Examkin

Simplifying and Factoring Algebraic Expressions

34 soru

Tüm türler34Çoktan seçmeli18Çoklu seçim8Sayısal8
Soru 21Soru →

For all non-zero real numbers xx and yy such that ∣x∣≠∣y∣|x| \neq |y| and x2+2xy+2y2≠0x^2 + 2xy + 2y^2 \neq 0, consider the algebraic expression:

Q(x,y)=x4+4y4x2+2xy+2y2+2x3y−2xy3x2−y2Q(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)Q(x, y)? Indicate all such expressions.

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Cevap: (x−y)2+2xy+y2(x - y)^2 + 2xy + y^2; \frac{x^3 + 2xy^2}{x}

Cevap

The expressions equivalent to Q(x,y)Q(x, y) are (x−y)2+2xy+y2(x - y)^2 + 2xy + y^2 and x3+2xy2x\frac{x^3 + 2xy^2}{x}.
First, simplify Q(x,y)Q(x, y) by factoring each term. Using Sophie Germain's identity on the numerator of the first term gives x4+4y4=(x2+2y2)2−(2xy)2=(x2+2xy+2y2)(x2−2xy+2y2)x^4 + 4y^4 = (x^2 + 2y^2)^2 - (2xy)^2 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2). Dividing this by (x2+2xy+2y2)(x^2 + 2xy + 2y^2) results in x2−2xy+2y2x^2 - 2xy + 2y^2. For the second term, factoring out 2xy2xy gives 2xy(x2−y2)x2−y2=2xy\frac{2xy(x^2 - y^2)}{x^2 - y^2} = 2xy. Adding the two simplified terms yields Q(x,y)=(x2−2xy+2y2)+2xy=x2+2y2Q(x, y) = (x^2 - 2xy + 2y^2) + 2xy = x^2 + 2y^2.

Evaluating the options for equivalence:
- The expression (x−y)2+2xy+y2(x - y)^2 + 2xy + y^2 expands to x2−2xy+y2+2xy+y2=x2+2y2x^2 - 2xy + y^2 + 2xy + y^2 = x^2 + 2y^2.
- The expression x3+2xy2x\frac{x^3 + 2xy^2}{x} factors as x(x2+2y2)x=x2+2y2\frac{x(x^2 + 2y^2)}{x} = x^2 + 2y^2.
Therefore, both of these expressions are equivalent to Q(x,y)Q(x, y).

Adım Adım Çözüm

1
Simplify the first algebraic fraction using Sophie Germain's identity.
x4+4y4x2+2xy+2y2=(x2+2y2)2−(2xy)2x2+2xy+2y2=(x2+2xy+2y2)(x2−2xy+2y2)x2+2xy+2y2=x2−2xy+2y2\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 x4+4y4x^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.
2x3y−2xy3x2−y2=2xy(x2−y2)x2−y2=2xy\frac{2x^3 y - 2xy^3}{x^2 - y^2} = \frac{2xy(x^2 - y^2)}{x^2 - y^2} = 2xy
The term (x2−y2)(x^2 - y^2) cancels out since x≠±yx \neq \pm y.
3
Combine the simplified terms to find the closed-form expression for Q(x,y)Q(x, y).
Q(x,y)=(x2−2xy+2y2)+2xy=x2+2y2Q(x, y) = (x^2 - 2xy + 2y^2) + 2xy = x^2 + 2y^2
The −2xy-2xy and +2xy+2xy terms sum to zero.
4
Test each option for equivalence to x2+2y2x^2 + 2y^2.
The option (x−y)2+2xy+y2(x - y)^2 + 2xy + y^2 expands to x2−2xy+y2+2xy+y2=x2+2y2x^2 - 2xy + y^2 + 2xy + y^2 = x^2 + 2y^2. The option x3+2xy2x\frac{x^3 + 2xy^2}{x} simplifies directly to x2+2y2x^2 + 2y^2.
Both expressions reduce identically to x2+2y2x^2 + 2y^2 for all non-zero xx and yy.

