Examkin

Elementary Algebra

302 soru

Tüm türler302Çoktan seçmeli155Sayısal125Eşleştirme15Boşluk doldurma7
Soru 41Soru →

For all real numbers xx and yy, when the expression −2x(x−3y)2−y2(4x−y)−3x2(2x−y)-2x(x - 3y)^2 - y^2(4x - y) - 3x^2(2x - y) is completely simplified, what is the coefficient of the x2yx^2y term?

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Cevap: 15

Cevap

15
Completely simplifying the given expression yields −8x3+15x2y−22xy2+y3-8x^3 + 15x^2y - 22xy^2 + y^3. The coefficient of the x2yx^2y term is 1515, which is obtained by combining the term 12x2y12x^2y (from the distribution of −2x-2x over the middle term of the squared binomial) and the term 3x2y3x^2y (from the distribution of −3x2-3x^2 over −y-y).

Adım Adım Çözüm

1
Expand the binomial squared expression (x−3y)2(x - 3y)^2.
(x−3y)2=x2−6xy+9y2(x - 3y)^2 = x^2 - 6xy + 9y^2
Following the order of operations, we must square the binomial before distributing the outer term.
2
Distribute −2x-2x to the trinomial result from the previous step.
−2x(x2−6xy+9y2)=−2x3+12x2y−18xy2-2x(x^2 - 6xy + 9y^2) = -2x^3 + 12x^2y - 18xy^2
Using the distributive property, we multiply coefficients and add exponents of like bases (noting that −2x×−6xy=12x2y-2x \times -6xy = 12x^2y).
3
Distribute −y2-y^2 to the binomial (4x−y)(4x - y).
−y2(4x−y)=−4xy2+y3-y^2(4x - y) = -4xy^2 + y^3
We multiply each term inside the parentheses by −y2-y^2, keeping track of the signs.
4
Distribute −3x2-3x^2 to the binomial (2x−y)(2x - y).
−3x2(2x−y)=−6x3+3x2y-3x^2(2x - y) = -6x^3 + 3x^2y
We distribute the −3x2-3x^2 factor, ensuring that multiplying two negative values results in a positive term (−3x2×−y=3x2y-3x^2 \times -y = 3x^2y).
5
Combine the coefficients of all like terms containing x2yx^2y.
12x2y+3x2y=15x2y12x^2y + 3x^2y = 15x^2y
We add the coefficients of the terms that share the exact variable part x2yx^2y to find the final coefficient.

Anahtar Kavram

Simplifying algebraic expressions by distributing terms, applying exponent rules, and combining like terms.
Soru 42Soru →

For all real values of aa and bb, which of the following is equivalent to the expression 2a(a2−3b)−3(a3−2ab+b2)−(4b2−a3)2a(a^2 - 3b) - 3(a^3 - 2ab + b^2) - (4b^2 - a^3)?

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Cevap: −7b2-7b^2

Cevap

−7b2-7b^2
Distributing the terms yields 2a3−6ab−3a3+6ab−3b2−4b2+a32a^3 - 6ab - 3a^3 + 6ab - 3b^2 - 4b^2 + a^3. Combining the like terms for a3a^3 gives 2a3−3a3+a3=02a^3 - 3a^3 + a^3 = 0. Combining the abab terms gives −6ab+6ab=0-6ab + 6ab = 0. Combining the b2b^2 terms gives −3b2−4b2=−7b2-3b^2 - 4b^2 = -7b^2. Thus, the simplified expression is −7b2-7b^2.

Adım Adım Çözüm

1
Distribute the factors outside the parentheses to each term inside the parentheses.
The terms expand to: 2a(a2)−2a(3b)−3(a3)−3(−2ab)−3(b2)−1(4b2)−1(−a3)=2a3−6ab−3a3+6ab−3b2−4b2+a32a(a^2) - 2a(3b) - 3(a^3) - 3(-2ab) - 3(b^2) - 1(4b^2) - 1(-a^3) = 2a^3 - 6ab - 3a^3 + 6ab - 3b^2 - 4b^2 + a^3
Distribution eliminates parentheses, making it possible to group and combine like terms.
2
Group like terms together based on their variable parts and powers.
(2a3−3a3+a3)+(−6ab+6ab)+(−3b2−4b2)(2a^3 - 3a^3 + a^3) + (-6ab + 6ab) + (-3b^2 - 4b^2)
Grouping like terms makes it easier to perform the arithmetic on the coefficients.
3
Combine the coefficients for each group of like terms.
0a3+0ab−7b2=−7b20a^3 + 0ab - 7b^2 = -7b^2
Simplifying the coefficient sums yields the final simplified form of the expression.

