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Factoring Polynomials

33 questions

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Question 21Question →

If the expression 6x2−11x−106x^2 - 11x - 10 is factored completely into the form (ax+b)(cx−d)(ax + b)(cx - d), where a,b,c,a, b, c, and dd are positive integers, what is the value of a+b+c+da + b + c + d?

Show answer & explanation

Answer: 12

Answer

The value of the sum of the coefficients and constants a+b+c+da + b + c + d is 1212.
The factored form of 6x2−11x−106x^2 - 11x - 10 is (3x+2)(2x−5)(3x + 2)(2x - 5). Matching this with (ax+b)(cx−d)(ax + b)(cx - d) where a,b,c,a, b, c, and dd are positive integers results in a=3a = 3, b=2b = 2, c=2c = 2, and d=5d = 5. The sum of these values is 3+2+2+5=123 + 2 + 2 + 5 = 12.

Step-by-Step Solution

1
Factor the quadratic trinomial 6x2−11x−106x^2 - 11x - 10 using the grouping method.
(3x+2)(2x−5)(3x + 2)(2x - 5)
Factoring splits the quadratic expression into its constituent linear binomial factors.
2
Equate the factored expression to the given form (ax+b)(cx−d)(ax + b)(cx - d) to find the values of a,b,c,a, b, c, and dd.
a=3a = 3, b=2b = 2, c=2c = 2, and d=5d = 5
Since the variables represent positive integers, we match the positive constant term to bb and the negative constant term to −d-d.
3
Sum the values of a,b,c,a, b, c, and dd.
1212
Calculating the final sum answers the target mathematical question.

Key Concept

Factoring quadratic polynomials with a leading coefficient greater than 1 using the grouping method.
Question 22Question →

The cubic polynomial 6x3−19x2+11x+66x^3 - 19x^2 + 11x + 6 can be factored completely over the integers in the form (x−a)(bx−c)(dx+e)(x - a)(bx - c)(dx + e), where aa, bb, cc, dd, and ee are positive integers. What is the value of a+b+c+d+ea + b + c + d + e?

Show answer & explanation

Answer: 11

Answer

The sum of the coefficients and constants from the factored form is 11.
By applying the Rational Root Theorem, we find the root x=2x = 2, which gives the factor (x−2)(x - 2). Dividing the original cubic expression by (x−2)(x - 2) yields 6x2−7x−36x^2 - 7x - 3. Factoring this quadratic expression by grouping yields (2x−3)(3x+1)(2x - 3)(3x + 1). Writing the completely factored form as (x−2)(2x−3)(3x+1)(x - 2)(2x - 3)(3x + 1) and comparing it to (x−a)(bx−c)(dx+e)(x - a)(bx - c)(dx + e) where a,b,c,d,ea, b, c, d, e are positive integers results in a=2a = 2, b=2b = 2, c=3c = 3, d=3d = 3, and e=1e = 1. The sum a+b+c+d+ea + b + c + d + e is equal to 11.

Step-by-Step Solution

1
Identify a rational root of 6x3−19x2+11x+66x^3 - 19x^2 + 11x + 6
x=2x = 2 is a root, meaning (x−2)(x - 2) is a factor.
Applying the Rational Root Theorem and testing potential integer root values.
2
Divide 6x3−19x2+11x+66x^3 - 19x^2 + 11x + 6 by (x−2)(x - 2)
The quotient is the quadratic polynomial 6x2−7x−36x^2 - 7x - 3.
To reduce the cubic polynomial to a quadratic expression that can be factored using standard trinomial methods.
3
Factor the quadratic trinomial 6x2−7x−36x^2 - 7x - 3
(2x−3)(3x+1)(2x - 3)(3x + 1)
Finding two numbers that multiply to −18-18 and add to −7-7 (which are −9-9 and 22) and factoring by grouping.
4
Match the factored form (x−2)(2x−3)(3x+1)(x - 2)(2x - 3)(3x + 1) to (x−a)(bx−c)(dx+e)(x - a)(bx - c)(dx + e)
a=2,b=2,c=3,d=3,e=1a = 2, b = 2, c = 3, d = 3, e = 1
Since a,b,c,d,a, b, c, d, and ee must be positive integers, the constant terms and signs uniquely determine the mapping of each factor.
5
Calculate the sum of a,b,c,d,a, b, c, d, and ee
2+2+3+3+1=112 + 2 + 3 + 3 + 1 = 11
To find the final numeric value requested by the question.