Anahtar Kavram

Simplifying complex algebraic expressions using polynomial factoring identities (Sophie Germain identity and difference of squares) and algebraic reduction.
Tahmini Süre:2m 0s
Soru 22Soru →
For all real numbers x≠−5x \neq -5, the algebraic expression
x3+125x2−5x+25−x2−25x+5\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?
Cevabı ve açıklamayı göster

Cevap: 10

Cevap

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+5x + 5 and the second term reduces to x−5x - 5. Subtracting (x−5)(x - 5) from (x+5)(x + 5) yields (x+5)−(x−5)=10(x + 5) - (x - 5) = 10, which is constant for all valid values of xx.

Adım Adım Çözüm

1
Factor the numerator of the first rational term using the sum of cubes identity.
x3+125=(x+5)(x2−5x+25)x^3 + 125 = (x + 5)(x^2 - 5x + 25)
The sum of cubes formula a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2) applies with a=xa = x and b=5b = 5.
2
Cancel the non-zero factor (x2−5x+25)(x^2 - 5x + 25) from the first fraction.
\frac{(x + 5)(x^2 - 5x + 25)}{x^2 - 5x + 25} = x + 5
The quadratic factor x2−5x+25x^2 - 5x + 25 has a negative discriminant ((−5)2−4(1)(25)=−75<0(-5)^2 - 4(1)(25) = -75 < 0), so it is never zero for any real number xx.
3
Factor the numerator of the second rational term using the difference of squares identity.
x2−25=(x−5)(x+5)x^2 - 25 = (x - 5)(x + 5)
The difference of squares formula a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b) applies with a=xa = x and b=5b = 5.
4
Cancel the common factor (x+5)(x + 5) from the second fraction.
\frac{(x - 5)(x + 5)}{x + 5} = x - 5
Given x≠−5x \neq -5, the factor x+5x + 5 is non-zero and can be canceled.
5
Subtract the two simplified terms.
(x+5)−(x−5)=x+5−x+5=10(x + 5) - (x - 5) = x + 5 - x + 5 = 10
Distribute the negative sign to both terms in (x−5)(x - 5) and combine like terms.

Anahtar Kavram

Factoring sum of cubes and difference of squares to simplify rational expressions
Soru 23Soru →

For all real numbers x≠2x \neq 2, which of the following expressions is equivalent to x4−16x3−2x2+4x−8\frac{x^4 - 16}{x^3 - 2x^2 + 4x - 8}?

Cevabı ve açıklamayı göster

Cevap: x+2x + 2

Cevap

x+2x + 2
Factoring the numerator x4−16x^4 - 16 as a difference of squares yields (x−2)(x+2)(x2+4)(x - 2)(x + 2)(x^2 + 4). Factoring the denominator x3−2x2+4x−8x^3 - 2x^2 + 4x - 8 by grouping yields (x−2)(x2+4)(x - 2)(x^2 + 4). Dividing the numerator by the denominator allows the shared factor (x−2)(x2+4)(x - 2)(x^2 + 4) to cancel, leaving x+2x + 2.

Adım Adım Çözüm

1
Factor the numerator x4−16x^4 - 16
x4−16=(x2−4)(x2+4)=(x−2)(x+2)(x2+4)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 x3−2x2+4x−8x^3 - 2x^2 + 4x - 8 by grouping
x3−2x2+4x−8=x2(x−2)+4(x−2)=(x−2)(x2+4)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)(x - 2).
3
Simplify the rational expression by canceling common factors
(x−2)(x+2)(x2+4)(x−2)(x2+4)=x+2\frac{(x - 2)(x + 2)(x^2 + 4)}{(x - 2)(x^2 + 4)} = x + 2
Cancel the non-zero common factor (x−2)(x2+4)(x - 2)(x^2 + 4) present in both numerator and denominator.

Anahtar Kavram

Simplifying rational expressions using difference of squares and factoring by grouping
Soru 24Soru →
For all real numbers xx, the algebraic expression
x6−27x4+3x2+9\frac{x^6 - 27}{x^4 + 3x^2 + 9}
can be simplified to the polynomial form ax2+bx+cax^2 + bx + c, where aa, bb, and cc are real constants. What is the value of a+b+ca + b + c?
Cevabı ve açıklamayı göster

Cevap: -2

Cevap

The simplified expression is x2−3x^2 - 3, which corresponds to polynomial coefficients a=1a = 1, b=0b = 0, and c=−3c = -3. The sum a+b+ca + b + c equals −2-2.
Factoring the numerator x6−27x^6 - 27 as a difference of cubes (x2)3−33(x^2)^3 - 3^3 produces (x2−3)(x4+3x2+9)(x^2 - 3)(x^4 + 3x^2 + 9). Canceling the non-zero factor (x4+3x2+9)(x^4 + 3x^2 + 9) from the numerator and denominator simplifies the expression to x2−3x^2 - 3. In standard form ax2+bx+cax^2 + bx + c, a=1a = 1, b=0b = 0, and c=−3c = -3. Adding these coefficients gives 1+0+(−3)=−21 + 0 + (-3) = -2.