Anahtar Kavram

Simplifying expressions by distributing terms and combining like terms
Soru 43Soru →

The quadratic equation x2−4x−12=0x^2 - 4x - 12 = 0 has two real solutions. What is the value of the positive solution to this equation?

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Cevap: 6

Cevap

The positive solution to the equation is 66.
Factoring the quadratic trinomial x2−4x−12=0x^2 - 4x - 12 = 0 yields (x−6)(x+2)=0(x - 6)(x + 2) = 0. Setting the individual binomial factors to zero gives the solutions x=6x = 6 and x=−2x = -2. The positive solution among these is 66.

Adım Adım Çözüm

1
Factor the quadratic equation
(x−6)(x+2)=0(x - 6)(x + 2) = 0
Factoring the trinomial x2−4x−12x^2 - 4x - 12 requires finding two integers whose product is −12-12 and whose sum is −4-4. These numbers are −6-6 and 22.
2
Apply the zero product property
x−6=0x - 6 = 0 or x+2=0x + 2 = 0
If the product of two factors is equal to zero, then at least one of the individual factors must equal zero.
3
Solve for the variable and identify the positive root
x=6x = 6 and x=−2x = -2
Solving the linear equations yields x=6x = 6 and x=−2x = -2. Since the question asks for the positive solution, we select 66.

Anahtar Kavram

Solving quadratic equations by factoring
Soru 44Soru →

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

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Cevap: 3x−53x - 5

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Factoring polynomials by grouping and difference of squares.
Tahmini Süre:2m 0s
Soru 45Soru →

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

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Cevap: x−2yx - 2y

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

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

Alternatif Yöntem

Instead of factoring out the GCF 2 first, you can group the terms directly: 2x4−2x2−8x2y2+8y2=2x2(x2−1)−8y2(x2−1)=(2x2−8y2)(x2−1)2x^4 - 2x^2 - 8x^2y^2 + 8y^2 = 2x^2(x^2 - 1) - 8y^2(x^2 - 1) = (2x^2 - 8y^2)(x^2 - 1). Then, factor out 2 from the first binomial to get 2(x2−4y2)(x2−1)2(x^2 - 4y^2)(x^2 - 1), and finally apply the difference of squares identity to both quadratic factors to obtain 2(x−2y)(x+2y)(x−1)(x+1)2(x - 2y)(x + 2y)(x - 1)(x + 1).
Tahmini Süre:1m 30s
Soru 46Soru →

A certain real number xx satisfies the condition that the square of 33 less than twice xx is equal to 88 times the quantity 33 minus xx. What is the ratio of the larger solution to the smaller solution of this equation?

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Cevap: −53-\frac{5}{3}

Cevap

The ratio of the larger solution to the smaller solution is −53-\frac{5}{3}.
The correct answer is −53-\frac{5}{3}. The verbal statement translates directly to (2x−3)2=8(3−x)(2x - 3)^2 = 8(3 - x). Expanding both sides yields 4x2−12x+9=24−8x4x^2 - 12x + 9 = 24 - 8x. Rearranging into standard form gives 4x2−4x−15=04x^2 - 4x - 15 = 0. Factoring by grouping yields (2x+3)(2x−5)=0(2x + 3)(2x - 5) = 0, giving the solutions x=−32x = -\frac{3}{2} and x=52x = \frac{5}{2}. The ratio of the larger root to the smaller root is 5/2−3/2=−53\frac{5/2}{-3/2} = -\frac{5}{3}.

Adım Adım Çözüm

1
Translate the verbal description into an algebraic equation.
(2x−3)2=8(3−x)(2x - 3)^2 = 8(3 - x)
'Twice xx' is 2x2x, '3 less than twice xx' is 2x−32x - 3, and its square is (2x−3)2(2x - 3)^2. This is equal to 8 times the quantity 3−x3 - x.
2
Expand both sides of the equation.
4x2−12x+9=24−8x4x^2 - 12x + 9 = 24 - 8x
Expanding the binomial (2x−3)2(2x - 3)^2 gives 4x2−12x+94x^2 - 12x + 9 and distributing the right side gives 24−8x24 - 8x.
3
Rearrange the equation to set it equal to zero.
4x2−4x−15=04x^2 - 4x - 15 = 0
Add 8x8x and subtract 2424 from both sides to gather all terms on one side of the equation.
4
Factor the quadratic equation by grouping.
(2x+3)(2x−5)=0(2x + 3)(2x - 5) = 0
Find two numbers that multiply to 4×(−15)=−604 \times (-15) = -60 and add to −4-4. These numbers are −10-10 and 66. Rewrite the middle term as −10x+6x-10x + 6x and factor: 2x(2x−5)+3(2x−5)=0  ⟹  (2x+3)(2x−5)=02x(2x - 5) + 3(2x - 5) = 0 \implies (2x + 3)(2x - 5) = 0.
5
Solve for the roots of the equation.
x=−32x = -\frac{3}{2} and x=52x = \frac{5}{2}
Set each factor equal to zero: 2x+3=0  ⟹  x=−322x + 3 = 0 \implies x = -\frac{3}{2} and 2x−5=0  ⟹  x=522x - 5 = 0 \implies x = \frac{5}{2}.
6
Identify the larger and smaller solutions and compute their ratio.
−53-\frac{5}{3}
The larger solution is 52\frac{5}{2} and the smaller solution is −32-\frac{3}{2}. Their ratio is 5/2−3/2=−53\frac{5/2}{-3/2} = -\frac{5}{3}.

