)(x - 3)(x + 3).
5
Rewrite the factors in the form (xa)(xb)(xc)(x - a)(x - b)(x - c) to identify the values of the constants.
(x3)(x(3))(x(5))(x - 3)(x - (-3))(x - (-5)) which gives the set of values {3,3,5}\{3, -3, -5\}.
Matching the signs of the given form (xconstant)(x - \text{constant}) is necessary to correctly identify the values of the constants.
6
Sort the values in ascending order to satisfy a<b<ca < b < c.
a=5a = -5, b=3b = -3, and c=3c = 3
The inequality constraint requires sorting the values from smallest to largest.
7
Calculate the value of the expression ab+ca - b + c.
5(3)+3=1-5 - (-3) + 3 = 1
Substitute the sorted values into the target expression.

Key Concept

Factoring a cubic polynomial by grouping and difference of squares, and identifying algebraic constants under inequality constraints.
Question 1746Question

When the polynomial 6x211x106x^2 - 11x - 10 is factored into the form (ax+b)(cx+d)(ax + b)(cx + d), where aa, bb, cc, and dd are integers such that aa and cc are positive and a>ca > c, what is the value of adbcad - bc?

Show answer & explanation

Answer: -19

Answer

The value of adbcad - bc is 19-19.
The correct answer is 19-19. Factoring the polynomial 6x211x106x^2 - 11x - 10 yields (3x+2)(2x5)(3x + 2)(2x - 5). Under the constraints that aa and cc are positive and a>ca > c, we must have a=3a = 3, b=2b = 2, c=2c = 2, and d=5d = -5. Evaluating the expression adbcad - bc gives (3)(5)(2)(2)=154=19(3)(-5) - (2)(2) = -15 - 4 = -19.

Step-by-Step Solution

1
Factor the quadratic expression 6x211x106x^2 - 11x - 10.
(3x+2)(2x5)(3x + 2)(2x - 5)
Factoring by grouping is used to rewrite the quadratic trinomial.
2
Apply the positive coefficient constraints and a>ca > c to identify the constants.
a=3a = 3, b=2b = 2, c=2c = 2, d=5d = -5
Since the leading coefficients must be positive and a>ca > c, we assign a=3a = 3 from the first factor and c=2c = 2 from the second factor.
3
Evaluate the expression adbcad - bc.
19-19
Substitute the values of aa, bb, cc, and dd to calculate the final numerical value.

Key Concept

Factoring quadratic trinomials of the form ax2+bx+cax^2 + bx + c where a>1a > 1
Question 1747Question

For what real values of xx is the rational expression g(x)=x33x22x2x+4x24g(x) = \frac{x^3 - 3x^2}{\frac{2}{x-2} - \frac{x+4}{x^2 - 4}} undefined? Determine these three values and enter them in the blanks below in order from least to greatest.

Fill in the blanks below

The values of xx for which g(x)g(x) is undefined, ordered from least to greatest, are x=x = , x=x = , and x=x = .
Show answer & explanation

Answer

The values of xx for which the expression is undefined, ordered from least to greatest, are 2-2, 00, and 22.
The rational expression is undefined where any constituent denominator is equal to zero, or where the entire main denominator is equal to zero. The constituent denominators are x2x-2 and x24x^2-4, which are zero when x=2x = 2 or x=2x = -2. The entire main denominator simplifies to x(x2)(x+2)\frac{x}{(x-2)(x+2)}, which is zero when x=0x = 0. Therefore, the three values of xx that make the expression undefined are 2-2, 00, and 22.

Step-by-Step Solution

1
Identify the values of xx that make the denominators of the individual rational terms equal to zero.
The denominators are x2x - 2 and x24x^2 - 4. Setting these to zero gives x=2x = 2 and x=±2x = \pm 2. Thus, x=2x = 2 and x=2x = -2 are values that make the individual terms undefined.
Any rational term is undefined if its denominator is equal to zero.
2
Find the values of xx that make the overall denominator of the main fraction equal to zero by first simplifying it.
The main denominator is 2x2x+4x24\frac{2}{x-2} - \frac{x+4}{x^2-4}. Finding a common denominator yields 2(x+2)(x+4)(x2)(x+2)=x(x2)(x+2)\frac{2(x+2) - (x+4)}{(x-2)(x+2)} = \frac{x}{(x-2)(x+2)}. Setting this equal to zero gives x=0x = 0.
The main rational expression is undefined if its entire denominator is equal to zero.
3
Combine all restricted values and list them in order from least to greatest.
The restricted values are x=2x = -2, x=0x = 0, and x=2x = 2.
These are all the distinct real numbers that cause any division by zero in the original expression.

