Simplifying and Factoring Algebraic Expressions Alıştırma Soruları — GRE General Test

For all real numbers

x

such that

x \neq 3

and

x \neq -5

, the algebraic expression

\frac{(x^2 - 9)^2 - 4(x - 3)^2}{(x - 3)(x + 5)}

can be simplified to the equivalent polynomial expression

x^2 + ax + b

, where

a

and

b

are constants. What is the value of

a + b

?

Cevabı ve açıklamayı göster

Cevap: -5

Cevap

The value of

a + b

is

-5

.

Factoring the numerator yields

(x - 3)^2(x + 1)(x + 5)

. Canceling the common factors

(x - 3)

and

(x + 5)

with the denominator simplifies the expression to

(x - 3)(x + 1) = x^2 - 2x - 3

. Comparing this to

x^2 + ax + b

identifies

a = -2

and

b = -3

, giving a sum of

a + b = -5

.

Adım Adım Çözüm

1

Factor the numerator by recognizing

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

(x^2 - 9)^2 - 4(x - 3)^2 = [(x - 3)(x + 3)]^2 - 4(x - 3)^2 = (x - 3)^2 (x + 3)^2 - 4(x - 3)^2

Applying the difference of squares identity inside the squared term allows factoring out common factors.

2

Factor out

(x - 3)^2

from the numerator

(x - 3)^2 \left[ (x + 3)^2 - 4 \right]

Extracting the greatest common algebraic factor simplifies the remaining expression.

3

Apply difference of squares to

(x + 3)^2 - 4

(x + 3)^2 - 2^2 = ((x + 3) - 2)((x + 3) + 2) = (x + 1)(x + 5)

Recognizing

(x + 3)^2 - 2^2

as

A^2 - B^2

yields factored linear terms directly.

4

Substitute the fully factored numerator back into the rational expression and simplify

\frac{(x - 3)^2 (x + 1)(x + 5)}{(x - 3)(x + 5)} = (x - 3)(x + 1)

Canceling non-zero common factors

(x - 3)

and

(x + 5)

simplifies the rational function.

5

Expand

(x - 3)(x + 1)

and determine

a + b

(x - 3)(x + 1) = x^2 - 2x - 3

, so

a = -2

and

b = -3

. Therefore,

a + b = -2 + (-3) = -5

.

Matching coefficients with

x^2 + ax + b

gives

a = -2

and

b = -3

.

Anahtar Kavram

Simplifying complex rational expressions through nested difference of squares factoring.

Soru 2Soru →

For all real numbers $x$ and $y$ such that $4x^2 + 6xy + 9y^2 \neq 0$ and $2x + 3y \neq 0$ , which of the following expressions are equivalent to $\frac{8x^3 - 27y^3}{4x^2 + 6xy + 9y^2}$ ? Select all such expressions.

Geçerli olan tümünü seçin

2x - 3y

-(3y - 2x)

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

2x + 3y

(2x - 3y)^2

Cevabı ve açıklamayı göster

Cevap: $2x - 3y$ ; $-(3y - 2x)$ ; \frac{4x^2 - 9y^2}{2x + 3y}

Cevap

The equivalent expressions are

2x - 3y

,

-(3y - 2x)

, and

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

.

Factoring

8x^3 - 27y^3

as

(2x - 3y)(4x^2 + 6xy + 9y^2)

allows canceling the denominator, leaving

2x - 3y

. The expression

2x - 3y

is directly correct. Rearranging terms in

-(3y - 2x)

yields

-3y + 2x = 2x - 3y

, making it correct. Factoring

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

as

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

simplifies to

2x - 3y

, making it also correct.

Adım Adım Çözüm

1

Factor the numerator of the rational expression using the difference of cubes formula.

8x^3 - 27y^3 = (2x)^3 - (3y)^3 = (2x - 3y)((2x)^2 + (2x)(3y) + (3y)^2) = (2x - 3y)(4x^2 + 6xy + 9y^2)

Recognizing

8x^3

as

(2x)^3

and

27y^3

as

(3y)^3

enables full polynomial factoring.

2

Simplify the fraction by canceling the common non-zero quadratic factor.

\frac{(2x - 3y)(4x^2 + 6xy + 9y^2)}{4x^2 + 6xy + 9y^2} = 2x - 3y

Since

4x^2 + 6xy + 9y^2 \neq 0

, dividing common factors reduces the expression to linear form.

3

Evaluate each provided choice for algebraic equivalence to

2x - 3y

.

The expressions

2x - 3y

,

-(3y - 2x) = 2x - 3y

, and

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

are all algebraically identical.

Re-expressing or factoring alternative choices proves their equivalence to the simplified expression.