Anahtar Kavram

Solving quadratic equations of the form ax2+bx+c=0ax^2 + bx + c = 0 by factoring over the integers.
Soru 47Soru →

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

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Cevap: 7

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Factoring quadratic trinomials with a leading coefficient greater than 1

Alternatif Yöntem

Factor the trinomial 2x2+7x+32x^2 + 7x + 3 using the AC method: multiply the leading coefficient (2) and the constant term (3) to get 6. Find two numbers that multiply to 6 and add to 7, which are 6 and 1. Rewrite the middle term: 2x2+6x+x+32x^2 + 6x + x + 3. Factor by grouping: 2x(x+3)+1(x+3)=(2x+1)(x+3)2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Compare this to (2x+a)(x+b)(2x + a)(x + b) to find a=1a = 1 and b=3b = 3, then compute a+2b=1+2(3)=7a + 2b = 1 + 2(3) = 7.
Tahmini Süre:45s
Soru 48Soru →

For all real values of xx and yy, the expression 3x(x−2y)−2(x2−4xy+y2)−x23x(x - 2y) - 2(x^2 - 4xy + y^2) - x^2 can be written in the form axy+by2axy + by^2, where aa and bb are constants. What is the value of aa?

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Cevap: 2

Cevap

The value of aa is 2.
The value of aa is 2 because distributing 3x(x−2y)3x(x - 2y) yields 3x2−6xy3x^2 - 6xy, and distributing −2(x2−4xy+y2)-2(x^2 - 4xy + y^2) yields −2x2+8xy−2y2-2x^2 + 8xy - 2y^2. Combining these with the −x2-x^2 term yields (3−2−1)x2+(−6+8)xy−2y2=2xy−2y2(3-2-1)x^2 + (-6+8)xy - 2y^2 = 2xy - 2y^2. Comparing this to axy+by2axy + by^2 shows that aa, the coefficient of the xyxy term, is 2.

Adım Adım Çözüm

1
Distribute 3x3x across the first parenthetical expression (x−2y)(x - 2y)
3x2−6xy3x^2 - 6xy
To clear the first set of parentheses by multiplying 3x3x by each term inside.
2
Distribute −2-2 across the second parenthetical expression (x2−4xy+y2)(x^2 - 4xy + y^2)
−2x2+8xy−2y2-2x^2 + 8xy - 2y^2
To clear the second set of parentheses. Note that multiplying −2-2 by −4xy-4xy yields a positive term +8xy+8xy due to the sign rules.
3
Write the full expression and group like terms
(3x2−2x2−x2)+(−6xy+8xy)−2y2(3x^2 - 2x^2 - x^2) + (-6xy + 8xy) - 2y^2
To group terms with identical variable parts so they can be combined.
4
Combine the coefficients of the grouped terms
2xy−2y22xy - 2y^2
Simplifying the groups: 3−2−1=03-2-1=0 for the x2x^2 terms, and −6+8=2-6+8=2 for the xyxy terms.
5
Compare the simplified expression to the form axy+by2axy + by^2 to find the coefficient aa
a=2a = 2
The coefficient of the xyxy term is 22, which corresponds to aa in the target expression.

Anahtar Kavram

Simplifying Expressions and Combining Like Terms
Tahmini Süre:1m 30s
Soru 49Soru →

What is the sum of all real values of xx that satisfy the equation (x−3)2+x(x+2)=15(x - 3)^2 + x(x + 2) = 15?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The sum of all real values of xx that satisfy the equation is 22.
Expanding the equation yields 2x2−4x+9=152x^2 - 4x + 9 = 15. Setting this to zero gives 2x2−4x−6=02x^2 - 4x - 6 = 0. Dividing by the common factor of 22 simplifies this to x2−2x−3=0x^2 - 2x - 3 = 0. Factoring the trinomial yields (x−3)(x+1)=0(x - 3)(x + 1) = 0, which gives the two solutions x=3x = 3 and x=−1x = -1. Summing these two solutions gives 3+(−1)=23 + (-1) = 2.