Key Concept

Factoring cubic polynomials by finding rational roots and factoring quadratic trinomials by grouping.
Estimated Time:2m 30s
Question 23Question →

An algebra student is asked to simplify a polynomial expression by factoring it completely. The student is given the expression 2x6−32x22x^6 - 32x^2. Which of the following is the completely factored form of the expression?

Show answer & explanation

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

Answer

The completely factored form of the expression is 2x2(x−2)(x+2)(x2+4)2x^2(x - 2)(x + 2)(x^2 + 4).
The correct answer is 2x2(x−2)(x+2)(x2+4)2x^2(x - 2)(x + 2)(x^2 + 4). First, find the greatest common factor (GCF) of 2x62x^6 and 32x232x^2, which is 2x22x^2. Factoring this out gives 2x2(x4−16)2x^2(x^4 - 16). Next, recognize that x4−16x^4 - 16 is a difference of squares since x4=(x2)2x^4 = (x^2)^2 and 16=4216 = 4^2. This factors into (x2−4)(x2+4)(x^2 - 4)(x^2 + 4). Finally, the term x2−4x^2 - 4 is also a difference of squares and factors into (x−2)(x+2)(x - 2)(x + 2), while the sum of squares x2+4x^2 + 4 cannot be factored further over the real numbers. Thus, the completely factored form is 2x2(x−2)(x+2)(x2+4)2x^2(x - 2)(x + 2)(x^2 + 4).

Step-by-Step Solution

1
Identify and factor out the greatest common factor (GCF) of the two terms in 2x6−32x22x^6 - 32x^2.
The GCF is 2x22x^2, so the expression becomes 2x2(x4−16)2x^2(x^4 - 16).
Factoring out the GCF simplifies the polynomial and reveals a difference of squares.
2
Recognize that x4−16x^4 - 16 is a difference of squares and factor it.
Since x4=(x2)2x^4 = (x^2)^2 and 16=4216 = 4^2, the expression factors as 2x2(x2−4)(x2+4)2x^2(x^2 - 4)(x^2 + 4).
The difference of squares formula a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b) applies directly to x4−16x^4 - 16.
3
Check if any remaining factors can be factored further, and factor them.
The term x2−4x^2 - 4 is another difference of squares (x2−22x^2 - 2^2), which factors into (x−2)(x+2)(x - 2)(x + 2). The term x2+4x^2 + 4 is a sum of squares and cannot be factored over the real numbers. The completely factored expression is 2x2(x−2)(x+2)(x2+4)2x^2(x - 2)(x + 2)(x^2 + 4).
To factor completely, all factorable terms must be broken down into their prime polynomial factors.

Key Concept

Factoring polynomials completely using the greatest common factor and the difference of squares.
Question 24Question →

Which of the following is the completely factored form of the expression 9x2−369x^2 - 36?

Show answer & explanation

Answer: 9(x−2)(x+2)9(x - 2)(x + 2)

Answer

The correct factored form is 9(x−2)(x+2)9(x - 2)(x + 2)
The expression 9x2−369x^2 - 36 has a greatest common factor of 99. Factoring out 99 yields 9(x2−4)9(x^2 - 4). The term inside the parentheses, x2−4x^2 - 4, is a difference of squares that can be factored as (x−2)(x+2)(x - 2)(x + 2). Combining these gives the completely factored form 9(x−2)(x+2)9(x - 2)(x + 2).

Step-by-Step Solution

1
Identify and factor out the greatest common factor (GCF) of the terms in the expression.
The GCF of 9x29x^2 and 3636 is 99. Factoring it out gives 9(x2−4)9(x^2 - 4).
Factoring out the GCF simplifies the remaining polynomial and is the first step in completely factoring an expression.
2
Recognize and factor the quadratic expression inside the parentheses.
The term x2−4x^2 - 4 is a difference of squares of the form a2−b2a^2 - b^2, where a=xa = x and b=2b = 2. Factoring it yields (x−2)(x+2)(x - 2)(x + 2).
A difference of squares a2−b2a^2 - b^2 always factors into (a−b)(a+b)(a - b)(a + b).
3
Combine the factored terms to write the completely factored expression.
Combining the GCF and the factored binomials gives 9(x−2)(x+2)9(x - 2)(x + 2).
This represents the expression written as a product of its prime polynomial factors.