Anahtar Kavram

Factoring difference of cubes

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

and difference of squares

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

Soru 3Soru →

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

\frac{a(a - b)}{a + b}

a - b

\frac{a(a + b)}{a - b}

\frac{a^2 + b^2}{a + b}

\frac{a - b}{a + b}

Cevabı ve açıklamayı göster

Cevap: $\frac{a(a - b)}{a + b}$

Cevap

\frac{a(a - b)}{a + b}

Factoring the numerator yields

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

, and factoring the denominator yields

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

. Dividing both the numerator and the denominator by the common factor

b(a + b)

produces

\frac{a(a - b)}{a + b}

.

Adım Adım Çözüm

1

Factor out the greatest common monomial factor from the numerator and denominator.

The numerator

a^3b - ab^3

becomes

ab(a^2 - b^2)

, and the denominator

a^2b + 2ab^2 + b^3

becomes

b(a^2 + 2ab + b^2)

.

Extracting common terms simplifies polynomials prior to applying polynomial identities.

2

Apply polynomial algebraic identities to factor the remaining terms.

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

and

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

. Thus, the expression becomes

\frac{ab(a - b)(a + b)}{b(a + b)^2}

.

The difference of squares and perfect square trinomial identities allow complete factorization.

3

Cancel non-zero common factors from the numerator and denominator.

Canceling

b

and one factor of

(a + b)

leaves

\frac{a(a - b)}{a + b}

.

Dividing both numerator and denominator by

b(a + b)

yields the simplified expression.

Anahtar Kavram

Simplifying rational expressions using common monomial factoring, difference of squares, and perfect square trinomial identities.

Tahmini Süre:1m 30s

Soru 4Soru →

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

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

Which of the following expressions are equivalent to $E(x, y)$ for all valid values of $x$ and $y$ ? Select all such expressions.

Geçerli olan tümünü seçin

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

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $\frac{x^2 - y^2}{x - y}$ ; $\frac{x^3 + y^3}{x^2 - xy + y^2}$

Cevap

The expressions equivalent to

E(x, y)

are

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

and

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

.

Simplifying

E(x, y)

by factoring both numerator and denominator yields

x + y

. The option with

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

simplifies directly to

x + y

by canceling

(x - y)

. The option with

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

uses the sum of cubes identity to factor the numerator into

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

, which also cancels down to

x + y

. Both of these options are mathematically identical to

E(x, y)

.

Adım Adım Çözüm

1

Factor the numerator of

E(x, y)

using the difference of squares identity twice.

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

Decomposing higher-power binomials into linear and quadratic factors enables algebraic simplification.

2

Factor the denominator of

E(x, y)

by grouping terms.

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

Grouping adjacent terms with shared factors allows factoring out

(x - y)

.

3

Simplify the full expression

E(x, y)

by canceling common non-zero factors.

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

Since

x \neq y

and

x^2 + y^2 \neq 0

, the factors

(x - y)

and

(x^2 + y^2)

cancel completely.

4

Evaluate the given options to determine which simplify to

x + y

.

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

and

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

, whereas

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

and

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

.

Matching each option's fully simplified form to

x + y

identifies all valid equivalent expressions.

Anahtar Kavram

Simplifying rational expressions by polynomial factoring (grouping, difference of squares, and sum of cubes).

Soru 5Soru →

For all real numbers $x$ such that $x \neq -3$ , which of the following expressions is equivalent to $\frac{2x^2 - 18}{4x + 12}$ ?

\frac{x - 3}{2}

\frac{x + 3}{2}

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

\frac{2x - 3}{4}

\frac{x - 9}{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{x - 3}{2}$

Cevap

The expression

\frac{x - 3}{2}

is equivalent to the given rational expression.

Factoring the numerator gives

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

and factoring the denominator gives

4(x + 3)

. Canceling the non-zero common terms

2(x + 3)

leaves

\frac{x - 3}{2}

.

Adım Adım Çözüm

1

Factor out the greatest common factor from the numerator and denominator.

Numerator:

2x^2 - 18 = 2(x^2 - 9)

; Denominator:

4x + 12 = 4(x + 3)

.

Factoring out common numerical coefficients simplifies the expression and reveals algebraic patterns.

2

Apply the difference of squares formula to factor

x^2 - 9

.

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

, so the numerator becomes

2(x - 3)(x + 3)

.

The algebraic identity

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

allows complete factoring of the numerator.

3

Cancel common factors shared by the numerator and denominator.

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

Since

x \neq -3

, the factor

(x + 3)

is non-zero and can be safely canceled along with reducing the constant ratio

\frac{2}{4}

to

\frac{1}{2}

.

