Simplifying and Factoring Algebraic Expressions Practice Questions — GRE General Test — Page 2

Question 21Question →

For all non-zero real numbers $x$ and $y$ such that $|x| \neq |y|$ and $x^2 + 2xy + 2y^2 \neq 0$ , consider the algebraic expression:

Q(x, y) = \frac{x^4 + 4y^4}{x^2 + 2xy + 2y^2} + \frac{2x^3 y - 2xy^3}{x^2 - y^2}

Which of the following expressions are equivalent to $Q(x, y)$ ? Indicate all such expressions.

Select all that apply

(x - y)^2 + 2xy + y^2

\frac{x^3 + 2xy^2}{x}

\frac{(x^2 + 2y^2)^2 - 4x^2 y^2}{x^2 - 2xy + 2y^2}

\frac{x^4 - 4y^4}{x^2 + 2y^2}

\frac{x^4 + 2x^2 y^2 + 2y^4}{x^2 + y^2}

Show answer & explanation

Answer: $(x - y)^2 + 2xy + y^2$ ; \frac{x^3 + 2xy^2}{x}

Answer

The expressions equivalent to

Q(x, y)

are

(x - y)^2 + 2xy + y^2

and

\frac{x^3 + 2xy^2}{x}

.

First, simplify

Q(x, y)

by factoring each term. Using Sophie Germain's identity on the numerator of the first term gives

x^4 + 4y^4 = (x^2 + 2y^2)^2 - (2xy)^2 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)

. Dividing this by

(x^2 + 2xy + 2y^2)

results in

x^2 - 2xy + 2y^2

. For the second term, factoring out

2xy

gives

\frac{2xy(x^2 - y^2)}{x^2 - y^2} = 2xy

. Adding the two simplified terms yields

Q(x, y) = (x^2 - 2xy + 2y^2) + 2xy = x^2 + 2y^2

.

Evaluating the options for equivalence:
- The expression

(x - y)^2 + 2xy + y^2

expands to

x^2 - 2xy + y^2 + 2xy + y^2 = x^2 + 2y^2

.
- The expression

\frac{x^3 + 2xy^2}{x}

factors as

\frac{x(x^2 + 2y^2)}{x} = x^2 + 2y^2

.
Therefore, both of these expressions are equivalent to

Q(x, y)

.

Step-by-Step Solution

1

Simplify the first algebraic fraction using Sophie Germain's identity.

\frac{x^4 + 4y^4}{x^2 + 2xy + 2y^2} = \frac{(x^2 + 2y^2)^2 - (2xy)^2}{x^2 + 2xy + 2y^2} = \frac{(x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)}{x^2 + 2xy + 2y^2} = x^2 - 2xy + 2y^2

Completing the square on

x^4 + 4y^4

allows it to be factored into the product of two quadratic expressions.

2

Simplify the second algebraic fraction by factoring out common factors.

\frac{2x^3 y - 2xy^3}{x^2 - y^2} = \frac{2xy(x^2 - y^2)}{x^2 - y^2} = 2xy

The term

(x^2 - y^2)

cancels out since

x \neq \pm y

.

3

Combine the simplified terms to find the closed-form expression for

Q(x, y)

.

Q(x, y) = (x^2 - 2xy + 2y^2) + 2xy = x^2 + 2y^2

The

-2xy

and

+2xy

terms sum to zero.

4

Test each option for equivalence to

x^2 + 2y^2

.

The option

(x - y)^2 + 2xy + y^2

expands to

x^2 - 2xy + y^2 + 2xy + y^2 = x^2 + 2y^2

. The option

\frac{x^3 + 2xy^2}{x}

simplifies directly to

x^2 + 2y^2

.

Both expressions reduce identically to

x^2 + 2y^2

for all non-zero

x

and

y

.

Key Concept

Simplifying complex algebraic expressions using polynomial factoring identities (Sophie Germain identity and difference of squares) and algebraic reduction.

Estimated Time:2m 0s

Question 22Question →

For all real numbers

x \neq -5

, the algebraic expression

\frac{x^3 + 125}{x^2 - 5x + 25} - \frac{x^2 - 25}{x + 5}

simplifies to a single constant value. What is the value of this constant?

