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Factoring Polynomials

33 questions

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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 →

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 3Question →

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

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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 4Question →

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

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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 5Question →

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

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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 6Question →

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?

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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 7Question →

The polynomial x3+5x2−9x−45x^3 + 5x^2 - 9x - 45 can be factored completely into three linear factors of the form (x−a)(x−b)(x−c)(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 a−b+ca - b + c?

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Answer: 1

Answer

The correct answer is 1.
Factoring the polynomial by grouping gives (x2−9)(x+5)(x^2 - 9)(x + 5), which simplifies to (x−3)(x+3)(x+5)(x - 3)(x + 3)(x + 5) after factoring the difference of squares. Writing this expression in the form (x−a)(x−b)(x−c)(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 a−b+ca - b + c gives −5−(−3)+3=1-5 - (-3) + 3 = 1.

Step-by-Step Solution

1
Group the terms of the polynomial x3+5x2−9x−45x^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).
(x2−9)(x+5)(x^2 - 9)(x + 5)
Both terms share the common factor (x+5)(x + 5).
4
Factor the quadratic term x2−9x^2 - 9 as a difference of squares.
(x−3)(x+3)(x+5)(x - 3)(x + 3)(x + 5)
x2−9x^2 - 9 is a difference of squares, which factors into (x−3)(x+3)(x - 3)(x + 3).
5
Rewrite the factors in the form (x−a)(x−b)(x−c)(x - a)(x - b)(x - c) to identify the values of the constants.
(x−3)(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 (x−constant)(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 a−b+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 8Question →

When the polynomial 6x2−11x−106x^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 ad−bcad - bc?

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Answer: -19

Answer

The value of ad−bcad - bc is −19-19.
The correct answer is −19-19. Factoring the polynomial 6x2−11x−106x^2 - 11x - 10 yields (3x+2)(2x−5)(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 ad−bcad - bc gives (3)(−5)−(2)(2)=−15−4=−19(3)(-5) - (2)(2) = -15 - 4 = -19.

Step-by-Step Solution

1
Factor the quadratic expression 6x2−11x−106x^2 - 11x - 10.
(3x+2)(2x−5)(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 ad−bcad - 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 9Question →

When the polynomial 6x3−7x2−16x+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?

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Answer: 5

Answer

The value of the sum of the coefficients is 5.
The polynomial factors completely over the integers as (x−2)(2x+3)(3x−2)(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 (x−2)(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+5x−66x^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)(3x−2)(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 (1x−2)(2x+3)(3x−2)(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 10Question →

For all real values of xx, the expression 9x2−369x^2 - 36 is equivalent to which of the following?

Show answer & explanation

Answer: 9(x−2)(x+2)9(x - 2)(x + 2)

Answer

The equivalent expression is 9(x−2)(x+2)9(x - 2)(x + 2)
Factoring out the greatest common factor of 99 from the expression 9x2−369x^2 - 36 gives 9(x2−4)9(x^2 - 4). The binomial x2−4x^2 - 4 is a difference of squares (x2−22x^2 - 2^2), which can be factored into (x−2)(x+2)(x - 2)(x + 2). Combining these parts results in the equivalent expression 9(x−2)(x+2)9(x - 2)(x + 2).

Step-by-Step Solution

1
Identify the greatest common factor (GCF) of the terms in the expression 9x2−369x^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(x2−4)9(x^2 - 4)
This separates the common numeric factor from the quadratic binomial.
3
Factor the remaining binomial expression x2−4x^2 - 4 inside the parentheses.
x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2)
The expression x2−4x^2 - 4 is a difference of squares (x2−22x^2 - 2^2), which follows the factoring pattern a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b).
4
Combine the factored components to write the final equivalent expression.
9(x−2)(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 11Question →

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?

Show answer & explanation

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 12Question →

If the polynomial 12x2+10x−812x^2 + 10x - 8 is factored completely into the form k(ax−1)(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?