Adım Adım Çözüm

1
Factor the numerator using the difference of cubes formula
x6−27=(x2)3−33=(x2−3)(x4+3x2+9)x^6 - 27 = (x^2)^3 - 3^3 = (x^2 - 3)(x^4 + 3x^2 + 9)
The expression x6−27x^6 - 27 matches the pattern u3−v3u^3 - v^3 with u=x2u = x^2 and v=3v = 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 x4+3x2+9>0x^4 + 3x^2 + 9 > 0 for all real numbers xx, the denominator is never zero, allowing direct cancellation of the common factor.
3
Match coefficients with ax2+bx+cax^2 + bx + c and calculate a+b+ca + b + c
a = 1, b = 0, c = -3 \implies a + b + c = 1 + 0 + (-3) = -2
Comparing x2−3=1x2+0x−3x^2 - 3 = 1x^2 + 0x - 3 to ax2+bx+cax^2 + bx + c determines the values of constants aa, bb, and cc.

Anahtar Kavram

Difference of Cubes Factoring Identity
Tahmini Süre:1m 30s
Soru 25Soru →
For all real numbers xx and yy such that x≠−yx \neq -y and x≠−2yx \neq -2y, which of the following expressions is equivalent to
x3+2x2y−xy2−2y3x2+3xy+2y2?\frac{x^3 + 2x^2y - xy^2 - 2y^3}{x^2 + 3xy + 2y^2}?
Cevabı ve açıklamayı göster

Cevap: x−yx - y

Cevap

x−yx - y
Factoring the numerator by grouping gives x2(x+2y)−y2(x+2y)=(x2−y2)(x+2y)=(x−y)(x+y)(x+2y)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)(x + y)(x + 2y). Dividing the numerator by the denominator cancels the common factors (x+y)(x + y) and (x+2y)(x + 2y), leaving x−yx - y.

Adım Adım Çözüm

1
Factor the numerator by grouping terms
x3+2x2y−xy2−2y3=x2(x+2y)−y2(x+2y)=(x2−y2)(x+2y)=(x−y)(x+y)(x+2y)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
x2+3xy+2y2=(x+y)(x+2y)x^2 + 3xy + 2y^2 = (x + y)(x + 2y)
Finding two terms whose sum is 3y3y and product is 2y22y^2 factors the quadratic in xx.
3
Simplify the rational expression by canceling non-zero common factors
(x−y)(x+y)(x+2y)(x+y)(x+2y)=x−y\frac{(x - y)(x + y)(x + 2y)}{(x + y)(x + 2y)} = x - y
Since x≠−yx \neq -y and x≠−2yx \neq -2y, the factors (x+y)(x + y) and (x+2y)(x + 2y) are non-zero and can be canceled.

Anahtar Kavram

Polynomial factoring by grouping, difference of squares, quadratic trinomial factoring, and simplifying rational algebraic expressions.
Tahmini Süre:1m 30s
Soru 26Soru →

For all real numbers xx such that x≠−2x \neq -2, x≠2x \neq 2, and x≠4x \neq 4, which of the following expressions are equivalent to x4−16x2x3−4x2−4x+16\frac{x^4 - 16x^2}{x^3 - 4x^2 - 4x + 16}? Select all such expressions.

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Cevap: x3+4x2x2−4\frac{x^3 + 4x^2}{x^2 - 4}; x2(x+4)(x−2)(x+2)\frac{x^2(x + 4)}{(x - 2)(x + 2)}

Cevap

The equivalent expressions are x3+4x2x2−4\frac{x^3 + 4x^2}{x^2 - 4} and x2(x+4)(x−2)(x+2)\frac{x^2(x + 4)}{(x - 2)(x + 2)}.
The given expression factors as x2(x−4)(x+4)(x2−4)(x−4)\frac{x^2(x - 4)(x + 4)}{(x^2 - 4)(x - 4)}. Canceling (x−4)(x - 4) leaves x2(x+4)x2−4\frac{x^2(x + 4)}{x^2 - 4}. Expanding the numerator gives x3+4x2x2−4\frac{x^3 + 4x^2}{x^2 - 4}, and factoring the denominator further gives x2(x+4)(x−2)(x+2)\frac{x^2(x + 4)}{(x - 2)(x + 2)}. Both represent valid equivalent forms of the expression.