Show answer & explanation

Answer: 10

Answer

The simplified value of the expression for all valid real numbers x is 10.

Factoring the numerators reveals that the first term reduces to

x + 5

and the second term reduces to

x - 5

. Subtracting

(x - 5)

from

(x + 5)

yields

(x + 5) - (x - 5) = 10

, which is constant for all valid values of

x

.

Step-by-Step Solution

1

Factor the numerator of the first rational term using the sum of cubes identity.

x^3 + 125 = (x + 5)(x^2 - 5x + 25)

The sum of cubes formula

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

applies with

a = x

and

b = 5

.

2

Cancel the non-zero factor

(x^2 - 5x + 25)

from the first fraction.

\frac{(x + 5)(x^2 - 5x + 25)}{x^2 - 5x + 25} = x + 5

The quadratic factor

x^2 - 5x + 25

has a negative discriminant (

(-5)^2 - 4(1)(25) = -75 < 0

), so it is never zero for any real number

x

.

3

Factor the numerator of the second rational term using the difference of squares identity.

x^2 - 25 = (x - 5)(x + 5)

The difference of squares formula

a^2 - b^2 = (a - b)(a + b)

applies with

a = x

and

b = 5

.

4

Cancel the common factor

(x + 5)

from the second fraction.

\frac{(x - 5)(x + 5)}{x + 5} = x - 5

Given

x \neq -5

, the factor

x + 5

is non-zero and can be canceled.

5

Subtract the two simplified terms.

(x + 5) - (x - 5) = x + 5 - x + 5 = 10

Distribute the negative sign to both terms in

(x - 5)

and combine like terms.

Key Concept

Factoring sum of cubes and difference of squares to simplify rational expressions

Question 23Question →

For all real numbers $x \neq 2$ , which of the following expressions is equivalent to $\frac{x^4 - 16}{x^3 - 2x^2 + 4x - 8}$ ?

x + 2

x - 2

x^2 + 4

\frac{x + 2}{x - 2}

\pm (x + 2)

Show answer & explanation

Answer: $x + 2$

Answer

x + 2

Factoring the numerator

x^4 - 16

as a difference of squares yields

(x - 2)(x + 2)(x^2 + 4)

. Factoring the denominator

x^3 - 2x^2 + 4x - 8

by grouping yields

(x - 2)(x^2 + 4)

. Dividing the numerator by the denominator allows the shared factor

(x - 2)(x^2 + 4)

to cancel, leaving

x + 2

.

Step-by-Step Solution

1

Factor the numerator

x^4 - 16

x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4)

Apply the difference of squares factorization identity twice.

2

Factor the denominator

x^3 - 2x^2 + 4x - 8

by grouping

x^3 - 2x^2 + 4x - 8 = x^2(x - 2) + 4(x - 2) = (x - 2)(x^2 + 4)

Group the terms in pairs and factor out the common binomial factor

(x - 2)

.

3

Simplify the rational expression by canceling common factors

\frac{(x - 2)(x + 2)(x^2 + 4)}{(x - 2)(x^2 + 4)} = x + 2

Cancel the non-zero common factor

(x - 2)(x^2 + 4)

present in both numerator and denominator.

Key Concept

Simplifying rational expressions using difference of squares and factoring by grouping

Question 24Question →

For all real numbers

x

, the algebraic expression

\frac{x^6 - 27}{x^4 + 3x^2 + 9}

can be simplified to the polynomial form

ax^2 + bx + c

, where

a

,

b

, and

c

are real constants. What is the value of

a + b + c

?

Show answer & explanation

Answer: -2

Answer

The simplified expression is

x^2 - 3

, which corresponds to polynomial coefficients

a = 1

,

b = 0

, and

c = -3

. The sum

a + b + c

equals

-2

.

Factoring the numerator

x^6 - 27

as a difference of cubes

(x^2)^3 - 3^3

produces

(x^2 - 3)(x^4 + 3x^2 + 9)

. Canceling the non-zero factor

(x^4 + 3x^2 + 9)

from the numerator and denominator simplifies the expression to

x^2 - 3

. In standard form

ax^2 + bx + c

,

a = 1

,

b = 0

, and

c = -3

. Adding these coefficients gives

1 + 0 + (-3) = -2

.