Key Concept

Finding the domain of nested rational expressions
Estimated Time:2m 0s
Question 1748Question

For all real numbers xx such that x0x \neq 0 and x1x \neq -1, which of the following expressions is equivalent to 3x+2x+1\frac{3}{x} + \frac{2}{x + 1}?

Show answer & explanation

Answer: 5x+3x(x+1)\frac{5x + 3}{x(x + 1)}

Answer

5x+3x(x+1)\frac{5x + 3}{x(x + 1)}
To add the rational expressions, find the common denominator x(x+1)x(x + 1). Multiplying the first term by x+1x+1\frac{x + 1}{x + 1} and the second term by xx\frac{x}{x} gives 3(x+1)+2xx(x+1)\frac{3(x + 1) + 2x}{x(x + 1)}. Expanding the numerator yields 3x+3+2x3x + 3 + 2x, which simplifies to 5x+35x + 3. Thus, the equivalent expression is 5x+3x(x+1)\frac{5x + 3}{x(x + 1)}.

Step-by-Step Solution

1
Identify a common denominator for the two rational expressions.
The common denominator for the denominators xx and x+1x + 1 is x(x+1)x(x + 1).
To add fractions with different denominators, they must be written with a common denominator.
2
Rewrite each fraction with the common denominator by multiplying the numerator and denominator of the first term by (x+1)(x + 1) and the second term by xx.
\frac{3(x + 1)}{x(x + 1)} + \frac{2x}{x(x + 1)}
This scales the fractions to have matching denominators without changing their value.
3
Add the numerators together over the common denominator and simplify.
\frac{3x + 3 + 2x}{x(x + 1)} = \frac{5x + 3}{x(x + 1)}
Combining like terms simplifies the numerator to its final form.

Key Concept

Adding rational expressions by finding a common denominator
Estimated Time:1m 0s
Question 1749Question

When the polynomial 6x37x216x+126x^3 - 7x^2 - 16x + 12 is factored completely into three linear factors of the form (ax+b)(cx+d)(ex+f)(ax + b)(cx + d)(ex + f), where aa, cc, and ee are positive integers, what is the value of a+b+c+d+e+fa + b + c + d + e + f?

Show answer & explanation

Answer: 5

Answer

The value of the sum of the coefficients is 5.
The polynomial factors completely over the integers as (x2)(2x+3)(3x2)(x - 2)(2x + 3)(3x - 2). The sum of the six coefficients is 1+(2)+2+3+3+(2)=51 + (-2) + 2 + 3 + 3 + (-2) = 5.

Step-by-Step Solution

1
Find one linear factor of the cubic polynomial using the Factor Theorem.
The root x=2x = 2 satisfies the equation, so (x2)(x - 2) is a factor.
Testing integer factors of the constant term 12 reveals that x=2x = 2 evaluates the polynomial to 0.
2
Perform synthetic division or polynomial long division to divide the cubic by the linear factor.
The quotient is the quadratic expression 6x2+5x66x^2 + 5x - 6.
This reduces the degree of the polynomial to allow quadratic factoring techniques.
3
Factor the quadratic quotient into two linear binomials.
The quadratic factors into (2x+3)(3x2)(2x + 3)(3x - 2).
Using the AC method, 6×(6)=366 \times (-6) = -36, and the factors of 36-36 that sum to 55 are 99 and 4-4.
4
Identify the coefficients and sum them.
The sum is 1+(2)+2+3+3+(2)=51 + (-2) + 2 + 3 + 3 + (-2) = 5.
The factors are (1x2)(2x+3)(3x2)(1x - 2)(2x + 3)(3x - 2), corresponding to the coefficients a=1,b=2,c=2,d=3,e=3,f=2a=1, b=-2, c=2, d=3, e=3, f=-2.

Key Concept

Complete factorization of cubic polynomials with integer coefficients using the Rational Root Theorem and quadratic factoring.
Question 1750Question

For each algebraic expression on the left, match it to its completely simplified equivalent expression on the right by distributing terms and combining like terms.