Key Concept

Factoring polynomials by first extracting the greatest common factor (GCF) and then applying the difference of squares identity.

Alternative Method

Instead of factoring out the greatest common factor first, the expression can be factored as a difference of squares directly: 9x2−36=(3x)2−62=(3x−6)(3x+6)9x^2 - 36 = (3x)^2 - 6^2 = (3x - 6)(3x + 6). Then, a common factor of 33 can be factored out from each binomial: 3(x−2)⋅3(x+2)=9(x−2)(x+2)3(x - 2) \cdot 3(x + 2) = 9(x - 2)(x + 2).
Estimated Time:45s
Question 25Question →

The cubic polynomial 2x3+5x2−8x−202x^3 + 5x^2 - 8x - 20 can be factored completely over the integers into the form (x−a)(x+b)(cx+d)(x - a)(x + b)(cx + d), 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: 11

Answer

The value of a+b+c+da + b + c + d is 11.
The polynomial 2x3+5x2−8x−202x^3 + 5x^2 - 8x - 20 can be factored completely by first grouping the terms as x2(2x+5)−4(2x+5)=(x2−4)(2x+5)x^2(2x + 5) - 4(2x + 5) = (x^2 - 4)(2x + 5). Factoring the difference of squares x2−4x^2 - 4 yields (x−2)(x+2)(2x+5)(x - 2)(x + 2)(2x + 5). Matching this to the given form (x−a)(x+b)(cx+d)(x - a)(x + b)(cx + d) gives the positive integers a=2a = 2, b=2b = 2, c=2c = 2, and d=5d = 5. The sum of these values is 2+2+2+5=112 + 2 + 2 + 5 = 11.

Step-by-Step Solution

1
Group the terms of the polynomial 2x3+5x2−8x−202x^3 + 5x^2 - 8x - 20.
(2x3+5x2)−(8x+20)(2x^3 + 5x^2) - (8x + 20)
Grouping the terms allows us to look for common factors within each pair of terms.
2
Factor out the greatest common factor (GCF) from each group.
x2(2x+5)−4(2x+5)x^2(2x + 5) - 4(2x + 5)
The GCF of the first group 2x3+5x22x^3 + 5x^2 is x2x^2, and the GCF of the second group 8x+208x + 20 is 44.
3
Factor out the common binomial factor (2x+5)(2x + 5).
(x2−4)(2x+5)(x^2 - 4)(2x + 5)
Both terms share the binomial factor (2x+5)(2x + 5).
4
Factor the difference of squares x2−4x^2 - 4.
(x−2)(x+2)(2x+5)(x - 2)(x + 2)(2x + 5)
The term x2−4x^2 - 4 is a difference of squares, which factors as (x−2)(x+2)(x - 2)(x + 2).
5
Compare the factored expression with the template (x−a)(x+b)(cx+d)(x - a)(x + b)(cx + d) to determine the values of aa, bb, cc, and dd.
a=2a = 2, b=2b = 2, c=2c = 2, and d=5d = 5
Comparing the terms yields x−a=x−2  ⟹  a=2x - a = x - 2 \implies a = 2, x+b=x+2  ⟹  b=2x + b = x + 2 \implies b = 2, and cx+d=2x+5  ⟹  c=2,d=5cx + d = 2x + 5 \implies c = 2, d = 5. All values are positive integers as required.
6
Calculate the sum a+b+c+da + b + c + d.
11
Substituting the values of the variables into the expression gives 2+2+2+5=112 + 2 + 2 + 5 = 11.

Key Concept

Factoring a cubic polynomial by grouping and then factoring the resulting difference of squares.

Alternative Method

Instead of factoring by grouping, we can use the Rational Root Theorem to find rational roots of the polynomial. The possible rational roots of 2x3+5x2−8x−20=02x^3 + 5x^2 - 8x - 20 = 0 are of the form ±pq\pm \frac{p}{q}, where pp is a factor of 2020 and qq is a factor of 22. Testing values shows that x=2x = 2 and x=−2x = -2 are roots, which corresponds to the linear factors (x−2)(x - 2) and (x+2)(x + 2). Dividing the original cubic by their product, (x2−4)(x^2 - 4), yields the remaining linear factor (2x+5)(2x + 5).
Estimated Time:1m 30s
Question 26Question →

Which of the following is the completely factored form of the expression 3x3−12x2+12x3x^3 - 12x^2 + 12x?