Adım Adım Çözüm

1
Expand both terms on the left side of the equation.
(x2−6x+9)+(x2+2x)=15(x^2 - 6x + 9) + (x^2 + 2x) = 15
Applying the binomial squaring formula (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 to (x−3)2(x - 3)^2 and distributing xx to both terms in x(x+2)x(x + 2) allows us to simplify the equation.
2
Combine like terms and set the quadratic equation to zero.
2x2−4x−6=02x^2 - 4x - 6 = 0
Grouping x2x^2 terms, xx terms, and constant terms on one side is necessary to format the quadratic equation as ax2+bx+c=0ax^2 + bx + c = 0 before factoring.
3
Divide the entire equation by the common factor of 22 to simplify factoring.
x2−2x−3=0x^2 - 2x - 3 = 0
Simplifying the quadratic equation makes it easier to find two binomial factors.
4
Factor the quadratic trinomial by finding two numbers that multiply to −3-3 and add to −2-2.
(x−3)(x+1)=0(x - 3)(x + 1) = 0
Since −3×1=−3-3 \times 1 = -3 and −3+1=−2-3 + 1 = -2, we can write the quadratic in factored form.
5
Set each factor to zero to solve for xx.
x=3x = 3 or x=−1x = -1
Applying the zero product property determines the two values of xx that satisfy the original equation.
6
Calculate the sum of the two solutions.
3+(−1)=23 + (-1) = 2
The question asks for the sum of all real values of xx that satisfy the equation.

Anahtar Kavram

Solving quadratic equations by rearranging terms, factoring trinomials, and applying the zero product property.
Soru 50Soru →

For all real values of xx and yy, which of the following is equivalent to the expression x(x−y)2−x2(x−2y)x(x - y)^2 - x^2(x - 2y)?

Cevabı ve açıklamayı göster

Cevap: xy2xy^2

Cevap

The simplified expression is xy2xy^2.
Expanding (x−y)2(x - y)^2 yields x2−2xy+y2x^2 - 2xy + y^2. Distributing xx to this expression results in x3−2x2y+xy2x^3 - 2x^2y + xy^2. Distributing −x2-x^2 to (x−2y)(x - 2y) yields −x3+2x2y-x^3 + 2x^2y. Combining these parts gives (x3−x3)+(−2x2y+2x2y)+xy2(x^3 - x^3) + (-2x^2y + 2x^2y) + xy^2, which simplifies completely to xy2xy^2.

Adım Adım Çözüm

1
Expand the squared binomial (x−y)2(x - y)^2 using the algebraic identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.
(x−y)2=x2−2xy+y2(x - y)^2 = x^2 - 2xy + y^2
Expanding the binomial is necessary before distributing the outer variable.
2
Distribute the term xx to each term in the expanded binomial, and distribute the term −x2-x^2 to each term inside the second parenthesis.
x(x2−2xy+y2)=x3−2x2y+xy2x(x^2 - 2xy + y^2) = x^3 - 2x^2y + xy^2 and −x2(x−2y)=−x3+2x2y-x^2(x - 2y) = -x^3 + 2x^2y
Distribution eliminates parentheses and prepares the expression for combining like terms.
3
Combine all like terms in the resulting expression: (x3−x3)+(−2x2y+2x2y)+xy2(x^3 - x^3) + (-2x^2y + 2x^2y) + xy^2.
xy2xy^2
Combining like terms simplifies the expression to its final equivalent form.

Anahtar Kavram

Simplifying algebraic expressions by expanding binomials, distributing variables, and combining like terms.
Tahmini Süre:1m 0s
Soru 51Soru →

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

Cevabı ve açıklamayı göster

Cevap: 14

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Factoring by grouping and the difference of squares
Soru 52Soru →

When the expression −(p−2q)3−p(p−3q)(2p+q)+3q2(p−q)-(p - 2q)^3 - p(p - 3q)(2p + q) + 3q^2(p - q) is completely simplified by combining like terms, what is the coefficient of p2qp^2q?

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

The coefficient of p2qp^2q in the fully simplified expression is 11.
Expanding the three components of the expression yields: −(p−2q)3=−p3+6p2q−12pq2+8q3-(p - 2q)^3 = -p^3 + 6p^2q - 12pq^2 + 8q^3; −p(p−3q)(2p+q)=−2p3+5p2q+3pq2-p(p - 3q)(2p + q) = -2p^3 + 5p^2q + 3pq^2; and 3q2(p−q)=3pq2−3q33q^2(p - q) = 3pq^2 - 3q^3. Combining these terms gives the simplified polynomial −3p3+11p2q−6pq2+5q3-3p^3 + 11p^2q - 6pq^2 + 5q^3. The coefficient of p2qp^2q is 11.