Adım Adım Çözüm

1
Factor the numerator of the given algebraic expression.
x4−16x2=x2(x2−16)=x2(x−4)(x+4)x^4 - 16x^2 = x^2(x^2 - 16) = x^2(x - 4)(x + 4)
Factor out the greatest common factor x2x^2, then apply the difference of squares identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b).
2
Factor the denominator by grouping terms.
x3−4x2−4x+16=x2(x−4)−4(x−4)=(x2−4)(x−4)=(x−2)(x+2)(x−4)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.
x2(x−4)(x+4)(x2−4)(x−4)=x2(x+4)x2−4=x3+4x2x2−4\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≠4x \neq 4, cancel the common factor (x−4)(x - 4) from both numerator and denominator.
4
Compare the simplified form to the given choices to identify all equivalent expressions.
Both x3+4x2x2−4\frac{x^3 + 4x^2}{x^2 - 4} and x2(x+4)(x−2)(x+2)\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.

Anahtar Kavram

Simplifying rational algebraic expressions by polynomial factoring and term grouping.
Soru 27Soru →
For all real numbers xx and yy such that x≠yx \neq y, which of the following expressions is equivalent to x3−x2y−xy2+y3x−y−(x−y)2\frac{x^3 - x^2y - xy^2 + y^3}{x - y} - (x - y)^2?
Cevabı ve açıklamayı göster

Cevap: 2xy−2y22xy - 2y^2

Cevap

The equivalent expression is 2xy−2y22xy - 2y^2.
Factoring the numerator by grouping gives x2(x−y)−y2(x−y)=(x2−y2)(x−y)x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y). Dividing by (x−y)(x - y) yields x2−y2x^2 - y^2. Expanding (x−y)2(x - y)^2 gives x2−2xy+y2x^2 - 2xy + y^2. Subtracting this expansion from x2−y2x^2 - y^2 results in (x2−y2)−(x2−2xy+y2)=2xy−2y2(x^2 - y^2) - (x^2 - 2xy + y^2) = 2xy - 2y^2.

Adım Adım Çözüm

1
Factor the numerator of the rational fraction by grouping terms
x3−x2y−xy2+y3=x2(x−y)−y2(x−y)=(x2−y2)(x−y)=(x+y)(x−y)2x^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)(x - y)
\frac{(x + y)(x - y)^2}{x - y} = (x + y)(x - y) = x^2 - y^2
Since x≠yx \neq y, x−y≠0x - y \neq 0, so common factors can be canceled.
3
Expand the subtracted term (x−y)2(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.

Anahtar Kavram

Simplifying algebraic expressions using factoring by grouping, difference of squares, and expanding binomial squares.
Soru 28Soru →
For all real numbers xx such that x≠−3x \neq -3, x≠0x \neq 0, and x≠3x \neq 3, which of the following expressions is equivalent to
x4−81x3+3x2÷x2+9x2\frac{x^4 - 81}{x^3 + 3x^2} \div \frac{x^2 + 9}{x^2}?
Cevabı ve açıklamayı göster

Cevap: x−3x - 3

Cevap

x−3x - 3
Factoring the numerator x4−81x^4 - 81 gives (x−3)(x+3)(x2+9)(x - 3)(x + 3)(x^2 + 9) and factoring the denominator x3+3x2x^3 + 3x^2 gives x2(x+3)x^2(x + 3). Multiplying by the reciprocal x2x2+9\frac{x^2}{x^2 + 9} allows the factors (x+3)(x + 3), (x2+9)(x^2 + 9), and x2x^2 to cancel completely, leaving x−3x - 3.