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Answer: 3x(x−2)23x(x - 2)^2

Answer

The completely factored form of the expression is 3x(x−2)23x(x - 2)^2.
To factor the expression 3x3−12x2+12x3x^3 - 12x^2 + 12x completely, we first look for the greatest common factor (GCF) of the three terms. The GCF of 3x33x^3, −12x2-12x^2, and 12x12x is 3x3x. Factoring out 3x3x yields 3x(x2−4x+4)3x(x^2 - 4x + 4). Next, we factor the quadratic trinomial inside the parentheses. The expression x2−4x+4x^2 - 4x + 4 is a perfect square trinomial that factors into (x−2)2(x - 2)^2. Putting it all together, the completely factored form is 3x(x−2)23x(x - 2)^2.

Step-by-Step Solution

1
Identify and factor out the Greatest Common Factor (GCF) from all three terms of the polynomial 3x3−12x2+12x3x^3 - 12x^2 + 12x.
3x(x2−4x+4)3x(x^2 - 4x + 4)
Each term in the polynomial is divisible by 33, and the lowest power of xx common to all terms is x1x^1. Thus, the GCF is 3x3x.
2
Factor the remaining quadratic trinomial x2−4x+4x^2 - 4x + 4 inside the parentheses.
(x−2)2(x - 2)^2
The trinomial x2−4x+4x^2 - 4x + 4 is a perfect square trinomial matching the form a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a - b)^2, where a=xa = x and b=2b = 2.
3
Combine the factored parts to write the final completely factored expression.
3x(x−2)23x(x - 2)^2
Bringing together the GCF and the factored trinomial yields the simplest, fully factored form.

Key Concept

Factoring a polynomial completely by first extracting the greatest common factor (GCF) and then factoring the remaining perfect square trinomial.

Alternative Method

Instead of factoring directly, you can expand the answer choices to see which one is equivalent to the original polynomial. For example, expanding the correct option: 3x(x−2)2=3x(x2−4x+4)=3x3−12x2+12x3x(x-2)^2 = 3x(x^2 - 4x + 4) = 3x^3 - 12x^2 + 12x. This matches the original expression.
Estimated Time:1m 0s
Question 27Question →

When the polynomial 12x2+11x−1512x^2 + 11x - 15 is factored completely into the form (ax+b)(cx+d)(ax + b)(cx + d), where aa, bb, cc, and dd are integers such that a>c>0a > c > 0, what is the value of the constant term dd?

Show answer & explanation

Answer: 5

Answer

The value of the constant term dd is 55.
Factoring the trinomial 12x2+11x−1512x^2 + 11x - 15 completely gives (4x−3)(3x+5)(4x - 3)(3x + 5). Applying the constraint a>c>0a > c > 0 means the factor with the larger xx-coefficient must be written first in the template (ax+b)(cx+d)(ax + b)(cx + d). This yields a=4a = 4, b=−3b = -3, c=3c = 3, and d=5d = 5. Thus, the constant term dd is 55.

Step-by-Step Solution

1
Find the factor pair for the AC method.
We need two numbers that multiply to 12×(−15)=−18012 \times (-15) = -180 and add up to 1111. The numbers are 2020 and −9-9.
This allows us to split the linear middle term to factor by grouping.
2
Rewrite the polynomial and factor by grouping.
12x2+20x−9x−15=4x(3x+5)−3(3x+5)=(4x−3)(3x+5)12x^2 + 20x - 9x - 15 = 4x(3x + 5) - 3(3x + 5) = (4x - 3)(3x + 5).
Grouping the first two terms and the last two terms reveals a common binomial factor of (3x+5)(3x + 5).
3
Apply the given inequality constraints to match the template.
Comparing (4x−3)(3x+5)(4x - 3)(3x + 5) to (ax+b)(cx+d)(ax + b)(cx + d) with a>c>0a > c > 0 yields a=4a = 4, b=−3b = -3, c=3c = 3, and d=5d = 5.
Since the lead coefficient 44 is greater than 33, the factor (4x−3)(4x - 3) must correspond to (ax+b)(ax + b).