Step-by-Step Solution

1

Factor the numerator using the difference of cubes formula

x^6 - 27 = (x^2)^3 - 3^3 = (x^2 - 3)(x^4 + 3x^2 + 9)

The expression

x^6 - 27

matches the pattern

u^3 - v^3

with

u = x^2

and

v = 3

.

2

Simplify the rational expression by canceling the common quadratic-biquadratic factor

\frac{(x^2 - 3)(x^4 + 3x^2 + 9)}{x^4 + 3x^2 + 9} = x^2 - 3

Since

x^4 + 3x^2 + 9 > 0

for all real numbers

x

, the denominator is never zero, allowing direct cancellation of the common factor.

3

Match coefficients with

ax^2 + bx + c

and calculate

a + b + c

a = 1, b = 0, c = -3 \implies a + b + c = 1 + 0 + (-3) = -2

Comparing

x^2 - 3 = 1x^2 + 0x - 3

to

ax^2 + bx + c

determines the values of constants

a

,

b

, and

c

.

Key Concept

Difference of Cubes Factoring Identity

Estimated Time:1m 30s

Question 25Question →

For all real numbers

x

and

y

such that

x \neq -y

and

x \neq -2y

, which of the following expressions is equivalent to

\frac{x^3 + 2x^2y - xy^2 - 2y^3}{x^2 + 3xy + 2y^2}?

x - y

x + y

x^2 - y^2

x^2 + y^2

x - 2y

Show answer & explanation

Answer: $x - y$

Answer

x - y

Factoring the numerator by grouping gives

x^2(x + 2y) - y^2(x + 2y) = (x^2 - y^2)(x + 2y) = (x - y)(x + y)(x + 2y)

. Factoring the denominator yields

(x + y)(x + 2y)

. Dividing the numerator by the denominator cancels the common factors

(x + y)

and

(x + 2y)

, leaving

x - y

.

Step-by-Step Solution

1

Factor the numerator by grouping terms

x^3 + 2x^2y - xy^2 - 2y^3 = x^2(x + 2y) - y^2(x + 2y) = (x^2 - y^2)(x + 2y) = (x - y)(x + y)(x + 2y)

Grouping pairs of terms allows factoring out common binomial factors.

2

Factor the quadratic denominator

x^2 + 3xy + 2y^2 = (x + y)(x + 2y)

Finding two terms whose sum is

3y

and product is

2y^2

factors the quadratic in

x

.

3

Simplify the rational expression by canceling non-zero common factors

\frac{(x - y)(x + y)(x + 2y)}{(x + y)(x + 2y)} = x - y

Since

x \neq -y

and

x \neq -2y

, the factors

(x + y)

and

(x + 2y)

are non-zero and can be canceled.

Key Concept

Polynomial factoring by grouping, difference of squares, quadratic trinomial factoring, and simplifying rational algebraic expressions.

Estimated Time:1m 30s

Question 26Question →

For all real numbers $x$ such that $x \neq -2$ , $x \neq 2$ , and $x \neq 4$ , which of the following expressions are equivalent to $\frac{x^4 - 16x^2}{x^3 - 4x^2 - 4x + 16}$ ? Select all such expressions.

Select all that apply

\frac{x^3 + 4x^2}{x^2 - 4}

\frac{x^2(x + 4)}{(x - 2)(x + 2)}

\frac{x^2(x - 4)}{x^2 - 4}

\frac{x^2 + 4x}{x - 2}

\frac{x^4 - 4x^2}{x^3 - 4}

Show answer & explanation

Answer: $\frac{x^3 + 4x^2}{x^2 - 4}$ ; $\frac{x^2(x + 4)}{(x - 2)(x + 2)}$

Answer

The equivalent expressions are

\frac{x^3 + 4x^2}{x^2 - 4}

and

\frac{x^2(x + 4)}{(x - 2)(x + 2)}

.

The given expression factors as

\frac{x^2(x - 4)(x + 4)}{(x^2 - 4)(x - 4)}

. Canceling

(x - 4)

leaves

\frac{x^2(x + 4)}{x^2 - 4}

. Expanding the numerator gives

\frac{x^3 + 4x^2}{x^2 - 4}

, and factoring the denominator further gives

\frac{x^2(x + 4)}{(x - 2)(x + 2)}

. Both represent valid equivalent forms of the expression.

