By LearningExpress Editors
Updated on Sep 26, 2011
Answers
The following explanations show one way in which each problem can be solved. You may have another method for solving these problems.
 d. The key word quotient means division so the problem becomes . Divide the coefficients: . When dividing like bases, subtract the exponents: ; simplify: . A variable in the numerator with a negative exponent is equal to the same variable in the denominator with the opposite sign—in this case, a positive sign on the exponent: .
 a. The translation of the question is . The key word product tells you to multiply 6x^{2} and 4xy^{2}. The result is then divided by 3x^{3}y. Use the commutative property in the numerator to arrange like variables and the coefficients together: . Multiply in the numerator. Remember that x^{2} · x = x^{2} · x^{1} = x^{2} + 1 = x^{3}: . Divide the coefficients; 24 ÷ 3 = 8: . Divide the variables by subtracting the exponents: 8x^{3}–3y^{2}–1; simplify. Recall that anything to the zero power is equal to 1: 8x^{0}y^{1} = 8y.
 b. Since the formula for the area of a square is A = s^{2}, then by substituting A = (a^{2}b^{3})^{2}. Multiply the outer exponent by each exponent inside the parentheses: a^{2×2}b^{3×2}. Simplify; a^{4}b^{6}.
 a. The statement in the question would translate to 3x^{2}(2x^{3}y)^{4}. The word quantity reminds you to put that part of the expression in parentheses. Evaluate the exponent by multiplying each number or variable inside the parentheses by the exponent outside the parentheses: 3x^{2}(24x^{3×4}y^{4}); simplify: 3x^{2}(16x^{12}y^{4}). Multiply the coefficients and add the exponents of like variables: 3(16x^{2}+12y^{4}); simplify: 48x^{14}y^{4}.
 d. Since the area of a rectangle is A = length times width, multiply the dimensions to find the area: 2x(4x + 5). Use the distributive property to multiply each term inside the parentheses by 2x: 2x × 4x + 2x × 5. Simplify by multiplying the coefficients of each term and adding the exponents of the like variables: 8x^{2} + 10x.
 b. The translated expression would be –9p3r(5p – 6r). Remember that the key word product means multiply. Use the distributive property to multiply each term inside the parentheses by –9p3r: –9p3r × 5p – (–9p3r) × 6r. Simplify by multiplying the coefficients of each term and adding the exponents of the like variables: –9 × 5p^{3+1}r – (–9 × 6p3r^{1+1}). Simplify: –45p4r – (–54p3r^{2}). Change subtraction to addition and change the sign of the following term to its opposite: –45p4r + (+54p^{3}r^{2}); this simplifies to: –45p4r + 54p^{3}r^{2}.
 c. The two numbers in terms of x would be x + 3 and x + 4 since increased by would tell you to add. Product tells you to multiply these two quantities: (x + 3)(x + 4). Use FOIL (First terms of each binomial multiplied, Outer terms in each multiplied, Inner terms of each multiplied, and Last term of each binomial multiplied) to multiply the binomials: (x · x) + (4 · x) + (3 · x) + (3 · 4); simplify each term: x^{2} + 4x + 3x + 12. Combine like terms: x^{2} + 7x + 12.
 d. Since the area of a rectangle is A = length times width, multiply the two expressions together: (2x – 1)(x + 6). Use FOIL (First terms of each binomial multiplied, Outer terms in each multiplied, Inner terms of each multiplied, and Last term of each binomial multiplied) to multiply the binomials: (2x · x) + (2x · 6) – (1 · x) – (1 · 6). Simplify: 2x^{2} + 12x – x – 6; combine like terms: 2x^{2} + 11x – 6.
 a. Use the formula distance = rate × time. By substitution, distance = (4x^{2} – 2) × (3x – 8). Use FOIL (First terms of each binomial multiplied, Outer terms in each multiplied, Inner terms of each multiplied, and Last term of each binomial multiplied) to multiply the binomials: (4x^{2} · 3x) – (8 · 4x^{2}) – (2 · 3x) – (2 · –8). Simplify each term: 12x^{3} – 32x^{2} – 6x + 16.