Adım Adım Çözüm

1
Rewrite division of rational expressions as multiplication by the reciprocal
x4−81x3+3x2⋅x2x2+9\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 x4−81x^4 - 81 using the difference of squares formula twice
x4−81=(x2−9)(x2+9)=(x−3)(x+3)(x2+9)x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)
Recognize x4−81x^4 - 81 as (x2)2−92(x^2)^2 - 9^2 and then factor x2−9x^2 - 9 further into (x−3)(x+3)(x - 3)(x + 3).
3
Factor out the greatest common factor from the denominator x3+3x2x^3 + 3x^2
x3+3x2=x2(x+3)x^3 + 3x^2 = x^2(x + 3)
Extract x2x^2 from both terms of the binomial.
4
Substitute the factored expressions into the product and cancel common factors
(x−3)(x+3)(x2+9)x2(x+3)⋅x2x2+9=x−3\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 + 3), (x2+9)(x^2 + 9), and x2x^2 cancel out completely from the numerator and denominator for all x≠−3,0,3x \neq -3, 0, 3.

Anahtar Kavram

Factoring polynomial expressions using difference of squares and common terms to simplify rational expressions.
Tahmini Süre:1m 30s
Soru 29Soru →

For all real numbers xx such that x≠−2x \neq -2, which of the following expressions are equivalent to x6−64x3+8\frac{x^6 - 64}{x^3 + 8}? Select all such expressions.

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Cevabı ve açıklamayı göster

Cevap: x3−8x^3 - 8; (x−2)(x2+2x+4)(x - 2)(x^2 + 2x + 4)

Cevap

The expressions equivalent to the given rational expression are x3−8x^3 - 8 and (x−2)(x2+2x+4)(x - 2)(x^2 + 2x + 4).
Factoring the numerator x6−64x^6 - 64 as a difference of squares yields (x3−8)(x3+8)(x^3 - 8)(x^3 + 8). Dividing by the denominator (x3+8)(x^3 + 8) leaves x3−8x^3 - 8. Further factoring x3−8x^3 - 8 as a difference of cubes gives (x−2)(x2+2x+4)(x - 2)(x^2 + 2x + 4). Both x3−8x^3 - 8 and (x−2)(x2+2x+4)(x - 2)(x^2 + 2x + 4) are valid equivalent expressions.

Adım Adım Çözüm

1
Factor the numerator using the difference of squares identity
x6−64=(x3)2−82=(x3−8)(x3+8)x^6 - 64 = (x^3)^2 - 8^2 = (x^3 - 8)(x^3 + 8)
The expression x6−64x^6 - 64 is a difference of two squares.
2
Simplify the fraction by canceling the common non-zero term (x3+8)(x^3 + 8)
\frac{(x^3 - 8)(x^3 + 8)}{x^3 + 8} = x^3 - 8
Since x≠−2x \neq -2, x3+8≠0x^3 + 8 \neq 0, so (x3+8)(x^3 + 8) can be canceled from numerator and denominator.
3
Factor x3−8x^3 - 8 using the difference of cubes identity
x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4)
Applying a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab + b^2) with a=xa = x and b=2b = 2.

Anahtar Kavram

Simplifying rational expressions by factoring polynomial numerators and denominators using algebraic identities such as difference of squares and difference of cubes.
Soru 30Soru →
For all real numbers xx such that x≠−3x \neq -3, x≠0x \neq 0, and x≠3x \neq 3, which of the following expressions is equivalent to x−2−9−1x−1+3−1⋅3x2+9xx−3?\frac{x^{-2} - 9^{-1}}{x^{-1} + 3^{-1}} \cdot \frac{3x^2 + 9x}{x - 3}?
Cevabı ve açıklamayı göster

Cevap: −x−3-x - 3

Cevap

−x−3-x - 3
The numerator of the first factor is a difference of squares (x−1−3−1)(x−1+3−1)(x^{-1} - 3^{-1})(x^{-1} + 3^{-1}). Dividing by x−1+3−1x^{-1} + 3^{-1} leaves x−1−3−1=1x−13=3−x3xx^{-1} - 3^{-1} = \frac{1}{x} - \frac{1}{3} = \frac{3 - x}{3x}. Factoring 3x3x out of the numerator of the second expression gives 3x(x+3)3x(x + 3). Multiplying these terms yields 3−x3x⋅3x(x+3)x−3\frac{3 - x}{3x} \cdot \frac{3x(x + 3)}{x - 3}. Since 3−x=−(x−3)3 - x = -(x - 3), the binomials (3−x)(3 - x) and (x−3)(x - 3) cancel to −1-1, and 3x3x cancels out completely, resulting in −(x+3)=−x−3-(x + 3) = -x - 3.