Anahtar Kavram

Simplifying rational algebraic expressions by factoring common numerical factors and applying the difference of squares identity.

Soru 6Soru →

For all non-zero real numbers $x$ and $y$ , which of the following expressions is equivalent to $\frac{12x^4y^2 - 8x^2y^3}{4x^2y}$ ?

3x^2y - 2y^2

3x^2y - 8x^2y^3

3x^6y^3 - 2x^4y^4

3x^2y + 2y^2

3x^2 - 2y

Cevabı ve açıklamayı göster

Cevap: $3x^2y - 2y^2$

Cevap

3x^2y - 2y^2

Dividing each term in the numerator by

4x^2y

yields

\frac{12x^4y^2}{4x^2y} - \frac{8x^2y^3}{4x^2y} = 3x^2y - 2y^2

, which matches the correct expression.

Adım Adım Çözüm

1

Split the rational algebraic expression into two separate fractions by distributing the denominator.

\frac{12x^4y^2}{4x^2y} - \frac{8x^2y^3}{4x^2y}

When dividing a polynomial by a monomial, each term of the numerator must be divided independently by the denominator.

2

Simplify the first term

\frac{12x^4y^2}{4x^2y}

.

\frac{12}{4} \cdot x^{4-2} \cdot y^{2-1} = 3x^2y

Divide the numerical coefficients and apply the exponent quotient rule

a^m / a^n = a^{m-n}

for each variable.

3

Simplify the second term

\frac{8x^2y^3}{4x^2y}

.

\frac{8}{4} \cdot x^{2-2} \cdot y^{3-1} = 2(1)y^2 = 2y^2

Divide the coefficients and apply the exponent quotient rule, noting that

x^0 = 1

for non-zero

x

.

4

Combine the simplified terms with the original subtraction operator.

3x^2y - 2y^2

Combine terms to form the final simplified expression.

Anahtar Kavram

Simplifying algebraic fractions by distributing a monomial denominator across numerator terms and applying exponent rules.

Tahmini Süre:45s

Soru 7Soru →

For all non-zero real numbers

a

and

b

such that

a \neq b

and

a \neq -b

, which of the following expressions is equivalent to

\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}} \div \frac{a^2 + ab + b^2}{a^3 - b^3}?

\frac{a^2 - b^2}{ab}

\frac{b^2 - a^2}{ab}

\frac{a - b}{a + b}

\frac{a + b}{ab(a - b)}

\frac{(a - b)^2}{ab}

Cevabı ve açıklamayı göster

Cevap: $\frac{a^2 - b^2}{ab}$

Cevap

\frac{a^2 - b^2}{ab}

Simplifying

\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}}

yields

\frac{a + b}{ab}

after canceling

(b - a)

from both numerator and denominator. Simplifying

\frac{a^2 + ab + b^2}{a^3 - b^3}

yields

\frac{1}{a - b}

using the difference of cubes identity. Dividing

\frac{a + b}{ab}

by

\frac{1}{a - b}

gives

\frac{a + b}{ab} \cdot (a - b) = \frac{a^2 - b^2}{ab}

.

Adım Adım Çözüm

1

Simplify the first rational expression

\frac{a^{-2} - b^{-2}}{a^{-1} - b^{-1}}

\frac{\frac{1}{a^2} - \frac{1}{b^2}}{\frac{1}{a} - \frac{1}{b}} = \frac{\frac{b^2 - a^2}{a^2 b^2}}{\frac{b - a}{ab}} = \frac{(b - a)(b + a)}{a^2 b^2} \cdot \frac{ab}{b - a} = \frac{a + b}{ab}

Convert negative exponents to fractions, find common denominators, and cancel the common non-zero factor

(b - a)

.

2

Simplify the second rational expression

\frac{a^2 + ab + b^2}{a^3 - b^3}

\frac{a^2 + ab + b^2}{(a - b)(a^2 + ab + b^2)} = \frac{1}{a - b}

Factor the denominator using the difference of cubes identity

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

.

3

Divide the simplified first expression by the simplified second expression

\frac{a + b}{ab} \div \frac{1}{a - b} = \frac{a + b}{ab} \cdot (a - b) = \frac{(a + b)(a - b)}{ab} = \frac{a^2 - b^2}{ab}

Multiply by the reciprocal of the second expression and apply the difference of squares identity.

Anahtar Kavram

Simplifying complex fractions and factoring using difference of squares and difference of cubes identities.

Soru 8Soru →

For all real numbers $a$ and $b$ , which of the following expressions are equivalent to $(a + b)^2 - (a - b)^2$ ? Select all that apply.