Step-by-Step Solution

1

Factor the numerator of the given algebraic expression.

x^4 - 16x^2 = x^2(x^2 - 16) = x^2(x - 4)(x + 4)

Factor out the greatest common factor

x^2

, then apply the difference of squares identity

a^2 - b^2 = (a - b)(a + b)

.

2

Factor the denominator by grouping terms.

x^3 - 4x^2 - 4x + 16 = x^2(x - 4) - 4(x - 4) = (x^2 - 4)(x - 4) = (x - 2)(x + 2)(x - 4)

Group the first two terms and last two terms, factor out common binomials, and expand the remaining difference of squares.

3

Simplify the full rational expression by canceling common factors.

\frac{x^2(x - 4)(x + 4)}{(x^2 - 4)(x - 4)} = \frac{x^2(x + 4)}{x^2 - 4} = \frac{x^3 + 4x^2}{x^2 - 4}

Since

x \neq 4

, cancel the common factor

(x - 4)

from both numerator and denominator.

4

Compare the simplified form to the given choices to identify all equivalent expressions.

Both

\frac{x^3 + 4x^2}{x^2 - 4}

and

\frac{x^2(x + 4)}{(x - 2)(x + 2)}

match the simplified algebraic forms.

Expanding the numerator or factoring the denominator yields these two equivalent representations.

Key Concept

Simplifying rational algebraic expressions by polynomial factoring and term grouping.

Question 27Question →

For all real numbers

x

and

y

such that

x \neq y

, which of the following expressions is equivalent to

\frac{x^3 - x^2y - xy^2 + y^3}{x - y} - (x - y)^2

?

2xy - 2y^2

2xy

-2y^2

2x^2 - 2xy

0

Show answer & explanation

Answer: $2xy - 2y^2$

Answer

The equivalent expression is

2xy - 2y^2

.

Factoring the numerator by grouping gives

x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y)

. Dividing by

(x - y)

yields

x^2 - y^2

. Expanding

(x - y)^2

gives

x^2 - 2xy + y^2

. Subtracting this expansion from

x^2 - y^2

results in

(x^2 - y^2) - (x^2 - 2xy + y^2) = 2xy - 2y^2

.

Step-by-Step Solution

1

Factor the numerator of the rational fraction by grouping terms

x^3 - x^2y - xy^2 + y^3 = x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y) = (x + y)(x - y)^2

Grouping terms allows factoring out common binomial components.

2

Simplify the fraction by dividing by

(x - y)

\frac{(x + y)(x - y)^2}{x - y} = (x + y)(x - y) = x^2 - y^2

Since

x \neq y

,

x - y \neq 0

, so common factors can be canceled.

3

Expand the subtracted term

(x - y)^2

(x - y)^2 = x^2 - 2xy + y^2

Applying the standard perfect square trinomial expansion formula.

4

Subtract the expanded expression from the simplified fraction result

(x^2 - y^2) - (x^2 - 2xy + y^2) = x^2 - y^2 - x^2 + 2xy - y^2 = 2xy - 2y^2

Distribute the negative sign across all terms inside the parentheses and combine like terms.

Key Concept

Simplifying algebraic expressions using factoring by grouping, difference of squares, and expanding binomial squares.

Question 28Question →

For all real numbers

x

such that

x \neq -3

,

x \neq 0

, and

x \neq 3

, which of the following expressions is equivalent to

\frac{x^4 - 81}{x^3 + 3x^2} \div \frac{x^2 + 9}{x^2}

?

x - 3

x + 3

x^2 - 3

\frac{x - 3}{x}

x - 9

Show answer & explanation

Answer: $x - 3$

Answer

x - 3

Factoring the numerator

x^4 - 81

gives

(x - 3)(x + 3)(x^2 + 9)

and factoring the denominator

x^3 + 3x^2

gives

x^2(x + 3)

. Multiplying by the reciprocal

\frac{x^2}{x^2 + 9}

allows the factors

(x + 3)

,

(x^2 + 9)

, and

x^2

to cancel completely, leaving

x - 3

.

Step-by-Step Solution

1

Rewrite division of rational expressions as multiplication by the reciprocal

\frac{x^4 - 81}{x^3 + 3x^2} \cdot \frac{x^2}{x^2 + 9}

Dividing by a fraction is equivalent to multiplying by its reciprocal.