Adım Adım Çözüm

1
Expand and negate the term −(p−2q)3-(p - 2q)^3
−p3+6p2q−12pq2+8q3-p^3 + 6p^2q - 12pq^2 + 8q^3
Using the binomial theorem expansion for (a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 and distributing the negative sign.
2
Multiply and distribute −p(p−3q)(2p+q)-p(p - 3q)(2p + q)
−2p3+5p2q+3pq2-2p^3 + 5p^2q + 3pq^2
First multiply the binomials (p−3q)(2p+q)=2p2−5pq−3q2(p - 3q)(2p + q) = 2p^2 - 5pq - 3q^2, and then multiply each term by −p-p.
3
Distribute 3q2(p−q)3q^2(p - q)
3pq2−3q33pq^2 - 3q^3
Multiply 3q23q^2 by both terms inside the binomial.
4
Combine the like terms of p2qp^2q
1111
Identify and sum all terms containing p2qp^2q: 6p2q+5p2q=11p2q6p^2q + 5p^2q = 11p^2q.

Anahtar Kavram

Simplifying algebraic expressions by expanding polynomials, applying the distributive property with negative signs, and combining like terms.
Soru 53Soru →

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

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Factoring polynomials using grouping, perfect square trinomials, and the difference of squares identity
Tahmini Süre:2m 0s
Soru 54Soru →

When the expression 2x(x2−3xy+2y2)−(x−2y)3−y(x2−5xy+4y2)2x(x^2 - 3xy + 2y^2) - (x - 2y)^3 - y(x^2 - 5xy + 4y^2) is fully simplified by combining like terms, what is the coefficient of the xy2xy^2 term?

Cevabı ve açıklamayı göster

Cevap: -3

Cevap

The coefficient of the xy2xy^2 term is −3-3.
Expanding the terms of the expression yields 2x3−6x2y+4xy22x^3 - 6x^2y + 4xy^2, −x3+6x2y−12xy2+8y3-x^3 + 6x^2y - 12xy^2 + 8y^3, and −x2y+5xy2−4y3-x^2y + 5xy^2 - 4y^3. Summing the coefficients of the xy2xy^2 terms gives 4−12+5=−34 - 12 + 5 = -3. Thus, the coefficient of xy2xy^2 is −3-3.

Adım Adım Çözüm

1
Expand the first term of the expression: 2x(x2−3xy+2y2)2x(x^2 - 3xy + 2y^2)
2x3−6x2y+4xy22x^3 - 6x^2y + 4xy^2
Distribute the monomial 2x2x to each term inside the parentheses: 2x⋅x2=2x32x \cdot x^2 = 2x^3, 2x⋅(−3xy)=−6x2y2x \cdot (-3xy) = -6x^2y, and 2x⋅2y2=4xy22x \cdot 2y^2 = 4xy^2.
2
Expand the cubed binomial (x−2y)3(x - 2y)^3 and distribute the negative sign
−x3+6x2y−12xy2+8y3-x^3 + 6x^2y - 12xy^2 + 8y^3
First, expand the binomial (x−2y)3=x3−3(x2)(2y)+3(x)(2y)2−(2y)3=x3−6x2y+12xy2−8y3(x - 2y)^3 = x^3 - 3(x^2)(2y) + 3(x)(2y)^2 - (2y)^3 = x^3 - 6x^2y + 12xy^2 - 8y^3. Then, distribute the negative sign to all terms inside the parentheses.
3
Expand the third term: −y(x2−5xy+4y2)-y(x^2 - 5xy + 4y^2)
−x2y+5xy2−4y3-x^2y + 5xy^2 - 4y^3
Distribute the negative monomial −y-y to each term inside the parentheses: −y⋅x2=−x2y-y \cdot x^2 = -x^2y, −y⋅(−5xy)=5xy2-y \cdot (-5xy) = 5xy^2, and −y⋅4y2=−4y3-y \cdot 4y^2 = -4y^3.
4
Group and combine all like terms
x3−x2y−3xy2+4y3x^3 - x^2y - 3xy^2 + 4y^3
Combine the coefficients of the corresponding terms: (2x3−x3)+(−6x2y+6x2y−x2y)+(4xy2−12xy2+5xy2)+(8y3−4y3)=x3−x2y−3xy2+4y3(2x^3 - x^3) + (-6x^2y + 6x^2y - x^2y) + (4xy^2 - 12xy^2 + 5xy^2) + (8y^3 - 4y^3) = x^3 - x^2y - 3xy^2 + 4y^3.