Show answer & explanation

Answer: 11

Answer

The value of k+a+b+ck + a + b + c is 1111.
To factor the polynomial 12x2+10x−812x^2 + 10x - 8 completely, we first factor out the greatest common factor of 22, yielding 2(6x2+5x−4)2(6x^2 + 5x - 4). Next, we factor the quadratic trinomial 6x2+5x−46x^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+8x−3x−4=2x(3x+4)−1(3x+4)=(2x−1)(3x+4)6x^2 + 8x - 3x - 4 = 2x(3x + 4) - 1(3x + 4) = (2x - 1)(3x + 4). The completely factored expression is 2(2x−1)(3x+4)2(2x - 1)(3x + 4). Comparing this with k(ax−1)(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+5x−4)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.
(2x−1)(3x+4)(2x - 1)(3x + 4)
Rewriting the trinomial as 6x2+8x−3x−46x^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(ax−1)(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(2x−1)(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 13Question →

A rectangular prism has a volume represented by the expression 3x3−5x2−12x+203x^3 - 5x^2 - 12x + 20 cubic centimeters. If the height of the prism is x−2x - 2 centimeters, which of the following expressions represents a possible length of the base of the prism, in centimeters, assuming the length and width are linear binomials with integer coefficients?

Show answer & explanation

Answer: 3x−53x - 5

Answer

The correct answer is the expression 3x−53x - 5.
The polynomial representing the volume can be factored by grouping: 3x3−5x2−12x+20=x2(3x−5)−4(3x−5)=(3x−5)(x2−4)3x^3 - 5x^2 - 12x + 20 = x^2(3x - 5) - 4(3x - 5) = (3x - 5)(x^2 - 4). Factoring the difference of squares yields (3x−5)(x−2)(x+2)(3x - 5)(x - 2)(x + 2). Since the height is x−2x - 2, the remaining dimensions of the base must be 3x−53x - 5 and x+2x + 2. Therefore, the expression 3x−53x - 5 is a possible length of the base.

Step-by-Step Solution

1
Group the terms of the cubic polynomial representing the volume: 3x3−5x2−12x+203x^3 - 5x^2 - 12x + 20.
(3x3−5x2)−(12x−20)(3x^3 - 5x^2) - (12x - 20)
Grouping terms allows us to factor the polynomial by grouping.
2
Factor out the greatest common factor (GCF) from each group.
x2(3x−5)−4(3x−5)x^2(3x - 5) - 4(3x - 5)
The GCF of 3x33x^3 and 5x25x^2 is x2x^2, and the GCF of 12x12x and 2020 is 44.
3
Factor out the common binomial factor (3x−5)(3x - 5).
(3x−5)(x2−4)(3x - 5)(x^2 - 4)
This rewrites the polynomial as a product of a linear binomial and a quadratic binomial.
4
Factor the quadratic term x2−4x^2 - 4 using the difference of squares identity.
(3x−5)(x−2)(x+2)(3x - 5)(x - 2)(x + 2)
The expression x2−4x^2 - 4 is a difference of squares, which factors into (x−2)(x+2)(x - 2)(x + 2).
5
Divide the factored volume by the height, x−2x - 2, to find the possible dimensions of the base.
The possible dimensions for the length and width of the base are 3x−53x - 5 and x+2x + 2.
Volume is the product of length, width, and height. Since the height is x−2x - 2, the remaining factors represent the length and width.

Key Concept

Factoring polynomials by grouping and difference of squares.
Estimated Time:2m 0s
Question 14Question →

Which of the following is a factor of the expression 2x4−8x2y2−2x2+8y22x^4 - 8x^2y^2 - 2x^2 + 8y^2 when it is factored completely?

Show answer & explanation

Answer: x−2yx - 2y

Answer

The correct answer is x−2yx - 2y because the completely factored form of the expression is 2(x−1)(x+1)(x−2y)(x+2y)2(x - 1)(x + 1)(x - 2y)(x + 2y), which contains x−2yx - 2y as a linear factor.
The expression 2x4−8x2y2−2x2+8y22x^4 - 8x^2y^2 - 2x^2 + 8y^2 can be factored by first pulling out the greatest common factor of 2, giving 2(x4−4x2y2−x2+4y2)2(x^4 - 4x^2y^2 - x^2 + 4y^2). Grouping the terms as x2(x2−4y2)−1(x2−4y2)x^2(x^2 - 4y^2) - 1(x^2 - 4y^2) produces 2(x2−1)(x2−4y2)2(x^2 - 1)(x^2 - 4y^2). Factoring the differences of squares yields the completely factored form 2(x−1)(x+1)(x−2y)(x+2y)2(x - 1)(x + 1)(x - 2y)(x + 2y). The expression x−2yx - 2y is one of these linear factors.