Click a left item, then click its matching right item

Items

2x(x23xy)3y(x2y2)(2x36x2y)2x(x^2 - 3xy) - 3y(x^2 - y^2) - (2x^3 - 6x^2y)
(2xy)38x(x23xy)y3(2x - y)^3 - 8x(x^2 - 3xy) - y^3
x(2x3y)2y(x2y)2(4x313x2y)x(2x - 3y)^2 - y(x - 2y)^2 - (4x^3 - 13x^2y)
2x2(x3y)(xy)3y2(3xy)2x^2(x - 3y) - (x - y)^3 - y^2(3x - y)

Matches

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Answer

The expressions match as follows: the first simplifies to 3x2y+3y3-3x^2y + 3y^3; the second simplifies to 12x2y+6xy22y312x^2y + 6xy^2 - 2y^3; the third simplifies to 13xy24y313xy^2 - 4y^3; and the fourth simplifies to x33x2y6xy2+2y3x^3 - 3x^2y - 6xy^2 + 2y^3.
Each expression is expanded fully by distributing multiplication and powers, then simplified by combining terms that have the exact same variable bases and exponents.

Step-by-Step Solution

1
Simplify the first expression by distributing coefficients and combining like terms.
3x2y+3y3-3x^2y + 3y^3
Expanding the expression gives 2x36x2y3x2y+3y32x3+6x2y2x^3 - 6x^2y - 3x^2y + 3y^3 - 2x^3 + 6x^2y. Grouping the like terms: (22)x3+(63+6)x2y+3y3(2 - 2)x^3 + (-6 - 3 + 6)x^2y + 3y^3, which simplifies to 3x2y+3y3-3x^2y + 3y^3.
2
Simplify the second expression using the binomial cube formula (ab)3=a33a2b+3ab2b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.
12x2y+6xy22y312x^2y + 6xy^2 - 2y^3
Expanding (2xy)3(2x - y)^3 yields 8x312x2y+6xy2y38x^3 - 12x^2y + 6xy^2 - y^3. Subtracting the remaining terms gives 8x312x2y+6xy2y38x3+24x2yy38x^3 - 12x^2y + 6xy^2 - y^3 - 8x^3 + 24x^2y - y^3. Combining like terms yields 12x2y+6xy22y312x^2y + 6xy^2 - 2y^3.
3
Simplify the third expression by squaring the binomials and distributing.
13xy24y313xy^2 - 4y^3
First expand the squares: (2x3y)2=4x212xy+9y2(2x - 3y)^2 = 4x^2 - 12xy + 9y^2 and (x2y)2=x24xy+4y2(x - 2y)^2 = x^2 - 4xy + 4y^2. Distributing the variables yields 4x312x2y+9xy2x2y+4xy24y34x3+13x2y4x^3 - 12x^2y + 9xy^2 - x^2y + 4xy^2 - 4y^3 - 4x^3 + 13x^2y. Combining like terms results in 13xy24y313xy^2 - 4y^3.
4
Simplify the fourth expression by expanding the cubic term and distributing signs.
x33x2y6xy2+2y3x^3 - 3x^2y - 6xy^2 + 2y^3
Expand (xy)3=x33x2y+3xy2y3(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3. Distribute all negative signs to get 2x36x2yx3+3x2y3xy2+y33xy2+y32x^3 - 6x^2y - x^3 + 3x^2y - 3xy^2 + y^3 - 3xy^2 + y^3. Combining like terms results in x33x2y6xy2+2y3x^3 - 3x^2y - 6xy^2 + 2y^3.

Key Concept

Simplifying Expressions and Combining Like Terms
Question 1751Question

What is the sum of the solutions to the quadratic equation x(x6)=16x(x - 6) = 16?

Show answer & explanation

Answer: 6

Answer

The sum of the solutions is 6.
Rearranging the equation to standard form gives x26x16=0x^2 - 6x - 16 = 0. Factoring the trinomial yields (x8)(x+2)=0(x - 8)(x + 2) = 0. Solving for xx by setting each factor to zero gives x=8x = 8 and x=2x = -2. The sum of these solutions is 8+(2)=68 + (-2) = 6.