2

Factor the numerator

x^4 - 81

using the difference of squares formula twice

x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)

Recognize

x^4 - 81

as

(x^2)^2 - 9^2

and then factor

x^2 - 9

further into

(x - 3)(x + 3)

.

3

Factor out the greatest common factor from the denominator

x^3 + 3x^2

x^3 + 3x^2 = x^2(x + 3)

Extract

x^2

from both terms of the binomial.

4

Substitute the factored expressions into the product and cancel common factors

\frac{(x - 3)(x + 3)(x^2 + 9)}{x^2(x + 3)} \cdot \frac{x^2}{x^2 + 9} = x - 3

The common terms

(x + 3)

,

(x^2 + 9)

, and

x^2

cancel out completely from the numerator and denominator for all

x \neq -3, 0, 3

.

Key Concept

Factoring polynomial expressions using difference of squares and common terms to simplify rational expressions.

Estimated Time:1m 30s

Question 29Question →

For all real numbers $x$ such that $x \neq -2$ , which of the following expressions are equivalent to $\frac{x^6 - 64}{x^3 + 8}$ ? Select all such expressions.

Select all that apply

x^3 - 8

(x - 2)(x^2 + 2x + 4)

(x - 2)^3

x^3 + 8

x^2 - 8

Show answer & explanation

Answer: $x^3 - 8$ ; $(x - 2)(x^2 + 2x + 4)$

Answer

The expressions equivalent to the given rational expression are

x^3 - 8

and

(x - 2)(x^2 + 2x + 4)

.

Factoring the numerator

x^6 - 64

as a difference of squares yields

(x^3 - 8)(x^3 + 8)

. Dividing by the denominator

(x^3 + 8)

leaves

x^3 - 8

. Further factoring

x^3 - 8

as a difference of cubes gives

(x - 2)(x^2 + 2x + 4)

. Both

x^3 - 8

and

(x - 2)(x^2 + 2x + 4)

are valid equivalent expressions.

Step-by-Step Solution

1

Factor the numerator using the difference of squares identity

x^6 - 64 = (x^3)^2 - 8^2 = (x^3 - 8)(x^3 + 8)

The expression

x^6 - 64

is a difference of two squares.

2

Simplify the fraction by canceling the common non-zero term

(x^3 + 8)

\frac{(x^3 - 8)(x^3 + 8)}{x^3 + 8} = x^3 - 8

Since

x \neq -2

,

x^3 + 8 \neq 0

, so

(x^3 + 8)

can be canceled from numerator and denominator.

3

Factor

x^3 - 8

using the difference of cubes identity

x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)

Applying

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

with

a = x

and

b = 2

.

Key Concept

Simplifying rational expressions by factoring polynomial numerators and denominators using algebraic identities such as difference of squares and difference of cubes.

Question 30Question →

For all real numbers

x

such that

x \neq -3

,

x \neq 0

, and

x \neq 3

, which of the following expressions is equivalent to

\frac{x^{-2} - 9^{-1}}{x^{-1} + 3^{-1}} \cdot \frac{3x^2 + 9x}{x - 3}?

-x - 3

x + 3

x - 3

\frac{x + 3}{3x}

-x + 3

Show answer & explanation

Answer: $-x - 3$

Answer

-x - 3

The numerator of the first factor is a difference of squares

(x^{-1} - 3^{-1})(x^{-1} + 3^{-1})

. Dividing by

x^{-1} + 3^{-1}

leaves

x^{-1} - 3^{-1} = \frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}

. Factoring

3x

out of the numerator of the second expression gives

3x(x + 3)

. Multiplying these terms yields

\frac{3 - x}{3x} \cdot \frac{3x(x + 3)}{x - 3}

. Since

3 - x = -(x - 3)

, the binomials

(3 - x)

and

(x - 3)

cancel to

-1

, and

3x

cancels out completely, resulting in

-(x + 3) = -x - 3

.

Step-by-Step Solution

1

Factor the numerator of the first expression as a difference of squares.

x^{-2} - 9^{-1} = (x^{-1})^2 - (3^{-1})^2 = (x^{-1} - 3^{-1})(x^{-1} + 3^{-1})

Expressing negative exponents as squares allows cancellation with the denominator.