Step-by-Step Solution

1
Identify and factor out the greatest common factor (GCF) of the terms in the polynomial.
The terms 2x42x^4, −8x2y2-8x^2y^2, −2x2-2x^2, and 8y28y^2 share a common factor of 2. Factoring out 2 yields: 2(x4−4x2y2−x2+4y2)2(x^4 - 4x^2y^2 - x^2 + 4y^2).
Factoring out the GCF simplifies the remaining polynomial expression, making it easier to factor further.
2
Group the terms inside the parentheses to perform factoring by grouping.
Group the terms as follows: 2[(x4−4x2y2)−(x2−4y2)]2[(x^4 - 4x^2y^2) - (x^2 - 4y^2)]. Factor out x2x^2 from the first group: 2[x2(x2−4y2)−1(x2−4y2)]2[x^2(x^2 - 4y^2) - 1(x^2 - 4y^2)]. Now, factor out the common binomial (x2−4y2)(x^2 - 4y^2) to get: 2(x2−1)(x2−4y2)2(x^2 - 1)(x^2 - 4y^2).
Grouping allows us to find common binomial factors within the terms of the polynomial.
3
Apply the difference of squares identity, a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b), to the remaining binomial factors.
For the factor (x2−1)(x^2 - 1), the difference of squares gives (x−1)(x+1)(x - 1)(x + 1). For the factor (x2−4y2)(x^2 - 4y^2), the difference of squares gives (x−2y)(x+2y)(x - 2y)(x + 2y). Substituting these back into the expression yields: 2(x−1)(x+1)(x−2y)(x+2y)2(x - 1)(x + 1)(x - 2y)(x + 2y).
Both quadratic factors are differences of squares and must be factored completely to find all linear factors.
4
Compare the complete factorization with the given choices to find the matching factor.
The linear factor x−2yx - 2y is present in the completely factored expression.
This confirms the correct option based on algebraic factorization.

Key Concept

Factoring polynomials completely using GCF, grouping, and the difference of squares identity.

Alternative Method

Instead of factoring out the GCF 2 first, you can group the terms directly: 2x4−2x2−8x2y2+8y2=2x2(x2−1)−8y2(x2−1)=(2x2−8y2)(x2−1)2x^4 - 2x^2 - 8x^2y^2 + 8y^2 = 2x^2(x^2 - 1) - 8y^2(x^2 - 1) = (2x^2 - 8y^2)(x^2 - 1). Then, factor out 2 from the first binomial to get 2(x2−4y2)(x2−1)2(x^2 - 4y^2)(x^2 - 1), and finally apply the difference of squares identity to both quadratic factors to obtain 2(x−2y)(x+2y)(x−1)(x+1)2(x - 2y)(x + 2y)(x - 1)(x + 1).
Estimated Time:1m 30s
Question 15Question →

The trinomial 2x2+7x+32x^2 + 7x + 3 can be factored into the product of two binomials of the form (2x+a)(x+b)(2x + a)(x + b), where aa and bb are integers. What is the value of the expression a+2ba + 2b?

Show answer & explanation

Answer: 7

Answer

The value of the expression a+2ba + 2b is 7.
Expanding the factored template (2x+a)(x+b)(2x + a)(x + b) yields 2x2+(a+2b)x+ab2x^2 + (a + 2b)x + ab. Comparing this to the given expression 2x2+7x+32x^2 + 7x + 3, the coefficient of xx on the left side is a+2ba + 2b, and on the right side is 7. Therefore, a+2b=7a + 2b = 7. Alternatively, factoring 2x2+7x+32x^2 + 7x + 3 yields (2x+1)(x+3)(2x + 1)(x + 3), where a=1a = 1 and b=3b = 3. Substituting these integers into a+2ba + 2b gives 1+2(3)=71 + 2(3) = 7.