Step-by-Step Solution

1
Distribute the variable on the left side of the equation.
x26x=16x^2 - 6x = 16
To solve a quadratic equation, we must first expand all products to identify the quadratic terms.
2
Subtract 16 from both sides of the equation to write it in standard form.
x26x16=0x^2 - 6x - 16 = 0
A quadratic equation must be set to zero before factoring.
3
Factor the quadratic trinomial.
(x8)(x+2)=0(x - 8)(x + 2) = 0
We need to find two numbers that multiply to 16-16 and add to 6-6. These numbers are 8-8 and 22.
4
Set each factor to zero and solve for xx.
x=8x = 8 and x=2x = -2
According to the zero product property, if the product of two factors is zero, at least one of the factors must be zero.
5
Add the two solutions to find their sum.
8+(2)=68 + (-2) = 6
The question asks for the sum of the solutions.

Key Concept

Solving quadratic equations by factoring after rearranging terms into standard form.
Question 1752Question

For all real values of xx, the expression 9x2369x^2 - 36 is equivalent to which of the following?

Show answer & explanation

Answer: 9(x2)(x+2)9(x - 2)(x + 2)

Answer

The equivalent expression is 9(x2)(x+2)9(x - 2)(x + 2)
Factoring out the greatest common factor of 99 from the expression 9x2369x^2 - 36 gives 9(x24)9(x^2 - 4). The binomial x24x^2 - 4 is a difference of squares (x222x^2 - 2^2), which can be factored into (x2)(x+2)(x - 2)(x + 2). Combining these parts results in the equivalent expression 9(x2)(x+2)9(x - 2)(x + 2).

Step-by-Step Solution

1
Identify the greatest common factor (GCF) of the terms in the expression 9x2369x^2 - 36.
The GCF of 9x29x^2 and 3636 is 99.
Factoring out the GCF simplifies the remaining polynomial expression.
2
Factor out the GCF of 99 from the original expression.
9(x24)9(x^2 - 4)
This separates the common numeric factor from the quadratic binomial.
3
Factor the remaining binomial expression x24x^2 - 4 inside the parentheses.
x24=(x2)(x+2)x^2 - 4 = (x - 2)(x + 2)
The expression x24x^2 - 4 is a difference of squares (x222x^2 - 2^2), which follows the factoring pattern a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b).
4
Combine the factored components to write the final equivalent expression.
9(x2)(x+2)9(x - 2)(x + 2)
Combining the GCF with the factored binomial factors yields the completely factored equivalent expression.

Key Concept

Factoring a polynomial by first removing a greatest common factor and then applying the difference of squares formula.
Question 1753Question

What is the sum of the distinct real solutions to the equation (2x1)2=(x+2)2(2x - 1)^2 = (x + 2)^2?

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Answer: 83\frac{8}{3}

Answer

The sum of the distinct real solutions is 83\frac{8}{3}.
To solve (2x1)2=(x+2)2(2x - 1)^2 = (x + 2)^2, we rearrange the equation to (2x1)2(x+2)2=0(2x - 1)^2 - (x + 2)^2 = 0. Using the difference of squares identity, we factor this into [(2x1)(x+2)][(2x1)+(x+2)]=0[(2x - 1) - (x + 2)][(2x - 1) + (x + 2)] = 0, which simplifies to (x3)(3x+1)=0(x - 3)(3x + 1) = 0. The solutions are x=3x = 3 and x=13x = -\frac{1}{3}. Summing these gives 3+(13)=833 + (-\frac{1}{3}) = \frac{8}{3}.

Step-by-Step Solution

1
Rearrange the equation by moving all terms to one side to set it to zero.
(2x1)2(x+2)2=0(2x - 1)^2 - (x + 2)^2 = 0
To solve a quadratic equation by factoring, it must first be set equal to zero.
2
Factor the expression using the difference of squares formula, a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b), where a=2x1a = 2x - 1 and b=x+2b = x + 2.
[(2x1)(x+2)][(2x1)+(x+2)]=0[(2x - 1) - (x + 2)][(2x - 1) + (x + 2)] = 0
Using the difference of squares allows us to factor the quadratic expression directly without fully expanding it.
3
Simplify the terms inside each set of brackets.
(2x1x2)(2x1+x+2)=0(x3)(3x+1)=0(2x - 1 - x - 2)(2x - 1 + x + 2) = 0 \Rightarrow (x - 3)(3x + 1) = 0
Simplifying the binomials reveals the two linear factors of the quadratic equation.
4
Set each linear factor to zero to find the distinct real solutions.
x3=0x=3x - 3 = 0 \Rightarrow x = 3 and 3x+1=0x=133x + 1 = 0 \Rightarrow x = -\frac{1}{3}
According to the zero product property, if the product of two factors is zero, at least one of the factors must be zero.
5
Calculate the sum of the distinct real solutions.
3+(13)=9313=833 + \left(-\frac{1}{3}\right) = \frac{9}{3} - \frac{1}{3} = \frac{8}{3}
The question asks for the sum of the solutions.