Geçerli olan tümünü seçin

4ab

(2a)(2b)

0

2a^2 + 2b^2

2ab

Cevabı ve açıklamayı göster

Cevap: $4ab$ ; $(2a)(2b)$

Cevap

The expressions

4ab

and

(2a)(2b)

are equivalent to

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

.

Expanding the squared expressions gives

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

and

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

. Subtracting the second expression from the first yields

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

. Since

(2a)(2b)

also equals

4ab

, both

4ab

and

(2a)(2b)

are mathematically equivalent to the original expression.

Adım Adım Çözüm

1

Expand

(a + b)^2

using the binomial square formula.

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

Squaring a sum produces a trinomial containing two squared terms and a middle product term.

2

Expand

(a - b)^2

using the binomial square formula.

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

Squaring a difference produces a trinomial with a negative middle product term.

3

Subtract the second expanded trinomial from the first.

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

Distributing the negative sign across

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

changes the signs inside the second set of parentheses.

4

Compare the simplified result

4ab

against all given choices.

Both

4ab

and

(2a)(2b) = 4ab

are identical to the calculated result.

Algebraic multiplication shows

(2a)(2b) = 4ab

.

Anahtar Kavram

Binomial expansion and combination of like terms

Soru 9Soru →

When the expression $\frac{3x^2 - 12}{x - 2}$ is completely simplified, it can be written in the form $ax + b$ for all $x \neq 2$ , where $a$ and $b$ are constants. What is the value of $a + b$ ?

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

The value of

a + b

is 9.

Factoring the numerator gives

3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)

. Canceling the common factor

(x - 2)

in the denominator yields

3(x + 2)

, which expands to

3x + 6

. Matching this expression to

ax + b

reveals

a = 3

and

b = 6

. Adding these constants gives

a + b = 9

.

Adım Adım Çözüm

1

Factor the numerator of the rational expression.

3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)

Factoring out the common numerical factor 3 and using the difference of squares identity

(u^2 - v^2 = (u-v)(u+v))

completely factors the numerator.

2

Simplify the expression by canceling common factors.

\frac{3(x - 2)(x + 2)}{x - 2} = 3(x + 2) = 3x + 6

Because

x \neq 2

, the factor

(x - 2)

is non-zero and can be canceled from both numerator and denominator.

3

Identify the values of

a

and

b

and compute

a + b

.

a = 3

,

b = 6

, so

a + b = 3 + 6 = 9

Comparing

3x + 6

to

ax + b

gives

a = 3

and

b = 6

.

Anahtar Kavram

Factoring algebraic expressions using common monomial factors and the difference of squares to simplify rational expressions.

Soru 10Soru →

For all real numbers $x$ such that $x \neq 5$ , which of the following expressions is equivalent to $\frac{x^2 - 25}{3x - 15}$ ?

\frac{x - 5}{3}

x + 5

\frac{x + 5}{3}

\frac{x - 25}{3}

\frac{x + 5}{15}

Cevabı ve açıklamayı göster

Cevap: $\frac{x + 5}{3}$

Cevap

The expression

\frac{x + 5}{3}

is equivalent to the given rational expression for all

x \neq 5

.

Factoring the numerator as a difference of squares gives

(x - 5)(x + 5)

, and factoring the denominator gives

3(x - 5)

. Dividing out the common factor

(x - 5)

simplifies the expression to

\frac{x + 5}{3}

.

Adım Adım Çözüm

1

Factor the numerator using the difference of squares identity.

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

The difference of squares

a^2 - b^2

factors into

(a - b)(a + b)

.

2

Factor out the greatest common factor from the denominator.

3x - 15 = 3(x - 5)

Both terms in

3x - 15

share a common factor of

3

.

3

Divide out the common binomial factor

(x - 5)

from the numerator and denominator.

\frac{(x - 5)(x + 5)}{3(x - 5)} = \frac{x + 5}{3}

Since

x \neq 5

, the factor

(x - 5)

is non-zero and can be cancelled.

Anahtar Kavram

Simplifying rational algebraic expressions by factoring difference of squares and common linear factors.

Soru 11Soru →

For all real numbers

x

such that

x \neq 3

and

x \neq -3

, the algebraic expression

\frac{\frac{x^4 - 81}{x^2 - 9} \cdot \frac{x^3 - 27}{x^2 + 3x + 9} - x(x - 3)^2}{x^2 - 9}

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

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The simplified expression evaluates to the constant value

3

.

Factoring the numerator components using difference of squares and difference of cubes simplifies the product term to

x^3 - 3x^2 + 9x - 27

. Subtracting

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

simplifies the entire numerator to

3x^2 - 27 = 3(x^2 - 9)

. Dividing by the denominator

(x^2 - 9)

cancels out the variable terms entirely, yielding the constant value 3.