Step-by-Step Solution

1
Expand the expression (2x+a)(x+b)(2x + a)(x + b) using the FOIL method.
2x2+2bx+ax+ab=2x2+(a+2b)x+ab2x^2 + 2bx + ax + ab = 2x^2 + (a + 2b)x + ab
Expanding the template allows direct comparison of its coefficients with the given trinomial.
2
Equate the coefficients of the expanded template to the given trinomial 2x2+7x+32x^2 + 7x + 3.
a+2b=7a + 2b = 7 and ab=3ab = 3
For the two polynomial expressions to be equivalent for all values of xx, their corresponding coefficients must be equal.
3
Identify the requested value directly from the system of equations.
7
The question asks for the value of a+2ba + 2b, which is precisely the coefficient of the linear xx term.

Key Concept

Factoring quadratic trinomials with a leading coefficient greater than 1

Alternative Method

Factor the trinomial 2x2+7x+32x^2 + 7x + 3 using the AC method: multiply the leading coefficient (2) and the constant term (3) to get 6. Find two numbers that multiply to 6 and add to 7, which are 6 and 1. Rewrite the middle term: 2x2+6x+x+32x^2 + 6x + x + 3. Factor by grouping: 2x(x+3)+1(x+3)=(2x+1)(x+3)2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Compare this to (2x+a)(x+b)(2x + a)(x + b) to find a=1a = 1 and b=3b = 3, then compute a+2b=1+2(3)=7a + 2b = 1 + 2(3) = 7.
Estimated Time:45s
Question 16Question →

If the expression x4−8x2+16−9y2x^4 - 8x^2 + 16 - 9y^2 is factored completely over the integers, the product of the factors can be written as (x2−a−by)(x2−c+dy)(x^2 - a - by)(x^2 - c + dy), where aa, bb, cc, and dd are positive integers. What is the value of a+b+c+da + b + c + d?

Show answer & explanation

Answer: 14

Answer

14
By grouping the first three terms, the expression x4−8x2+16x^4 - 8x^2 + 16 is recognized as (x2−4)2(x^2 - 4)^2. Substituting this back into the original expression gives (x2−4)2−(3y)2(x^2 - 4)^2 - (3y)^2. Applying the difference of squares identity, this factors into (x2−4−3y)(x2−4+3y)(x^2 - 4 - 3y)(x^2 - 4 + 3y). Comparing this result to (x2−a−by)(x2−c+dy)(x^2 - a - by)(x^2 - c + dy) where a,b,c,da, b, c, d are positive integers yields a=4a = 4, b=3b = 3, c=4c = 4, and d=3d = 3. Summing these values gives 4+3+4+3=144 + 3 + 4 + 3 = 14.

Step-by-Step Solution

1
Group the first three terms of the polynomial.
x4−8x2+16=(x2−4)2x^4 - 8x^2 + 16 = (x^2 - 4)^2
To recognize the perfect square trinomial structure in terms of x2x^2.
2
Rewrite the original expression using the grouped terms.
(x2−4)2−9y2=(x2−4)2−(3y)2(x^2 - 4)^2 - 9y^2 = (x^2 - 4)^2 - (3y)^2
To express the polynomial as a difference of squares.
3
Factor the expression using the difference of squares formula A2−B2=(A−B)(A+B)A^2 - B^2 = (A - B)(A + B).
(x2−4−3y)(x2−4+3y)(x^2 - 4 - 3y)(x^2 - 4 + 3y)
To obtain the completely factored form over the integers.
4
Compare the factored expression to the given template (x2−a−by)(x2−c+dy)(x^2 - a - by)(x^2 - c + dy) where a,b,c,da, b, c, d are positive integers.
a=4a = 4, b=3b = 3, c=4c = 4, d=3d = 3
To identify the values of the constants that satisfy the positivity constraint.
5
Calculate the sum of the identified values.
a+b+c+d=4+3+4+3=14a + b + c + d = 4 + 3 + 4 + 3 = 14
To answer the question.

Key Concept

Factoring by grouping and the difference of squares
Question 17Question →

For all real values of xx and yy, the expression 4x2−9y2−12y−44x^2 - 9y^2 - 12y - 4 can be factored into the form (2x+ay+b)(2x−cy−d)(2x + ay + b)(2x - cy - d), where a,b,ca, b, c, and dd are positive integers. What is the value of a+b+c+da + b + c + d?