Key Concept

Solving quadratic equations by factoring, specifically utilizing the difference of squares method after rearranging terms.
Question 1754Question

Solve the equation 2x+7x=4\sqrt{2x + 7} - x = -4 for xx. What is the value of the real solution?

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Answer: 9

Answer

The only real solution to the equation is 9.
Isolating the radical yields 2x+7=x4\sqrt{2x + 7} = x - 4. Squaring both sides results in the quadratic equation 2x+7=x28x+162x + 7 = x^2 - 8x + 16, which simplifies to x210x+9=0x^2 - 10x + 9 = 0. Factoring gives (x9)(x1)=0(x - 9)(x - 1) = 0, yielding potential solutions of 9 and 1. Checking these solutions in the original equation reveals that 9 is valid, while 1 is extraneous. Therefore, the correct real solution is 9.

Step-by-Step Solution

1
Isolate the radical on one side of the equation.
2x+7=x4\sqrt{2x + 7} = x - 4
Before squaring both sides, the radical term must be isolated to simplify the algebraic manipulation.
2
Square both sides of the equation.
2x+7=(x4)22x + 7 = (x - 4)^2
Squaring both sides eliminates the square root.
3
Expand the squared binomial.
2x+7=x28x+162x + 7 = x^2 - 8x + 16
Applying the identity (ab)2=a22ab+b2(a - b)^2 = a^2 - 2ab + b^2 is necessary to write the equation in polynomial form.
4
Set the quadratic equation to zero.
x210x+9=0x^2 - 10x + 9 = 0
Subtracting 2x2x and 77 from both sides allows us to solve the quadratic equation.
5
Factor the quadratic equation.
(x9)(x1)=0(x - 9)(x - 1) = 0, giving potential solutions x=9x = 9 or x=1x = 1.
Finding the roots of the quadratic equation provides the candidate solutions.
6
Verify candidates in the original equation.
The solution x=9x = 9 is valid, while x=1x = 1 is extraneous.
Squaring both sides can introduce extraneous solutions, so each candidate must be checked in the original equation.

Key Concept

Solving radical equations and verifying for extraneous solutions
Question 1755Question

A local community center surveyed a group of 2020 students about the number of hours they volunteered last month. The table below displays the results of the survey, where aa and bb represent the number of students in their respective categories.

Hours VolunteeredNumber of Students
22aa
55bb
8844
121233
151522

If the mean number of hours volunteered per student for this group is 6.66.6, what is the median number of hours volunteered per student for these 2020 students?

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Answer: 5.0

Answer

The median number of hours volunteered per student is 5.0.
The correct answer is 5.0. To find this, we first establish two equations from the given information: the total count of students (a+b+9=20    a+b=11a + b + 9 = 20 \implies a + b = 11) and the mean of the data (2a+5b+32+36+3020=6.6    2a+5b=34\frac{2a + 5b + 32 + 36 + 30}{20} = 6.6 \implies 2a + 5b = 34). Solving this system of equations gives a=7a = 7 and b=4b = 4. To find the median of the 20 values, we look at the 10th and 11th sorted values. Since there are seven 2s (positions 1–7) and four 5s (positions 8–11), both the 10th and 11th values are 5, making the median 5.0.

Step-by-Step Solution

1
Set up an equation for the total number of students.
a+b=11a + b = 11
The total number of students in the survey is 20. Summing the frequencies gives a+b+4+3+2=20a + b + 4 + 3 + 2 = 20, which simplifies to a+b=11a + b = 11.
2
Set up an equation for the mean of the dataset.
2a+5b=342a + 5b = 34
The mean of the data is 6.6. Using the formula for the weighted mean: 2a+5b+8(4)+12(3)+15(2)20=6.6\frac{2a + 5b + 8(4) + 12(3) + 15(2)}{20} = 6.6. Multiplying both sides by 20 gives 2a+5b+98=1322a + 5b + 98 = 132, which simplifies to 2a+5b=342a + 5b = 34.
3
Solve the system of linear equations for aa and bb.
a=7a = 7 and b=4b = 4
Multiply the first equation by 2 to get 2a+2b=222a + 2b = 22. Subtract this from 2a+5b=342a + 5b = 34 to find 3b=12    b=43b = 12 \implies b = 4. Substitute b=4b = 4 back into the first equation to find a=7a = 7.
4
Find the median of the 20 sorted values.
5.0
With 20 data points, the median is the average of the 10th and 11th data values in ascending order. Since there are seven 2s followed by four 5s, the 10th and 11th values are both 5. The average of 5 and 5 is 5.0.