Anahtar Kavram

Simplifying Expressions and Combining Like Terms
Tahmini Süre:3m 0s
Soru 55Soru →

For a certain positive number yy, the product of yy and the quantity 2y+52y + 5 is equal to 1212. What is the value of yy?

Cevabı ve açıklamayı göster

Cevap: 1.5

Cevap

The positive value of yy is 1.51.5.
The correct answer is 1.51.5. By translating the word problem, we obtain y(2y+5)=12y(2y + 5) = 12. Expanding this gives 2y2+5y=122y^2 + 5y = 12, and subtracting 1212 from both sides yields the standard quadratic equation 2y2+5y−12=02y^2 + 5y - 12 = 0. Factoring this expression gives (2y−3)(y+4)=0(2y - 3)(y + 4) = 0. Setting the first factor to zero yields y=1.5y = 1.5, which is positive and therefore satisfies the given condition.

Adım Adım Çözüm

1
Write the equation representing the relationship.
y(2y+5)=12y(2y + 5) = 12
To translate the verbal description into an algebraic equation.
2
Distribute yy and set the equation equal to zero.
2y2+5y−12=02y^2 + 5y - 12 = 0
To put the quadratic equation into standard form ay2+by+c=0ay^2 + by + c = 0 so it can be factored.
3
Factor the quadratic equation.
(2y−3)(y+4)=0(2y - 3)(y + 4) = 0
To find the factors that multiply to give the quadratic expression.
4
Solve for yy and apply the constraint.
y=1.5y = 1.5
Setting the factors to zero gives y=1.5y = 1.5 and y=−4y = -4. Since the problem states yy is positive, we select the positive root.

Anahtar Kavram

Solving a non-monic quadratic equation by factoring after translating a verbal description into algebra.
Soru 56Soru →

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

Cevabı ve açıklamayı göster

Cevap: 6x6x

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Factoring quartic polynomials using quadratic substitution and the difference of squares identity.
Soru 57Soru →

When the expression 2x2(3x−y)−3y(x2−2y)−(x3−3xy2)2x^2(3x - y) - 3y(x^2 - 2y) - (x^3 - 3xy^2) is fully simplified by combining like terms, which of the following represents the resulting expression?

Cevabı ve açıklamayı göster

Cevap: 5x3−5x2y+3xy2+6y25x^3 - 5x^2y + 3xy^2 + 6y^2

Cevap

5x3−5x2y+3xy2+6y25x^3 - 5x^2y + 3xy^2 + 6y^2
The correct expression is obtained by systematically distributing the coefficients outside the parentheses and then grouping and combining the coefficients of the like terms: (6x3−x3)+(−2x2y−3x2y)+3xy2+6y2=5x3−5x2y+3xy2+6y2(6x^3 - x^3) + (-2x^2y - 3x^2y) + 3xy^2 + 6y^2 = 5x^3 - 5x^2y + 3xy^2 + 6y^2.

Adım Adım Çözüm

1
Distribute 2x22x^2 to each term in the first parenthetical expression: (3x−y)(3x - y)
6x3−2x2y6x^3 - 2x^2y
Applying the distributive property of multiplication over subtraction.
2
Distribute −3y-3y to each term in the second parenthetical expression: (x2−2y)(x^2 - 2y)
−3x2y+6y2-3x^2y + 6y^2
Applying the distributive property and multiplying negative coefficients (−3×−2=6-3 \times -2 = 6).
3
Distribute the negative sign to each term in the third parenthetical expression: (x3−3xy2)(x^3 - 3xy^2)
−x3+3xy2-x^3 + 3xy^2
Distributing −1-1 across the parentheses to remove them.
4
Combine the expanded expressions and group the like terms
(6x3−x3)+(−2x2y−3x2y)+3xy2+6y2(6x^3 - x^3) + (-2x^2y - 3x^2y) + 3xy^2 + 6y^2
Grouping together terms that have the same variables raised to the same powers.
5
Simplify by performing the operations on the coefficients of the like terms
5x3−5x2y+3xy2+6y25x^3 - 5x^2y + 3xy^2 + 6y^2
Combining coefficients: 6−1=56 - 1 = 5 for the x3x^3 terms, and −2−3=−5-2 - 3 = -5 for the x2yx^2y terms.

Anahtar Kavram

Simplifying algebraic expressions by distributing coefficients and combining like terms.