Show answer & explanation

Answer: 10

Answer

The value of a+b+c+da + b + c + d is 10.
Grouping the yy terms gives 4x2−(9y2+12y+4)4x^2 - (9y^2 + 12y + 4). Factoring the quadratic within the parentheses gives 4x2−(3y+2)24x^2 - (3y + 2)^2. Applying the difference of squares identity (u2−v2)=(u+v)(u−v)(u^2 - v^2) = (u + v)(u - v) leads to (2x+(3y+2))(2x−(3y+2))=(2x+3y+2)(2x−3y−2)(2x + (3y + 2))(2x - (3y + 2)) = (2x + 3y + 2)(2x - 3y - 2). Matching this expression to the template (2x+ay+b)(2x−cy−d)(2x + ay + b)(2x - cy - d) reveals that a=3a = 3, b=2b = 2, c=3c = 3, and d=2d = 2, all of which are positive integers. The sum of these values is 3+2+3+2=103 + 2 + 3 + 2 = 10.

Step-by-Step Solution

1
Group the terms containing yy
4x2−(9y2+12y+4)4x^2 - (9y^2 + 12y + 4)
Grouping the terms allows us to identify a perfect square trinomial pattern.
2
Factor the trinomial inside the parentheses
4x2−(3y+2)24x^2 - (3y + 2)^2
The expression 9y2+12y+49y^2 + 12y + 4 is a perfect square trinomial of the form (3y)2+2(3y)(2)+22(3y)^2 + 2(3y)(2) + 2^2.
3
Apply the difference of squares identity
(2x+(3y+2))(2x−(3y+2))(2x + (3y + 2))(2x - (3y + 2))
The expression is in the form u2−v2u^2 - v^2, where u=2xu = 2x and v=3y+2v = 3y + 2. Using u2−v2=(u+v)(u−v)u^2 - v^2 = (u + v)(u - v) factors the expression.
4
Simplify the factored binomials by distributing signs
(2x+3y+2)(2x−3y−2)(2x + 3y + 2)(2x - 3y - 2)
Simplifying the expressions removes inner parentheses, allowing comparison with the target template.
5
Compare with the template (2x+ay+b)(2x−cy−d)(2x + ay + b)(2x - cy - d) to find the constants
a=3a = 3, b=2b = 2, c=3c = 3, d=2d = 2
Matching the terms directly gives ay=3y⇒a=3ay = 3y \Rightarrow a = 3, b=2b = 2, −cy=−3y⇒c=3-cy = -3y \Rightarrow c = 3, and −d=−2⇒d=2-d = -2 \Rightarrow d = 2.
6
Calculate the sum of the positive integers
10
The question asks for the value of a+b+c+da + b + c + d, which is 3+2+3+2=103 + 2 + 3 + 2 = 10.

Key Concept

Factoring polynomials using grouping, perfect square trinomials, and the difference of squares identity
Estimated Time:2m 0s
Question 18Question →

When the polynomial 4x4−37x2+94x^4 - 37x^2 + 9 is factored completely into linear factors of the form ax+bax + b, where aa and bb are integers and a>0a > 0, which of the following expressions represents the sum of these linear factors?

Show answer & explanation

Answer: 6x6x

Answer

The sum of the linear factors is 6x6x.
Factoring the polynomial 4x4−37x2+94x^4 - 37x^2 + 9 by substituting u=x2u = x^2 yields (4u−1)(u−9)(4u - 1)(u - 9), which becomes (4x2−1)(x2−9)(4x^2 - 1)(x^2 - 9). Applying the difference of squares identity to both terms results in the four linear factors (2x−1)(2x - 1), (2x+1)(2x + 1), (x−3)(x - 3), and (x+3)(x + 3). The sum of these factors is 6x6x.