Key Concept

Determining the median of a grouped frequency distribution by solving for missing frequencies using the total count and the mean.
Question 1756Question

If kk is a positive constant and the expression 4x2+kx+94x^2 + kx + 9 can be written in the form (ax+b)2(ax + b)^2 for some integers aa and bb, what is the value of kk?

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Answer: 12

Answer

The value of kk is 12.
A perfect square trinomial is of the form (ax+b)2=a2x2+2abx+b2(ax + b)^2 = a^2x^2 + 2abx + b^2. Comparing this to 4x2+kx+94x^2 + kx + 9, we have a2=4a^2 = 4 and b2=9b^2 = 9. Taking the positive roots, a=2a = 2 and b=3b = 3. The coefficient of the middle term is k=2ab=2(2)(3)=12k = 2ab = 2(2)(3) = 12. Since kk is a positive constant, the correct value is 12.

Step-by-Step Solution

1
Identify the standard form of a perfect square trinomial.
A perfect square trinomial can be written as (ax+b)2=a2x2+2abx+b2(ax + b)^2 = a^2x^2 + 2abx + b^2.
This allows us to equate the coefficients of the given expression 4x2+kx+94x^2 + kx + 9 to the expanded form.
2
Solve for the values of a|a| and b|b| by equating the coefficients of x2x^2 and the constant term.
a2=4    a=2a^2 = 4 \implies |a| = 2 and b2=9    b=3b^2 = 9 \implies |b| = 3.
The square of the first term's coefficient is a2a^2 and the square of the last term's coefficient is b2b^2.
3
Calculate the middle term coefficient k=2abk = 2ab using the positive values since kk is a positive constant.
k=2×2×3=12k = 2 \times 2 \times 3 = 12.
The middle term of (ax+b)2(ax + b)^2 is 2abx2abx, so the coefficient kk must be 2ab2ab.

Key Concept

Factoring perfect square trinomials
Estimated Time:1m 0s
Question 1757Question

When the expression 3m(m2n)2(2mn)(m23mn+2n2)4n(m2mn)3m(m - 2n)^2 - (2m - n)(m^2 - 3mn + 2n^2) - 4n(m^2 - mn) is fully simplified by combining like terms, what is the coefficient of the m2nm^2n term?

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Answer: 9-9

Answer

9-9
The correct answer is 9-9. This is found by carefully expanding each term and distributing the negative signs: the first term expands to 3m312m2n+12mn23m^3 - 12m^2n + 12mn^2; the second term, when subtracted, becomes 2m3+7m2n7mn2+2n3-2m^3 + 7m^2n - 7mn^2 + 2n^3; and the third term simplifies to 4m2n+4mn2-4m^2n + 4mn^2. Summing the coefficients of the m2nm^2n terms gives 12+74=9-12 + 7 - 4 = -9.

Step-by-Step Solution

1
Expand the first term of the expression, 3m(m2n)23m(m - 2n)^2.
First, square the binomial: (m2n)2=m24mn+4n2(m - 2n)^2 = m^2 - 4mn + 4n^2. Next, distribute 3m3m to get 3m312m2n+12mn23m^3 - 12m^2n + 12mn^2.
To remove parentheses from the first term before combining like terms.
2
Expand the product in the second term, (2mn)(m23mn+2n2)(2m - n)(m^2 - 3mn + 2n^2).
2m(m23mn+2n2)n(m23mn+2n2)=2m36m2n+4mn2m2n+3mn22n32m(m^2 - 3mn + 2n^2) - n(m^2 - 3mn + 2n^2) = 2m^3 - 6m^2n + 4mn^2 - m^2n + 3mn^2 - 2n^3. Combining like terms within this product yields 2m37m2n+7mn22n32m^3 - 7m^2n + 7mn^2 - 2n^3.
To expand the binomial-trinomial product before applying the subtraction.
3
Subtract the expanded second term and expand the third term, 4n(m2mn)-4n(m^2 - mn).
Subtracting the second term gives 2m3+7m2n7mn2+2n3-2m^3 + 7m^2n - 7mn^2 + 2n^3. Distributing the negative sign in the third term gives 4m2n+4mn2-4m^2n + 4mn^2.
To distribute the negative signs across the remaining parenthetical expressions.
4
Identify and combine all the m2nm^2n terms to find the final coefficient.
The m2nm^2n terms are 12m2n-12m^2n (from the first term), +7m2n+7m^2n (from the subtracted second term), and 4m2n-4m^2n (from the third term). Combining these yields (12+74)m2n=9m2n(-12 + 7 - 4)m^2n = -9m^2n.
To determine the final coefficient of the m2nm^2n term.