Adım Adım Çözüm

1

Simplify the first rational component using the difference of squares identity

\frac{x^4 - 81}{x^2 - 9} = \frac{(x^2 - 9)(x^2 + 9)}{x^2 - 9} = x^2 + 9

Since

x \neq \pm 3

,

x^2 - 9 \neq 0

, allowing direct cancellation of

(x^2 - 9)

.

2

Simplify the second rational component using the difference of cubes identity

\frac{x^3 - 27}{x^2 + 3x + 9} = \frac{(x - 3)(x^2 + 3x + 9)}{x^2 + 3x + 9} = x - 3

Applying

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

where

a = x

and

b = 3

allows cancellation of the quadratic factor.

3

Multiply the simplified expressions

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

Distribute each term of the binomials to get the expanded polynomial.

4

Expand the subtracted term in the numerator

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

Expand

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

and distribute

x

.

5

Subtract the two expanded expressions to simplify the entire numerator

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

Combine like terms;

x^3

and

9x

terms cancel out, leaving

3x^2 - 27

.

6

Divide the simplified numerator by the main denominator

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

Cancel the common factor

(x^2 - 9)

from numerator and denominator.

Anahtar Kavram

Simplifying complex algebraic expressions via polynomial factoring (difference of squares and difference of cubes) and combining like terms.

Soru 12Soru →

If $a$ and $b$ are distinct real numbers, what is the simplified form of the algebraic expression $\frac{a^3 - b^3 - a(a^2 - b^2) + b(a - b)^2}{a - b}$ ?

ab + 2b^2

-ab

ab

a^2 + b^2

a(a + b)

Cevabı ve açıklamayı göster

Cevap: $ab$

Cevap

ab

Factoring the common binomial term

(a - b)

from each term in the numerator yields

(a - b)[(a^2 + ab + b^2) - a(a + b) + b(a - b)]

. Expanding inside the brackets gives

a^2 + ab + b^2 - a^2 - ab + ab - b^2

, which reduces completely to

ab

. Dividing

(a - b)(ab)

by

(a - b)

leaves the simplified expression

ab

.

Adım Adım Çözüm

1

Apply standard algebraic factoring identities to each component of the numerator.

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

,

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

, and

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

.

Rewriting each grouping with the common binomial factor

(a - b)

allows for structured factoring of the numerator.

2

Factor out

(a - b)

from the entire numerator.

Numerator

= (a - b) \left[ (a^2 + ab + b^2) - a(a + b) + b(a - b) \right]

.

Extracting the common factor isolates the remaining polynomial terms inside brackets.

3

Expand and collect like terms within the bracketed expression.

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

.

Canceling opposite terms (

a^2 - a^2 = 0

,

ab - ab = 0

,

b^2 - b^2 = 0

) leaves only

ab

inside the brackets.

4

Divide the factored numerator by the denominator

(a - b)

.

\frac{(a - b)(ab)}{a - b} = ab

.

Since

a

and

b

are distinct (

a \neq b

),

a - b \neq 0

, allowing non-zero cancellation.

Anahtar Kavram

Simplifying complex algebraic expressions by factoring common binomial terms and applying algebraic identities

Soru 13Soru →

Given non-zero real numbers $x$ and $y$ where $|x| \neq |y|$ , simplify the complex rational expression:

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

Which of the following represents the completely simplified expression?

1

x - y

x + y

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

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

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

1

Both the entire complex numerator and the entire complex denominator independently simplify to

x - y

. Consequently, dividing the numerator

x - y

by the denominator

x - y

gives

1

.

Adım Adım Çözüm

1

Simplify the first term of the main numerator

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

Factor the sum of cubes in the numerator and the difference of squares in the denominator, then cancel the common factor

(x + y)

.

2

Simplify the second term of the main numerator

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

Factor out the greatest common factor

xy

from the numerator and cancel one factor of

(x - y)

.

3

Subtract the simplified terms in the main numerator

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

Combine the numerators over the common denominator

(x - y)

, factor the perfect square trinomial

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

, and simplify.

4

Simplify the main denominator

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

Factor the numerator using difference of squares twice, factor the denominator by grouping, and cancel common factors

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

.

5

Divide the main numerator by the main denominator

\frac{x - y}{x - y} = 1

Divide the simplified main numerator (

x - y

) by the simplified main denominator (

x - y

).

Anahtar Kavram

Multi-step algebraic expression simplification using special factoring identities (sum/difference of cubes, difference of squares, quadratic trinomials, and factoring by grouping).