Step-by-Step Solution

1
Substitute u=x2u = x^2 to rewrite the quartic polynomial as a quadratic expression.
4u2−37u+94u^2 - 37u + 9
This simplifies the polynomial from degree 4 to degree 2, making it easier to factor.
2
Factor the quadratic expression by grouping or finding two numbers that multiply to 3636 and add to −37-37.
(4u−1)(u−9)(4u - 1)(u - 9)
The numbers are −36-36 and −1-1. Rewriting and grouping gives 4u(u−9)−1(u−9)=(4u−1)(u−9)4u(u - 9) - 1(u - 9) = (4u - 1)(u - 9).
3
Substitute x2x^2 back in place of uu and factor the resulting difference of squares binomials.
(2x−1)(2x+1)(x−3)(x+3)(2x - 1)(2x + 1)(x - 3)(x + 3)
Since 4x2−1=(2x)2−124x^2 - 1 = (2x)^2 - 1^2 and x2−9=x2−32x^2 - 9 = x^2 - 3^2, both binomials can be factored completely using the difference of squares identity.
4
Sum the four linear factors.
6x6x
Combining the like terms gives (2x−1)+(2x+1)+(x−3)+(x+3)=(2x+2x+x+x)+(−1+1−3+3)=6x+0=6x(2x - 1) + (2x + 1) + (x - 3) + (x + 3) = (2x + 2x + x + x) + (-1 + 1 - 3 + 3) = 6x + 0 = 6x.

Key Concept

Factoring quartic polynomials using quadratic substitution and the difference of squares identity.
Question 19Question →

Which of the following is the completely factored form of the expression x3−9xx^3 - 9x?

Show answer & explanation

Answer: x(x−3)(x+3)x(x - 3)(x + 3)

Answer

x(x−3)(x+3)x(x - 3)(x + 3)
The expression x3−9xx^3 - 9x can be factored by first finding the greatest common factor of the terms. Since both terms share a factor of xx, factoring out xx yields x(x2−9)x(x^2 - 9). The binomial x2−9x^2 - 9 is a difference of squares because it can be written as x2−32x^2 - 3^2. Applying the difference of squares formula, a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b), gives (x−3)(x+3)(x - 3)(x + 3). Therefore, the completely factored form is x(x−3)(x+3)x(x - 3)(x + 3).

Step-by-Step Solution

1
Identify and factor out the greatest common factor (GCF) of the terms in the expression.
x(x2−9)x(x^2 - 9)
Both x3x^3 and 9x9x share a common factor of xx.
2
Factor the remaining binomial expression inside the parentheses using the difference of squares formula.
x(x−3)(x+3)x(x - 3)(x + 3)
The expression x2−9x^2 - 9 is a difference of squares, which factors as (a−b)(a+b)(a - b)(a + b) where a=xa = x and b=3b = 3.

Key Concept

Factoring out the greatest common factor and factoring a difference of squares.
Question 20Question →

The area of a rectangle is represented by the expression 18x3−50x18x^3 - 50x square inches. If the width of the rectangle is 2x2x inches, which of the following expressions represents the length of the rectangle, in inches?

Show answer & explanation

Answer: (3x−5)(3x+5)(3x - 5)(3x + 5)

Answer

The length of the rectangle is represented by the expression (3x−5)(3x+5)(3x - 5)(3x + 5).
The correct answer is the expression (3x−5)(3x+5)(3x - 5)(3x + 5) because the length is found by dividing the area, 18x3−50x18x^3 - 50x, by the width, 2x2x. Simplifying this quotient yields 9x2−259x^2 - 25. The expression 9x2−259x^2 - 25 is a difference of squares and factors completely into (3x−5)(3x+5)(3x - 5)(3x + 5).

Step-by-Step Solution

1
Set up the equation for the length of the rectangle by dividing the area by the width.
Length=18x3−50x2x\text{Length} = \frac{18x^3 - 50x}{2x}
Since the area of a rectangle is the product of its length and width, the length can be found by dividing the area by the width.
2
Perform the polynomial division by dividing each term of the numerator by the denominator 2x2x.
18x3−50x2x=18x32x−50x2x=9x2−25\frac{18x^3 - 50x}{2x} = \frac{18x^3}{2x} - \frac{50x}{2x} = 9x^2 - 25
Dividing a polynomial by a monomial requires dividing every term of the polynomial by that monomial.
3
Factor the resulting expression 9x2−259x^2 - 25 using the difference of squares identity, a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b).
9x2−25=(3x)2−52=(3x−5)(3x+5)9x^2 - 25 = (3x)^2 - 5^2 = (3x - 5)(3x + 5)
The quadratic expression is a difference of two perfect squares, which can be factored into a product of binomial conjugates.

Key Concept

Factoring a difference of squares after dividing a polynomial by a monomial
Estimated Time:1m 30s
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