Key Concept

Simplifying Expressions and Combining Like Terms
Estimated Time:2m 0s
Question 1758Question

If the polynomial 12x2+10x812x^2 + 10x - 8 is factored completely into the form k(ax1)(bx+c)k(ax - 1)(bx + c), where kk, aa, bb, and cc are positive integers, what is the value of k+a+b+ck + a + b + c?

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Answer: 11

Answer

The value of k+a+b+ck + a + b + c is 1111.
To factor the polynomial 12x2+10x812x^2 + 10x - 8 completely, we first factor out the greatest common factor of 22, yielding 2(6x2+5x4)2(6x^2 + 5x - 4). Next, we factor the quadratic trinomial 6x2+5x46x^2 + 5x - 4 by finding two numbers that multiply to 6×(4)=246 \times (-4) = -24 and add to 55. These numbers are 88 and 3-3. Splitting the linear term and factoring by grouping gives 6x2+8x3x4=2x(3x+4)1(3x+4)=(2x1)(3x+4)6x^2 + 8x - 3x - 4 = 2x(3x + 4) - 1(3x + 4) = (2x - 1)(3x + 4). The completely factored expression is 2(2x1)(3x+4)2(2x - 1)(3x + 4). Comparing this with k(ax1)(bx+c)k(ax - 1)(bx + c) where k,a,b,ck, a, b, c are positive integers, we determine that k=2k = 2, a=2a = 2, b=3b = 3, and c=4c = 4. Summing these values gives 2+2+3+4=112 + 2 + 3 + 4 = 11.

Step-by-Step Solution

1
Factor out the greatest common factor (GCF) from the terms of the polynomial.
2(6x2+5x4)2(6x^2 + 5x - 4)
Factoring out the greatest common factor simplifies the coefficients, making the quadratic trinomial easier to factor.
2
Find two integers that multiply to ac=6×(4)=24ac = 6 \times (-4) = -24 and add to b=5b = 5.
The two numbers are 88 and 3-3.
These integers are needed to split the linear term in order to factor the quadratic by grouping.
3
Rewrite the middle term and factor the trinomial by grouping.
(2x1)(3x+4)(2x - 1)(3x + 4)
Rewriting the trinomial as 6x2+8x3x46x^2 + 8x - 3x - 4 allows grouping of the first two terms 2x(3x+4)2x(3x + 4) and the last two terms 1(3x+4)-1(3x + 4) to extract the common binomial factor.
4
Combine the factors and match the coefficients to the form k(ax1)(bx+c)k(ax - 1)(bx + c).
k=2k = 2, a=2a = 2, b=3b = 3, and c=4c = 4
The completely factored expression is 2(2x1)(3x+4)2(2x - 1)(3x + 4). Matching this to the given template where all constants are positive integers yields k=2k = 2, a=2a = 2, b=3b = 3, and c=4c = 4.
5
Calculate the sum of the constants.
1111
Adding the values gives 2+2+3+4=112 + 2 + 3 + 4 = 11.

Key Concept

Factoring quadratic trinomials of the form ax2+bx+cax^2 + bx + c after removing a greatest common factor.
Question 1759Question

A rectangular garden is surrounded by a uniform gravel path that is 11 foot wide. The length of the garden is 33 feet less than twice its width. If the total area of the garden and the path combined is 117117 square feet, what is the width of the garden, in feet?

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Answer: 7

Answer

The width of the garden is 7 feet.
By representing the garden's width as ww, the length is 2w32w - 3. The combined dimensions including the 1-foot uniform path on all sides are w+2w + 2 and 2w12w - 1. Setting their product equal to the combined area of 117 square feet gives (w+2)(2w1)=117(w+2)(2w-1) = 117, which simplifies to the quadratic equation 2w2+3w119=02w^2 + 3w - 119 = 0. Factoring this equation yields (2w+17)(w7)=0(2w + 17)(w - 7) = 0, giving the solutions w=8.5w = -8.5 and w=7w = 7. Since width must be positive, the width of the garden is 77 feet.