Alternatif Yöntem

To check your work, substitute simple values for xx and yy, such as x=1x = 1 and y=1y = 1, into the original expression and the simplified expression. Evaluating the original expression: 2(1)2(3(1)−1)−3(1)(12−2(1))−(13−3(1)(1)2)=2(2)−3(−1)−(1−3)=4+3−(−2)=92(1)^2(3(1) - 1) - 3(1)(1^2 - 2(1)) - (1^3 - 3(1)(1)^2) = 2(2) - 3(-1) - (1 - 3) = 4 + 3 - (-2) = 9. Evaluating the correct simplified expression: 5(1)3−5(1)2(1)+3(1)(1)2+6(1)2=5−5+3+6=95(1)^3 - 5(1)^2(1) + 3(1)(1)^2 + 6(1)^2 = 5 - 5 + 3 + 6 = 9. Since both evaluations yield 9, this confirms the simplification.
Tahmini Süre:1m 0s
Soru 58Soru →

A business analyst models the daily cost, CC, and daily revenue, RR, of a manufacturing process using the following expressions:

C=2x2(x−3y)−y(x2−2y2)C = 2x^2(x - 3y) - y(x^2 - 2y^2)
R=5x3−x(y−2x)2+3y3R = 5x^3 - x(y - 2x)^2 + 3y^3

where xx represents the number of units of product X sold, and yy represents the number of units of product Y sold. The daily profit, PP, is defined as P=R−CP = R - C. When the profit expression is fully simplified by combining like terms, which of the following expressions represents the daily profit PP?

Cevabı ve açıklamayı göster

Cevap: $-x^3 + 11x^2y - xy^2 + y^3

Cevap

−x3+11x2y−xy2+y3-x^3 + 11x^2y - xy^2 + y^3
The expression −x3+11x2y−xy2+y3-x^3 + 11x^2y - xy^2 + y^3 is correct. Calculating profit requires subtracting the fully simplified cost from the fully simplified revenue. Simplifying CC yields 2x3−7x2y+2y32x^3 - 7x^2y + 2y^3. Expanding and simplifying RR yields x3+4x2y−xy2+3y3x^3 + 4x^2y - xy^2 + 3y^3. Subtracting these two expressions and correctly distributing the negative sign gives −x3+11x2y−xy2+y3-x^3 + 11x^2y - xy^2 + y^3.

Adım Adım Çözüm

1
Expand and simplify the cost expression C=2x2(x−3y)−y(x2−2y2)C = 2x^2(x - 3y) - y(x^2 - 2y^2).
C=2x3−6x2y−x2y+2y3=2x3−7x2y+2y3C = 2x^3 - 6x^2y - x^2y + 2y^3 = 2x^3 - 7x^2y + 2y^3
Distribute the terms outside the parentheses and combine the like terms −6x2y-6x^2y and −x2y-x^2y.
2
Expand and simplify the revenue expression R=5x3−x(y−2x)2+3y3R = 5x^3 - x(y - 2x)^2 + 3y^3.
R=5x3−x(y2−4xy+4x2)+3y3=5x3−xy2+4x2y−4x3+3y3=x3+4x2y−xy2+3y3R = 5x^3 - x(y^2 - 4xy + 4x^2) + 3y^3 = 5x^3 - xy^2 + 4x^2y - 4x^3 + 3y^3 = x^3 + 4x^2y - xy^2 + 3y^3
Square the binomial (y−2x)2=y2−4xy+4x2(y - 2x)^2 = y^2 - 4xy + 4x^2, distribute the −x-x, and combine the like terms 5x35x^3 and −4x3-4x^3.
3
Subtract the cost expression from the revenue expression to find P=R−CP = R - C.
P=(x3+4x2y−xy2+3y3)−(2x3−7x2y+2y3)=x3+4x2y−xy2+3y3−2x3+7x2y−2y3=−x3+11x2y−xy2+y3P = (x^3 + 4x^2y - xy^2 + 3y^3) - (2x^3 - 7x^2y + 2y^3) = x^3 + 4x^2y - xy^2 + 3y^3 - 2x^3 + 7x^2y - 2y^3 = -x^3 + 11x^2y - xy^2 + y^3
Distribute the subtraction negative sign to all terms in the cost expression and group like terms together to obtain the final simplified expression.