Key Concept

Factoring quadratic trinomials of the form Ax2+Bx+CAx^2 + Bx + C using the grouping (AC) method.
Question 28Question →

Which of the following is a factor of the polynomial 4x2−12xy+9y2−254x^2 - 12xy + 9y^2 - 25?

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Answer: 2x−3y−52x - 3y - 5

Answer

The expression 2x−3y−52x - 3y - 5 is a factor of the polynomial.
The polynomial can be factored by first grouping the first three terms as a perfect square trinomial: 4x2−12xy+9y2=(2x−3y)24x^2 - 12xy + 9y^2 = (2x - 3y)^2. This simplifies the expression to (2x−3y)2−25(2x - 3y)^2 - 25. Since 25=5225 = 5^2, this is a difference of squares of the form A2−B2A^2 - B^2, which factors into (A−B)(A+B)(A - B)(A + B). Substituting A=2x−3yA = 2x - 3y and B=5B = 5 gives the factored form (2x−3y−5)(2x−3y+5)(2x - 3y - 5)(2x - 3y + 5). Thus, the option representing 2x−3y−52x - 3y - 5 is a factor.

Step-by-Step Solution

1
Group the first three terms of the polynomial and recognize the perfect square trinomial pattern.
4x2−12xy+9y2=(2x−3y)24x^2 - 12xy + 9y^2 = (2x - 3y)^2
The term 4x24x^2 is (2x)2(2x)^2, 9y29y^2 is (3y)2(3y)^2, and the middle term −12xy-12xy is −2(2x)(3y)-2(2x)(3y), which matches the perfect square trinomial identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.
2
Rewrite the original expression using the factored trinomial and write 25 as a perfect square.
(2x−3y)2−52(2x - 3y)^2 - 5^2
Substituting the factored trinomial and expressing 25 as 525^2 sets up the expression as a difference of squares in the form A2−B2A^2 - B^2.
3
Apply the difference of squares factoring formula A2−B2=(A−B)(A+B)A^2 - B^2 = (A - B)(A + B).
(2x−3y−5)(2x−3y+5)(2x - 3y - 5)(2x - 3y + 5)
Substituting A=2x−3yA = 2x - 3y and B=5B = 5 into the formula yields the completely factored polynomial.

Key Concept

Factoring by grouping using perfect square trinomials and difference of squares
Estimated Time:1m 0s
Question 29Question →

If the polynomial 2x2+kx−122x^2 + kx - 12 can be factored as the product of two binomials with integer coefficients, what is the greatest possible integer value of kk?

Show answer & explanation

Answer: 23

Answer

The greatest possible integer value of kk is 23.
The greatest possible value of kk is 23. This is found by setting the factored form of the expression as (2x+a)(x+b)=2x2+(2b+a)x+ab(2x + a)(x + b) = 2x^2 + (2b + a)x + ab. We find that ab=−12ab = -12 and k=2b+ak = 2b + a. The possible integer pairs for (a,b)(a, b) that multiply to −12-12 are tested to find the maximum sum of 2b+a2b + a. The maximum is achieved when b=12b = 12 and a=−1a = -1, yielding 2(12)−1=232(12) - 1 = 23.

Step-by-Step Solution

1
Set up the algebraic representation for the binomial factors with integer coefficients.
The polynomial must factor into the form (2x+a)(x+b)(2x + a)(x + b) where aa and bb are integers.
Since the leading coefficient is 2 (a prime number), the coefficients of the linear terms in the binomial factors must be 2 and 1 to obtain a product of 2x22x^2.
2
Expand the product of the binomials to relate the parameters aa and bb to the original expression.
(2x+a)(x+b)=2x2+(2b+a)x+ab(2x + a)(x + b) = 2x^2 + (2b + a)x + ab, which implies ab=−12ab = -12 and k=2b+ak = 2b + a.
By equating the coefficients of corresponding terms in the expanded expression and the original polynomial, we establish relationships for the constant term and the linear coefficient.
3
Analyze the factors of −12-12 to find the integer values of aa and bb that maximize the linear term coefficient.
Choosing b=12b = 12 and a=−1a = -1 gives ab=−12ab = -12 and results in k=2(12)+(−1)=23k = 2(12) + (-1) = 23.
To maximize 2b+a2b + a, we choose the largest possible positive factor of −12-12 for the term multiplied by 2, which is 12, paired with the corresponding negative factor −1-1 for aa.