Tahmini Süre:2m 0s

Soru 14Soru →

For all real numbers $x$ and $y$ such that $x^2 + xy + y^2 \neq 0$ , which of the following expressions are equivalent to $\frac{x^6 - y^6}{x^2 + xy + y^2}$ ? Select all such expressions.

Geçerli olan tümünü seçin

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

(x - y)(x^3 + y^3)

x^4 - x^3 y + x y^3 - y^4

(x + y)(x^3 - y^3)

x^4 - y^4

Cevabı ve açıklamayı göster

Cevap: $(x^2 - y^2)(x^2 - xy + y^2)$ ; $(x - y)(x^3 + y^3)$ ; $x^4 - x^3 y + x y^3 - y^4$

Cevap

The equivalent expressions are

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

,

(x - y)(x^3 + y^3)

, and

x^4 - x^3y + xy^3 - y^4

.

Factoring the numerator

x^6 - y^6

as a difference of squares yields

(x^3 - y^3)(x^3 + y^3)

. Applying the difference of cubes identity

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

enables cancellation of the non-zero denominator

x^2 + xy + y^2

, leaving

(x - y)(x^3 + y^3)

. Expanding this product gives

x^4 - x^3y + xy^3 - y^4

. Furthermore, factoring

x^3 + y^3

as

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

and regrouping

(x - y)(x + y)

gives

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

. Consequently, the three valid equivalent forms are

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

,

(x - y)(x^3 + y^3)

, and

x^4 - x^3y + xy^3 - y^4

.

Adım Adım Çözüm

1

Factor the numerator

x^6 - y^6

as a difference of squares.

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

Applying the difference of squares identity

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

with

a = x^3

and

b = y^3

.

2

Apply the difference of cubes identity to

x^3 - y^3

.

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

This reveals the non-zero quadratic factor present in the denominator.

3

Substitute into the original fraction and cancel the common factor

(x^2 + xy + y^2)

.

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

Since

x^2 + xy + y^2 \neq 0

, dividing numerator and denominator by

(x^2 + xy + y^2)

simplifies the expression to

(x - y)(x^3 + y^3)

.

4

Expand

(x - y)(x^3 + y^3)

to check for equivalent expanded polynomial forms.

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

Distributing terms verifies polynomial equivalence.

5

Factor

x^3 + y^3

and regroup to find another equivalent factored representation.

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

Applying the sum of cubes identity

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

and combining

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

.

Anahtar Kavram

Factoring higher-degree algebraic expressions using difference of squares and sum/difference of cubes identities

Soru 15Soru →

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

x^2 + y^2

(x + y)^2

(x - y)^2

x^2 + 2xy - y^2

x + y

Cevabı ve açıklamayı göster

Cevap: $(x + y)^2$

Cevap

The simplified expression is

(x + y)^2

.

Expanding and grouping the terms in the numerator gives

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

. Canceling the common factor

(x^2 - y^2)

from both the numerator and the denominator leaves

(x + y)^2

.

Adım Adım Çözüm

1

Expand the numerator terms

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

Apply the distributive property to remove parentheses in the numerator.

2

Group the terms in the numerator to factor

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

Use difference of squares on

x^4 - y^4

and factor out the greatest common factor

2xy

from the remaining terms.

3

Factor out the common term

(x^2 - y^2)

from the numerator

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

Recognize that

x^2 + 2xy + y^2

is the perfect square binomial

(x + y)^2

.

4

Simplify the rational expression by canceling common factors

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

Divide numerator and denominator by

(x^2 - y^2)

, which is non-zero since

x \neq \pm y

.

Anahtar Kavram

Factoring high-degree algebraic expressions by grouping terms, recognizing difference of squares, and applying perfect square binomial identities.

Tahmini Süre:2m 0s

Soru 16Soru →

If

a

,

b

, and

c

are pairwise distinct real numbers, which of the following expressions is equivalent to

\frac{a^3(b - c) + b^3(c - a) + c^3(a - b)}{(a - b)(b - c)(c - a)}?

-(a + b + c)

a + b + c

a^2 + b^2 + c^2

ab + bc + ca

-(ab + bc + ca)

Cevabı ve açıklamayı göster

Cevap: $-(a + b + c)$

Cevap

-(a + b + c)

Factoring the numerator by using the Factor Theorem and cyclic symmetry reveals that

a^3(b - c) + b^3(c - a) + c^3(a - b) = -(a - b)(b - c)(c - a)(a + b + c)

. Dividing this by the denominator

(a - b)(b - c)(c - a)

cancels the pairwise difference terms, leaving

-(a + b + c)

.