Anahtar Kavram

Simplifying multi-variable algebraic expressions by distributing coefficients and combining like terms.
Soru 59Soru →

Match each algebraic expression on the left with its fully simplified equivalent expression on the right. All variables represent real numbers.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

3a(a2−2ab)−(a3−5a2b)3a(a^2 - 2ab) - (a^3 - 5a^2b)
a2(2a−b)−2a(a2−ab)a^2(2a - b) - 2a(a^2 - ab)
−(a3+3a2b)+3a2(a+b)-(a^3 + 3a^2b) + 3a^2(a + b)

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

The first expression 3a(a2−2ab)−(a3−5a2b)3a(a^2 - 2ab) - (a^3 - 5a^2b) matches 2a3−a2b2a^3 - a^2b. The second expression a2(2a−b)−2a(a2−ab)a^2(2a - b) - 2a(a^2 - ab) matches a2ba^2b. The third expression −(a3+3a2b)+3a2(a+b)-(a^3 + 3a^2b) + 3a^2(a + b) matches 2a32a^3.
Each expression on the left is simplified by distributing the coefficients and combining the like terms. The first expression simplifies to 2a3−a2b2a^3 - a^2b. The second expression simplifies to a2ba^2b. The third expression simplifies to 2a32a^3. These match the corresponding simplified expressions on the right.

Adım Adım Çözüm

1
Simplify the first expression 3a(a2−2ab)−(a3−5a2b)3a(a^2 - 2ab) - (a^3 - 5a^2b)
2a3−a2b2a^3 - a^2b
Multiply 3a3a by both terms in the first parentheses to get 3a3−6a2b3a^3 - 6a^2b. Distribute the negative sign to both terms in the second parentheses to get −a3+5a2b-a^3 + 5a^2b. Combine the a3a^3 terms to get 2a32a^3 and the a2ba^2b terms to get −a2b-a^2b.
2
Simplify the second expression a2(2a−b)−2a(a2−ab)a^2(2a - b) - 2a(a^2 - ab)
a2ba^2b
Multiply a2a^2 by both terms in the first parentheses to get 2a3−a2b2a^3 - a^2b. Multiply −2a-2a by both terms in the second parentheses to get −2a3+2a2b-2a^3 + 2a^2b. Combine the a3a^3 terms to get 00 and the a2ba^2b terms to get a2ba^2b.
3
Simplify the third expression −(a3+3a2b)+3a2(a+b)-(a^3 + 3a^2b) + 3a^2(a + b)
2a32a^3
Distribute the negative sign to get −a3−3a2b-a^3 - 3a^2b. Multiply 3a23a^2 by both terms in the second parentheses to get 3a3+3a2b3a^3 + 3a^2b. Combine the a3a^3 terms to get 2a32a^3 and the a2ba^2b terms to get 00.

Anahtar Kavram

Simplifying Expressions and Combining Like Terms
Tahmini Süre:1m 30s
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Match each algebraic expression on the left with its fully simplified equivalent expression on the right.

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Öğeler

−−(a2−2ab)+a(a−3b)--(a^2 - 2ab) + a(a - 3b)
2(a2b−ab)−ab(2a−3)2(a^2b - ab) - ab(2a - 3)
a2(a−b)−a(a2−ab)a^2(a - b) - a(a^2 - ab)

Eşleşmeler

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Cevap

The expression −−(a2−2ab)+a(a−3b)--(a^2 - 2ab) + a(a - 3b) simplifies to −ab-ab; the expression 2(a2b−ab)−ab(2a−3)2(a^2b - ab) - ab(2a - 3) simplifies to abab; and the expression a2(a−b)−a(a2−ab)a^2(a - b) - a(a^2 - ab) simplifies to 00.
Each expression is correctly simplified by distributing the external factors across parenthetical terms and combining the resulting like terms.

Adım Adım Çözüm

1
Simplify the first expression by distributing coefficients and combining like terms.
−−a2+2ab+a2−3b=−ab--a^2 + 2ab + a^2 - 3b = -ab
Distributing the negative sign yields −−a2+2ab--a^2 + 2ab and distributing aa yields a2−3aba^2 - 3ab. Adding them eliminates the a2a^2 terms, leaving −ab-ab.
2
Simplify the second expression by distributing coefficients and combining like terms.
2a2b−2ab−2a2b+3ab=ab2a^2b - 2ab - 2a^2b + 3ab = ab
Distributing 22 yields 2a2b−2ab2a^2b - 2ab and distributing −ab-ab yields −2a2b+3ab-2a^2b + 3ab. Adding them eliminates the 2a2b2a^2b terms, leaving abab.
3
Simplify the third expression by distributing coefficients and combining like terms.
a3−a2b−a3+a2b=0a^3 - a^2b - a^3 + a^2b = 0
Distributing a2a^2 yields a3−a2ba^3 - a^2b and distributing −a-a yields −a3+a2b-a^3 + a^2b. Adding them eliminates all terms, leaving 00.

Anahtar Kavram

Simplifying algebraic expressions containing multiple variables and higher powers by applying the distributive property and combining like terms.
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Elementary Algebra Alıştırma Soruları — ACT — Sayfa 3 | Examkin