2

Divide by

(x^{-1} + 3^{-1})

and convert negative exponents into a single rational term.

\frac{(x^{-1} - 3^{-1})(x^{-1} + 3^{-1})}{x^{-1} + 3^{-1}} = x^{-1} - 3^{-1} = \frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}

Finding a common denominator simplifies the first factor.

3

Factor the numerator of the second expression

3x^2 + 9x

.

3x^2 + 9x = 3x(x + 3)

Factoring out the greatest common factor

3x

enables further simplification.

4

Multiply the simplified expressions and reduce.

\left(\frac{3 - x}{3x}\right) \cdot \left(\frac{3x(x + 3)}{x - 3}\right) = \frac{-(x - 3)}{3x} \cdot \frac{3x(x + 3)}{x - 3} = -(x + 3) = -x - 3

Canceling

3x

and noting that

(3 - x)/(x - 3) = -1

leaves the linear expression

-x - 3

.

Key Concept

Simplifying algebraic expressions containing negative exponents, rational fractions, and opposite-sign binomial factors

Question 31Question →

When the algebraic expression $\frac{4x^3 - 16x}{2x^2 + 4x}$ is simplified for all $x \neq 0$ and $x \neq -2$ , it reduces to the linear polynomial $ax + b$ , where $a$ and $b$ are constants. What is the value of $a + b$ ?

Show answer & explanation

Answer: -2

Answer

The correct numerical answer is -2.

Factoring the numerator yields

4x(x - 2)(x + 2)

and factoring the denominator yields

2x(x + 2)

. Canceling the common factors

2x

and

(x + 2)

leaves

2(x - 2) = 2x - 4

. Comparing

2x - 4

to

ax + b

gives

a = 2

and

b = -4

. Summing these values gives

a + b = 2 + (-4) = -2

.

Step-by-Step Solution

1

Factor out the greatest common factor and apply the difference of squares formula to the numerator.

4x^3 - 16x = 4x(x^2 - 4) = 4x(x - 2)(x + 2)

Fully factoring the numerator allows identification of all linear factors.

2

Factor out the greatest common factor from the denominator.

2x^2 + 4x = 2x(x + 2)

Extracting

2x

reveals the common terms shared with the numerator.

3

Divide the numerator by the denominator by canceling identical non-zero factors

2x

and

(x + 2)

.

\frac{4x(x - 2)(x + 2)}{2x(x + 2)} = \frac{4}{2}(x - 2) = 2(x - 2) = 2x - 4

Simplifying rational expressions requires canceling common factors present in both numerator and denominator.

4

Compare the simplified expression

2x - 4

to the form

ax + b

to find

a

and

b

, then compute their sum.

a = 2

and

b = -4 \implies a + b = 2 + (-4) = -2

Matching corresponding terms identifies the values of the target constants.

Key Concept

Simplifying rational expressions by factoring out common terms and applying the difference of squares identity.

Question 32Question →

For all real numbers $x$ and $y$ such that $y \neq 0$ and $3x^2 + y^2 \neq 0$ , which of the following expressions is equivalent to $\frac{(x + y)^3 - (x - y)^3}{2y(3x^2 + y^2)}$ ?

1

\frac{y^2}{3x^2 + y^2}

\frac{3x^2}{3x^2 + y^2}

\frac{3x + y}{3x^2 + y^2}

\frac{1}{3}

Show answer & explanation

Answer: $1$

Answer

1

Expanding the numerator yields

(x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) = 6x^2y + 2y^3 = 2y(3x^2 + y^2)

. Since the numerator and denominator are identical non-zero expressions, the quotient simplifies to 1.

Step-by-Step Solution

1

Expand

(x + y)^3

and

(x - y)^3

using the binomial theorem.

(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

and

(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

Expanding the binomial expressions allows combination of like terms in the numerator.

2

Subtract

(x - y)^3

from

(x + y)^3

.

(x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) = 6x^2y + 2y^3

The terms

x^3

and

3xy^2

subtract to zero, while the remaining terms double.

3

Factor out common terms from the simplified numerator.

6x^2y + 2y^3 = 2y(3x^2 + y^2)

Factoring out

2y

reveals a factor identical to the denominator.