Adım Adım Çözüm

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)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)(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 3x2+9x3x^2 + 9x.
3x^2 + 9x = 3x(x + 3)
Factoring out the greatest common factor 3x3x 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 3x3x and noting that (3−x)/(x−3)=−1(3 - x)/(x - 3) = -1 leaves the linear expression −x−3-x - 3.

Anahtar Kavram

Simplifying algebraic expressions containing negative exponents, rational fractions, and opposite-sign binomial factors
Soru 31Soru →

When the algebraic expression 4x3−16x2x2+4x\frac{4x^3 - 16x}{2x^2 + 4x} is simplified for all x≠0x \neq 0 and x≠−2x \neq -2, it reduces to the linear polynomial ax+bax + b, where aa and bb are constants. What is the value of a+ba + b?

Cevabı ve açıklamayı göster

Cevap: -2

Cevap

The correct numerical answer is -2.
Factoring the numerator yields 4x(x−2)(x+2)4x(x - 2)(x + 2) and factoring the denominator yields 2x(x+2)2x(x + 2). Canceling the common factors 2x2x and (x+2)(x + 2) leaves 2(x−2)=2x−42(x - 2) = 2x - 4. Comparing 2x−42x - 4 to ax+bax + b gives a=2a = 2 and b=−4b = -4. Summing these values gives a+b=2+(−4)=−2a + b = 2 + (-4) = -2.

Adım Adım Çözüm

1
Factor out the greatest common factor and apply the difference of squares formula to the numerator.
4x3−16x=4x(x2−4)=4x(x−2)(x+2)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.
2x2+4x=2x(x+2)2x^2 + 4x = 2x(x + 2)
Extracting 2x2x reveals the common terms shared with the numerator.
3
Divide the numerator by the denominator by canceling identical non-zero factors 2x2x and (x+2)(x + 2).
4x(x−2)(x+2)2x(x+2)=42(x−2)=2(x−2)=2x−4\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−42x - 4 to the form ax+bax + b to find aa and bb, then compute their sum.
a=2a = 2 and b=−4  ⟹  a+b=2+(−4)=−2b = -4 \implies a + b = 2 + (-4) = -2
Matching corresponding terms identifies the values of the target constants.

Anahtar Kavram

Simplifying rational expressions by factoring out common terms and applying the difference of squares identity.
Soru 32Soru →

For all real numbers xx and yy such that y≠0y \neq 0 and 3x2+y2≠03x^2 + y^2 \neq 0, which of the following expressions is equivalent to (x+y)3−(x−y)32y(3x2+y2)\frac{(x + y)^3 - (x - y)^3}{2y(3x^2 + y^2)}?

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Cevap: 11

Cevap

11
Expanding the numerator yields (x3+3x2y+3xy2+y3)−(x3−3x2y+3xy2−y3)=6x2y+2y3=2y(3x2+y2)(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.

Adım Adım Çözüm

1
Expand (x+y)3(x + y)^3 and (x−y)3(x - y)^3 using the binomial theorem.
(x+y)3=x3+3x2y+3xy2+y3(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 and (x−y)3=x3−3x2y+3xy2−y3(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(x - y)^3 from (x+y)3(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 x3x^3 and 3xy23xy^2 subtract to zero, while the remaining terms double.
3
Factor out common terms from the simplified numerator.
6x2y+2y3=2y(3x2+y2)6x^2y + 2y^3 = 2y(3x^2 + y^2)
Factoring out 2y2y reveals a factor identical to the denominator.
4
Divide the numerator by the denominator 2y(3x2+y2)2y(3x^2 + y^2).
2y(3x2+y2)2y(3x2+y2)=1\frac{2y(3x^2 + y^2)}{2y(3x^2 + y^2)} = 1
Since y≠0y \neq 0 and 3x2+y2≠03x^2 + y^2 \neq 0, identical non-zero factors cancel out.

Anahtar Kavram

Binomial expansion of cubic expressions and factoring algebraic expressions.
Tahmini Süre:1m 30s
Soru 33Soru →

If xx and yy are non-zero real numbers such that x≠yx \neq y, x2−y2x−y=12\frac{x^2 - y^2}{x - y} = 12, and x2y−xy2xy=4\frac{x^2y - xy^2}{xy} = 4, what is the value of x2+y2x^2 + y^2?