Key Concept

Factoring quadratic trinomials of the form ax2+bx+cax^2 + bx + c with a>1a > 1
Question 30Question →

Which of the following represents the completely factored form of the expression 4x4−16x24x^4 - 16x^2?

Show answer & explanation

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

Answer

The completely factored form of the expression is 4x2(x−2)(x+2)4x^2(x - 2)(x + 2).
The correct answer is found by first identifying the greatest common factor of the terms in the polynomial 4x4−16x24x^4 - 16x^2, which is 4x24x^2. Factoring this out results in 4x2(x2−4)4x^2(x^2 - 4). Next, the binomial x2−4x^2 - 4 is recognized as a difference of squares and is factored into (x−2)(x+2)(x - 2)(x + 2). Putting these together gives the completely factored form 4x2(x−2)(x+2)4x^2(x - 2)(x + 2).

Step-by-Step Solution

1
Identify and factor out the greatest common factor (GCF) of the terms in the expression 4x4−16x24x^4 - 16x^2.
The GCF of 4x44x^4 and 16x216x^2 is 4x24x^2. Factoring it out yields 4x2(x2−4)4x^2(x^2 - 4).
Factoring out the greatest common factor simplifies the polynomial and reveals remaining factorable patterns.
2
Factor the remaining binomial expression inside the parentheses, x2−4x^2 - 4.
Since x2−4x^2 - 4 is a difference of squares (x2−22x^2 - 2^2), it factors into (x−2)(x+2)(x - 2)(x + 2).
A difference of squares of the form a2−b2a^2 - b^2 always factors into (a−b)(a+b)(a - b)(a + b).
3
Combine the factored parts to write the completely factored expression.
4x2(x−2)(x+2)4x^2(x - 2)(x + 2)
This combines the GCF and the factored difference of squares to represent the original expression in its simplest factored parts.

Key Concept

Factoring a polynomial completely by first extracting the greatest common factor (GCF) and then applying the difference of squares formula.
Estimated Time:1m 0s
Question 31Question →

If the polynomial x4−13x2+36x^4 - 13x^2 + 36 is factored completely into the product of four linear binomials of the form (x−r1)(x−r2)(x−r3)(x−r4)(x - r_1)(x - r_2)(x - r_3)(x - r_4), where r1<r2<r3<r4r_1 < r_2 < r_3 < r_4, what is the value of the expression r1+2r2+3r3+4r4r_1 + 2r_2 + 3r_3 + 4r_4?

Show answer & explanation

Answer: 11

Answer

The value of the expression is 11.
Factoring the quartic polynomial x4−13x2+36x^4 - 13x^2 + 36 as a quadratic in x2x^2 yields (x2−9)(x2−4)(x^2 - 9)(x^2 - 4). Applying the difference of squares identity to each factor gives (x−3)(x+3)(x−2)(x+2)(x - 3)(x + 3)(x - 2)(x + 2). Matching these to the template (x−r1)(x−r2)(x−r3)(x−r4)(x - r_1)(x - r_2)(x - r_3)(x - r_4) with the condition r1<r2<r3<r4r_1 < r_2 < r_3 < r_4 yields r1=−3r_1 = -3, r2=−2r_2 = -2, r3=2r_3 = 2, and r4=3r_4 = 3. Evaluating the linear combination −3+2(−2)+3(2)+4(3)-3 + 2(-2) + 3(2) + 4(3) results in 1111.

Step-by-Step Solution

1
Substitute u=x2u = x^2 and factor the quadratic trinomial.
u2−13u+36=(u−9)(u−4)⇒(x2−9)(x2−4)u^2 - 13u + 36 = (u - 9)(u - 4) \Rightarrow (x^2 - 9)(x^2 - 4)
To reduce the degree of the polynomial and make it easier to factor.
2
Apply the difference of squares formula to each binomial factor.
(x−3)(x+3)(x−2)(x+2)(x - 3)(x + 3)(x - 2)(x + 2)
Both x2−9x^2 - 9 and x2−4x^2 - 4 are differences of squares.
3
Determine the roots r1,r2,r3,r4r_1, r_2, r_3, r_4 in ascending order.
r1=−3r_1 = -3, r2=−2r_2 = -2, r3=2r_3 = 2, r4=3r_4 = 3
We write each factor as (x−ri)(x - r_i) to find the roots, and then sort them from least to greatest according to the inequality constraint.
4
Calculate the value of the requested expression.
1111
Substitute the sorted values into the linear combination: (−3)+2(−2)+3(2)+4(3)=−3−4+6+12=11(-3) + 2(-2) + 3(2) + 4(3) = -3 - 4 + 6 + 12 = 11.