Adım Adım Çözüm

1

Analyze the numerator for polynomial factors using cyclic symmetry

Let

P(a, b, c) = a^3(b - c) + b^3(c - a) + c^3(a - b)

. If

a = b

, then

P(b, b, c) = b^3(b - c) + b^3(c - b) + 0 = 0

. By the Factor Theorem,

(a - b)

is a factor. By cyclic symmetry,

(b - c)

and

(c - a)

are also factors.

Identifying linear factors reduces the polynomial simplification problem.

2

Determine the degree and form of the remaining factor

P(a, b, c)

is a homogeneous polynomial of degree 4, while

(a - b)(b - c)(c - a)

has degree 3. Therefore, the remaining factor must be a homogeneous symmetric polynomial of degree 1, which takes the form

k(a + b + c)

for some constant

k

.

Homogeneous degree properties dictate the algebraic structure of the quotient.

3

Find the constant

k

by substituting test values

Let

a = 0

,

b = 1

, and

c = 2

. Evaluating

P(0, 1, 2) = 0 + 1^3(2 - 0) + 2^3(0 - 1) = 2 - 8 = -6

. The factor product gives

(0 - 1)(1 - 2)(2 - 0) = (-1)(-1)(2) = 2

. Setting

2 \cdot k(0 + 1 + 2) = -6 \implies 6k = -6 \implies k = -1

.

Evaluating at convenient integer values determines the missing constant scalar.

4

Divide the factored numerator by the denominator

\frac{-(a - b)(b - c)(c - a)(a + b + c)}{(a - b)(b - c)(c - a)} = -(a + b + c)

.

Canceling common non-zero factors yields the simplified expression.

Anahtar Kavram

Factoring Cyclic Symmetric Polynomials

Soru 17Soru →

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

P(x, y) = \left(\frac{x^3 + y^3 - x^2 y - xy^2}{x^4 - y^4}\right) \cdot \left(\frac{x^3 + y^3}{x^2 - xy + y^2}\right)

Which of the following expressions are equivalent to $P(x, y)$ for all valid values of $x$ and $y$ ? Indicate all such expressions.

Geçerli olan tümünü seçin

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

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $\frac{x^4 - y^4}{(x^2 + y^2)^2}$ ; $1 - \frac{2y^2}{x^2 + y^2}$

Cevap

The expressions equivalent to

P(x, y)

are

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

and

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

.

The given expression simplifies to

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

. The option

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

simplifies directly to

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

after factoring the numerator. The option

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

simplifies to

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

when combined over a common denominator.

Adım Adım Çözüm

1

Factor the numerator of the first rational term by grouping terms.

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

Grouping allows rewriting four polynomial terms into product of linear/quadratic factors.

2

Factor the denominator of the first rational term as a difference of squares.

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

Decomposing

x^4 - y^4

into simpler factors reveals common factors with the numerator.

3

Simplify the first rational term by canceling common factors

(x - y)(x + y)

.

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

Canceling non-zero common factors simplifies the fraction.

4

Factor the numerator of the second rational term using the sum of cubes formula.

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

Applying the sum of cubes identity exposes the irreducible quadratic factor present in the denominator.

5

Simplify the second rational term and multiply the result by the simplified first term.

P(x, y) = \left(\frac{x - y}{x^2 + y^2}\right) \cdot (x + y) = \frac{(x - y)(x + y)}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}

Multiplying the simplified forms yields the simplest explicit representation of

P(x, y)

.

6

Verify equivalence of the options against

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

.

The option

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

, and the option

1 - \frac{2y^2}{x^2 + y^2} = \frac{x^2 + y^2 - 2y^2}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}

. Both match

P(x, y)

.

Transforming algebraic expressions under common denominators or factoring confirms equivalence.

Anahtar Kavram

Simplifying complex algebraic expressions using factoring by grouping, difference of squares, sum of cubes, and common denominator manipulation.

Soru 18Soru →

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

x + y + 1

x - y + 1

x + y - 1

(x + y)^2 + 1

x^2 + xy + y^2 + 1

Cevabı ve açıklamayı göster

Cevap: $x + y + 1$

Cevap

x + y + 1

Factoring the numerator by grouping yields

(x - y)(x + y)(x + y + 1)

. Factoring the denominator gives

(x - y)(x + y)

. Canceling the non-zero common factor

(x - y)(x + y)

leaves the simplified expression

x + y + 1

.

Adım Adım Çözüm

1

Group terms in the numerator to identify common factor pairs.

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

Grouping cubic terms, quadratic cross-terms, and difference of squares separately allows factoring out fundamental algebraic patterns.

2

Apply standard algebraic formulas to each grouped term.

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

Using difference of cubes

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

and difference of squares

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

reveals a common

(x-y)

factor across all terms.

3

Factor out

(x - y)

from the numerator and simplify the remaining polynomial.