Step-by-Step Solution

1
Define variables for the garden's dimensions and the combined dimensions including the path.
Garden width = ww, garden length = 2w32w - 3. Combined width = w+2w + 2, combined length = 2w12w - 1.
The path surrounds the garden uniformly, adding 11 foot of width to each of the four sides (adding 22 feet total to both overall width and overall length).
2
Write the area equation for the combined area.
(w+2)(2w1)=117(w + 2)(2w - 1) = 117
The total area of the garden and path combined is given as 117117 square feet.
3
Expand and rearrange the equation into standard quadratic form aw2+bw+c=0aw^2 + bw + c = 0.
2w2+3w119=02w^2 + 3w - 119 = 0
Expanding (w+2)(2w1)(w + 2)(2w - 1) gives 2w2+3w22w^2 + 3w - 2. Subtracting 117117 from both sides yields the standard form.
4
Factor the quadratic equation over the integers.
(2w+17)(w7)=0(2w + 17)(w - 7) = 0
We find two numbers that multiply to 2×(119)=2382 \times (-119) = -238 and sum to 33. These numbers are 1717 and 14-14. Rewriting the middle term and factoring by grouping yields (2w+17)(w7)=0(2w + 17)(w - 7) = 0.
5
Solve for ww and select the mathematically and physically valid solution.
w=7w = 7 (discarding the negative root w=8.5w = -8.5)
A physical measurement like width must be positive.

Key Concept

Solving quadratic word problems by setting up a quadratic equation and solving it by factoring.
Estimated Time:2m 0s
Question 1760Question

For the imaginary unit ii, if the complex number zz is defined by z=(1+2i)22iz = \frac{(1 + 2i)^2}{2 - i}, what is the real part of zz?

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Answer: 2-2

Answer

The real part of the complex number zz is 2-2.
To find the real part of the complex number, we first simplify the expression by expanding the squared binomial in the numerator, which yields 3+4i-3 + 4i. Next, we rationalize the fraction by multiplying both the numerator and the denominator by the complex conjugate of the denominator, 2+i2 + i. This multiplication yields 10+5i5\frac{-10 + 5i}{5}. Dividing both the real and imaginary terms by 55 results in the standard form 2+i-2 + i. Thus, the real part of this complex number is 2-2.

Step-by-Step Solution

1
Expand the squared binomial in the numerator of the expression for zz.
(1+2i)2=12+2(1)(2i)+(2i)2=1+4i+4i2(1 + 2i)^2 = 1^2 + 2(1)(2i) + (2i)^2 = 1 + 4i + 4i^2
Apply the algebraic identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2.
2
Simplify the expanded numerator using the definition of the imaginary unit.
1+4i+4(1)=3+4i1 + 4i + 4(-1) = -3 + 4i
Since i2=1i^2 = -1, the term 4i24i^2 simplifies to 4-4. Combining this with 11 gives the real part 3-3.
3
Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator to rationalize it.
z=3+4i2i2+i2+i=(3+4i)(2+i)(2i)(2+i)z = \frac{-3 + 4i}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{(-3 + 4i)(2 + i)}{(2 - i)(2 + i)}
Multiplying the denominator by its complex conjugate, 2+i2 + i, eliminates the imaginary unit from the denominator.
4
Expand and simplify the numerator and denominator.
z=63i+8i+4i24i2=6+5i44(1)=10+5i5z = \frac{-6 - 3i + 8i + 4i^2}{4 - i^2} = \frac{-6 + 5i - 4}{4 - (-1)} = \frac{-10 + 5i}{5}
Using the distributive property in the numerator gives 6+5i+4i2-6 + 5i + 4i^2. Since i2=1i^2 = -1, this simplifies to 10+5i-10 + 5i. In the denominator, (2i)(2+i)=4i2=5(2-i)(2+i) = 4 - i^2 = 5.
5
Divide both terms of the simplified numerator by the denominator to express zz in standard form a+bia + bi.
z=2+iz = -2 + i
Dividing the real part 10-10 by 55 yields the real part 2-2, and dividing the imaginary part 5i5i by 55 yields the imaginary part ii.

Key Concept

Division of complex numbers using the complex conjugate
Estimated Time:2m 0s
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