Key Concept

Factoring a quartic polynomial of quadratic form followed by factoring differences of squares.
Question 32Question →

Which of the following is the completely factored form of the expression 3x(x−2)2−12x3x(x - 2)^2 - 12x?

Show answer & explanation

Answer: 3x2(x−4)3x^2(x - 4)

Answer

The completely factored form of the expression is 3x2(x−4)3x^2(x - 4).
The correct answer is 3x2(x−4)3x^2(x-4). Factoring out the greatest common factor 3x3x from the terms 3x(x−2)23x(x-2)^2 and −12x-12x gives 3x[(x−2)2−4]3x[(x-2)^2-4]. The expression within the brackets is a difference of squares that can be factored as [(x−2)−2][(x−2)+2][(x-2)-2][(x-2)+2], which simplifies to x(x−4)x(x-4). Multiplying this by the GCF 3x3x yields 3x2(x−4)3x^2(x-4).

Step-by-Step Solution

1
Factor out the greatest common factor (GCF), 3x3x, from both terms of the expression 3x(x−2)2−12x3x(x - 2)^2 - 12x.
3x[(x−2)2−4]3x[(x - 2)^2 - 4]
Both terms 3x(x−2)23x(x-2)^2 and 12x12x share the common factors 33 and xx.
2
Factor the difference of squares inside the bracket, (x−2)2−4(x-2)^2 - 4, using the formula a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b) where a=x−2a = x-2 and b=2b = 2.
[(x−2)−2][(x−2)+2]=(x−4)(x)[(x-2)-2][(x-2)+2] = (x-4)(x)
Since 4=224 = 2^2, the terms inside the brackets form a difference of squares.
3
Combine the factored parts and simplify the expression by multiplying the variable terms.
3x⋅x(x−4)=3x2(x−4)3x \cdot x(x-4) = 3x^2(x-4)
Multiplying 3x3x by xx requires adding their exponents (1+1=21 + 1 = 2).

Key Concept

Factoring polynomials using the greatest common factor (GCF) and difference of squares.
Question 33Question →

If the quadratic expression 6x2−7x−56x^2 - 7x - 5 is factored completely into the product of two linear binomials of the form (ax+b)(cx+d)(ax + b)(cx + d), where aa, bb, cc, and dd are integers such that a>c>0a > c > 0, what is the value of the expression ad−bcad - bc?

Show answer & explanation

Answer: 13

Answer

The value of the expression ad−bcad - bc is 13.

Step-by-Step Solution

1
Factor the quadratic expression 6x2−7x−56x^2 - 7x - 5.
(3x−5)(2x+1)(3x - 5)(2x + 1)
Find two numbers that multiply to 6×(−5)=−306 \times (-5) = -30 and add to −7-7, which are −10-10 and 33. Rewrite the middle term and factor by grouping: 6x2−10x+3x−5=2x(3x−5)+1(3x−5)=(3x−5)(2x+1)6x^2 - 10x + 3x - 5 = 2x(3x - 5) + 1(3x - 5) = (3x - 5)(2x + 1).
2
Determine the values of the coefficients aa, bb, cc, and dd.
a=3a = 3, b=−5b = -5, c=2c = 2, and d=1d = 1
The expression is factored into the form (ax+b)(cx+d)(ax + b)(cx + d) where a>c>0a > c > 0. Comparing the factors (3x−5)(3x - 5) and (2x+1)(2x + 1), we see the coefficients of xx are 33 and 22. Since 3>2>03 > 2 > 0, we have a=3a = 3 and c=2c = 2. This leaves b=−5b = -5 and d=1d = 1.
3
Calculate the value of ad−bcad - bc.
13
Substitute a=3a = 3, b=−5b = -5, c=2c = 2, and d=1d = 1 into the expression: ad−bc=(3)(1)−(−5)(2)=3+10=13ad - bc = (3)(1) - (-5)(2) = 3 + 10 = 13.

Key Concept

Factoring quadratic polynomials of the form ax2+bx+cax^2 + bx + c with a>1a > 1.
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