N = (x - y)\left[(x^2 + xy + y^2) + xy + (x + y)\right] = (x - y)\left[x^2 + 2xy + y^2 + x + y\right]

Combining like terms inside the bracket simplifies the expression.

4

Recognize the perfect square trinomial inside the expression.

N = (x - y)\left[(x + y)^2 + (x + y)\right] = (x - y)(x + y)(x + y + 1)

Rewriting

x^2 + 2xy + y^2

as

(x + y)^2

allows factoring out

(x + y)

.

5

Divide the factored numerator by the denominator.

\frac{(x - y)(x + y)(x + y + 1)}{(x - y)(x + y)} = x + y + 1

Since

|x| \neq |y|

, both

(x - y)

and

(x + y)

are non-zero and can be canceled.

Anahtar Kavram

Factoring multivariable polynomials using grouping, difference of cubes, and difference of squares formulas.

Soru 19Soru →

For pairwise distinct real numbers $x$ , $y$ , and $z$ , consider the algebraic expression:

E(x, y, z) = \frac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y - z)^3 + (z - x)^3}

If $x = 5$ , $y = 3$ , and $z = 1$ , what is the numerical value of $E(5, 3, 1)$ ?

Cevabı ve açıklamayı göster

Cevap: 192

Cevap

192

Using the identity that

a^3 + b^3 + c^3 = 3abc

whenever

a + b + c = 0

, both the numerator and denominator can be factored directly. Factoring the difference of squares in the numerator yields

3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)

. Dividing this by the factored denominator

3(x - y)(y - z)(z - x)

simplifies the expression to

(x + y)(y + z)(z + x)

. Substituting

x = 5

,

y = 3

, and

z = 1

yields

(8)(4)(6) = 192

.

Adım Adım Çözüm

1

Use the conditional cubic identity

a + b + c = 0 \implies a^3 + b^3 + c^3 = 3abc

on the denominator.

Denominator becomes

3(x - y)(y - z)(z - x)

.

The sum of the three terms

(x - y) + (y - z) + (z - x)

equals 0.

2

Apply the same identity to the numerator.

Numerator becomes

3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)

.

The sum of the squared difference terms

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

also equals 0.

3

Factor each difference of squares in the numerator.

Numerator becomes

3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)

.

Using the difference of squares identity

u^2 - v^2 = (u - v)(u + v)

on each term.

4

Simplify the fraction by dividing the common factors in the numerator and denominator.

E(x, y, z) = (x + y)(y + z)(z + x)

.

The factors

3

,

(x - y)

,

(y - z)

, and

(z - x)

cancel out completely.

5

Evaluate the simplified product for

x = 5

,

y = 3

, and

z = 1

.

(5 + 3)(3 + 1)(1 + 5) = 8 \times 4 \times 6 = 192

.

Direct evaluation after algebraic simplification.

Anahtar Kavram

Simplifying rational expressions involving sum of cubes identity

a^3 + b^3 + c^3 = 3abc

when

a + b + c = 0

and difference of squares factoring.

Soru 20Soru →

For all real numbers

x

and

y

such that

x \neq y

,

x \neq -y

, and

x^2 + y^2 \neq 0

, which of the following expressions is equivalent to

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

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

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

\frac{x + y - 2xy}{1 + x + y}

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

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

Cevabı ve açıklamayı göster

Cevap: $\frac{x^2 + y^2}{x + y}$

Cevap

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

Factoring

x^4 - y^4

as

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

and

x^3 - x^2 y + x y^2 - y^3

as

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

simplifies the first term to

x + y

. Combining

x + y

with

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

over the common denominator

(x + y)

gives

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

.

Adım Adım Çözüm

1

Factor the numerator and denominator of the first rational expression

Numerator:

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

. Denominator:

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

.

Factoring allows for cancellation of common factors in rational expressions.

2

Simplify the first rational expression by canceling common terms

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

.

Since

x \neq y

and

x^2 + y^2 \neq 0

, the common terms

(x - y)

and

(x^2 + y^2)

are non-zero and can be divided out.

3

Subtract the second expression using a common denominator

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

.

Combining terms under the common denominator

(x + y)

enables algebraic reduction.

4

Expand the squared binomial in the numerator and simplify like terms

\frac{x^2 + 2xy + y^2 - 2xy}{x + y} = \frac{x^2 + y^2}{x + y}.

Expanding

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

allows the

+2xy

and

-2xy

terms to cancel out.

Anahtar Kavram

Simplifying rational expressions using polynomial factoring (difference of squares and grouping) and common denominators.

Tahmini Süre:2m 0s

Simplifying and Factoring Algebraic Expressions

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