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Cevap: 80

Cevap

The value of x2+y2x^2 + y^2 is 80.
Simplifying the first equation by factoring the difference of squares (x−y)(x+y)x−y=12\frac{(x-y)(x+y)}{x-y} = 12 gives x+y=12x + y = 12. Simplifying the second equation by factoring out xyxy yields xy(x−y)xy=4\frac{xy(x-y)}{xy} = 4, which gives x−y=4x - y = 4. Solving this system yields x=8x = 8 and y=4y = 4. Squaring and summing these values gives 82+42=64+16=808^2 + 4^2 = 64 + 16 = 80. Alternatively, using the algebraic identity (x+y)2+(x−y)22=122+422=144+162=80\frac{(x+y)^2 + (x-y)^2}{2} = \frac{12^2 + 4^2}{2} = \frac{144 + 16}{2} = 80 directly yields the correct answer.

Adım Adım Çözüm

1
Simplify the first given algebraic expression using the difference of squares identity.
x+y=12x + y = 12
Factoring x2−y2x^2 - y^2 gives (x−y)(x+y)(x - y)(x + y). Since x≠yx \neq y, we can cancel the non-zero common factor (x−y)(x - y) from the numerator and denominator.
2
Simplify the second given algebraic expression by factoring out the greatest common factor.
x−y=4x - y = 4
Factoring xyxy from x2y−xy2x^2y - xy^2 yields xy(x−y)xy(x - y). Since x,y≠0x, y \neq 0, we cancel xyxy from the numerator and denominator.
3
Solve for the individual values of xx and yy.
x=8x = 8 and y=4y = 4
Adding (x+y=12)(x + y = 12) and (x−y=4)(x - y = 4) gives 2x=162x = 16, so x=8x = 8. Subtracting the equations gives 2y=82y = 8, so y=4y = 4.
4
Calculate the target expression x2+y2x^2 + y^2.
80
x2+y2=82+42=64+16=80x^2 + y^2 = 8^2 + 4^2 = 64 + 16 = 80.

Anahtar Kavram

Factoring Algebraic Expressions (Difference of Squares and GCF Extraction)
Soru 34Soru →

For all real numbers xx and yy such that x≠2yx \neq 2y and x≠−2yx \neq -2y, which of the following expressions is equivalent to x3+2x2y−4xy2−8y3x2−4y2\frac{x^3 + 2x^2y - 4xy^2 - 8y^3}{x^2 - 4y^2}?

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Cevap: x+2yx + 2y

Cevap

The simplified expression is x+2yx + 2y.
Grouping terms in the numerator gives x2(x+2y)−4y2(x+2y)=(x2−4y2)(x+2y)x^2(x + 2y) - 4y^2(x + 2y) = (x^2 - 4y^2)(x + 2y). Dividing this by the denominator (x2−4y2)(x^2 - 4y^2) cancels out the identical non-zero factor (x2−4y2)(x^2 - 4y^2), leaving the linear binomial x+2yx + 2y.

Adım Adım Çözüm

1
Group the four terms in the numerator in pairs to factor by grouping
x3+2x2y−4xy2−8y3=x2(x+2y)−4y2(x+2y)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 x2x^2 and −4y2-4y^2 respectively.
2
Factor out the common binomial factor (x+2y)(x + 2y) from the numerator
x2(x+2y)−4y2(x+2y)=(x2−4y2)(x+2y)x^2(x + 2y) - 4y^2(x + 2y) = (x^2 - 4y^2)(x + 2y)
Both terms share the common factor (x+2y)(x + 2y).
3
Divide the factored numerator by the denominator (x2−4y2)(x^2 - 4y^2)
(x2−4y2)(x+2y)x2−4y2=x+2y\frac{(x^2 - 4y^2)(x + 2y)}{x^2 - 4y^2} = x + 2y
Since x≠±2yx \neq \pm 2y, x2−4y2≠0x^2 - 4y^2 \neq 0, allowing the common polynomial factor (x2−4y2)(x^2 - 4y^2) to be canceled.

Anahtar Kavram

Factoring Four-Term Polynomials by Grouping and Rational Expression Simplification
ÖncekiSayfa 2 / 2
Simplifying and Factoring Algebraic Expressions Alıştırma Soruları — GRE General Test — Sayfa 2 | Examkin