4

Divide the numerator by the denominator

2y(3x^2 + y^2)

.

\frac{2y(3x^2 + y^2)}{2y(3x^2 + y^2)} = 1

Since

y \neq 0

and

3x^2 + y^2 \neq 0

, identical non-zero factors cancel out.

Key Concept

Binomial expansion of cubic expressions and factoring algebraic expressions.

Estimated Time:1m 30s

Question 33Question →

If $x$ and $y$ are non-zero real numbers such that $x \neq y$ , $\frac{x^2 - y^2}{x - y} = 12$ , and $\frac{x^2y - xy^2}{xy} = 4$ , what is the value of $x^2 + y^2$ ?

Show answer & explanation

Answer: 80

Answer

The value of

x^2 + y^2

is 80.

Simplifying the first equation by factoring the difference of squares

\frac{(x-y)(x+y)}{x-y} = 12

gives

x + y = 12

. Simplifying the second equation by factoring out

xy

yields

\frac{xy(x-y)}{xy} = 4

, which gives

x - y = 4

. Solving this system yields

x = 8

and

y = 4

. Squaring and summing these values gives

8^2 + 4^2 = 64 + 16 = 80

. Alternatively, using the algebraic identity

\frac{(x+y)^2 + (x-y)^2}{2} = \frac{12^2 + 4^2}{2} = \frac{144 + 16}{2} = 80

directly yields the correct answer.

Step-by-Step Solution

1

Simplify the first given algebraic expression using the difference of squares identity.

x + y = 12

Factoring

x^2 - y^2

gives

(x - y)(x + y)

. Since

x \neq y

, we can cancel the non-zero common factor

(x - y)

from the numerator and denominator.

2

Simplify the second given algebraic expression by factoring out the greatest common factor.

x - y = 4

Factoring

xy

from

x^2y - xy^2

yields

xy(x - y)

. Since

x, y \neq 0

, we cancel

xy

from the numerator and denominator.

3

Solve for the individual values of

x

and

y

.

x = 8

and

y = 4

Adding

(x + y = 12)

and

(x - y = 4)

gives

2x = 16

, so

x = 8

. Subtracting the equations gives

2y = 8

, so

y = 4

.

4

Calculate the target expression

x^2 + y^2

.

80

x^2 + y^2 = 8^2 + 4^2 = 64 + 16 = 80

.

Key Concept

Factoring Algebraic Expressions (Difference of Squares and GCF Extraction)

Question 34Question →

For all real numbers $x$ and $y$ such that $x \neq 2y$ and $x \neq -2y$ , which of the following expressions is equivalent to $\frac{x^3 + 2x^2y - 4xy^2 - 8y^3}{x^2 - 4y^2}$ ?

x - 2y

x + 4y

x + 2y

\frac{x + 2y}{x - 2y}

x^2 + 2y

Show answer & explanation

Answer: $x + 2y$

Answer

The simplified expression is

x + 2y

.

Grouping terms in the numerator gives

x^2(x + 2y) - 4y^2(x + 2y) = (x^2 - 4y^2)(x + 2y)

. Dividing this by the denominator

(x^2 - 4y^2)

cancels out the identical non-zero factor

(x^2 - 4y^2)

, leaving the linear binomial

x + 2y

.

Step-by-Step Solution

1

Group the four terms in the numerator in pairs to factor by grouping

x^3 + 2x^2y - 4xy^2 - 8y^3 = x^2(x + 2y) - 4y^2(x + 2y)

Grouping the first two terms and the last two terms allows factoring out

x^2

and

-4y^2

respectively.

2

Factor out the common binomial factor

(x + 2y)

from the numerator

x^2(x + 2y) - 4y^2(x + 2y) = (x^2 - 4y^2)(x + 2y)

Both terms share the common factor

(x + 2y)

.

3

Divide the factored numerator by the denominator

(x^2 - 4y^2)

\frac{(x^2 - 4y^2)(x + 2y)}{x^2 - 4y^2} = x + 2y

Since

x \neq \pm 2y

,

x^2 - 4y^2 \neq 0

, allowing the common polynomial factor

(x^2 - 4y^2)

to be canceled.

Key Concept

Factoring Four-Term Polynomials by Grouping and Rational Expression Simplification

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