 d. Since the formula for the volume of a prism is V = Bh, where B is the area of the base and h is the height of the prism, V = (x – 3)(x^{2} + 4x + 1). Use the distributive property to multiply the first term of the binomial, x, by each term of the trinomial, and then the second term of the binomial, –3, by each term of the trinomial: x(x^{2} + 4x + 1) – 3(x^{2} + 4x + 1). Then distribute: (x · x^{2}) + (x · 4x) + (x · 1) – (3 · x^{2}) – (3 · 4x) – (3 · 1). Simplify by multiplying within each term: x^{3} + 4x^{2} + x – 3x^{2} – 12x – 3. Use the commutative property to arrange like terms next to each other. Remember that 1x = x: x^{3} + 4x^{2} – 3x^{2} + x –12x – 3; combine like terms: x^{3} + x^{2} – 11x – 3.
 b. Since the formula for the volume of a rectangular prism is V = l × w × h, multiply the dimensions together: (x + 1)(x – 2)(x + 4). Use FOIL (First terms of each binomial multiplied, Outer terms in each multiplied, Inner terms of each multiplied, and Last term of each binomial multiplied) to multiply the first two binomials: (x + 1)(x – 2); (x · x) + x(–2) + (1 · x) + 1(–2). Simplify by multiplying within each term: x^{2} – 2x + 1x – 2; combine like terms: x^{2} – x – 2. Multiply the third factor by this result: (x + 4)(x^{2} – x – 2). To do this, use the distributive property to multiply the first term of the binomial, x, by each term of the trinomial, and then the second term of the binomial, 4, by each term of the trinomial: x(x^{2} – x – 2) + 4(x^{2} – x – 2). Distribute: (x · x^{2}) + (x · –x) + (x · –2) + (4 · x^{2}) + (4 · –x) + (4 · –2). Simplify by multiplying in each term: x^{3} – x^{2} – 2x + 4x^{2} – 4x – 8. Use the commutative property to arrange like terms next to each other: x^{3} – x^{2} + 4x^{2} – 2x – 4x – 8; combine like terms: x^{3} + 3x^{2} – 6x – 8.
 c. Since area of a rectangle is found by multiplying length by width, we need to find the factors that multiply out to yield x^{2} – 36. Because x^{2} and 36 are both perfect squares (x^{2} = x · x and 36 = 6 · 6), the product, x^{2} – 36, is called a difference of two perfect squares, and its factors are the sum and difference of the square roots of its terms. Therefore, because the square root of x^{2} = x and the square root of 36 = 6, x^{2} – 36 = (x + 6)(x – 6).
 b. To find the base and the height of the parallelogram, find the factors of this binomial. First look for factors that both terms have in common; 2x^{2} and 10x both have a factor of 2 and x. Factor out the greatest common factor, 2x, from each term. 2x^{2} – 10x; 2x(x – 5). To check an answer like this, multiply through using the distributive property. 2x(x –5); (2x · x) – (2x · 5); simplify and look for a result that is the same as the original question. This question checked: 2x^{2} – 10x.
 d. Since the formula for the area of a rectangle is A = length times width, find the two factors of x^{2} + 2x + 1 to get the dimensions. First check to see if there is a common factor in each of the terms or if it is the difference between two perfect squares, and it is neither of these. The next step would be to factor the trinomial into two binomials. To do this, you will be doing a method that resembles FOIL backwards (First terms of each binomial multiplied, Outer terms in each multiplied, Inner terms of each multiplied, and Last term of each binomial multiplied.) First results in x^{2}, so the first terms must be: (x)(x); Outer added to the Inner combines to 2x, and the Last is 1, so you need to find two numbers that add to +2 and multiply to +1. These two numbers would have to be +1 and +1: (x + 1)(x + 1). Since the factors of the trinomial are the same, this is an example of a perfect square trinomial, meaning that the farmer's rectangular field was, more specifically, a square field. To check to make sure these are the factors, multiply them by using FOIL (First terms of each binomial multiplied, Outer terms in each multiplied, Inner terms of each multiplied, and Last term of each binomial multiplied; (x · x) + (1 · x) + (1 · x) + (1 · 1); multiply in each term: x^{2} + 1x + 1x + 1; combine like terms: x^{2} + 2x + 1.
 a. Since area of a rectangle is length × width, look for the factors of the trinomial to find the two dimensions. First check to see if there is a common factor in each of the terms or if it is the difference between two perfect squares, and it is neither of these. The next step would be to factor the trinomial into two binomials. To do this, you will be doing a method that resembles FOIL backwards. (First terms of each binomial multiplied, Outer terms in each multiplied, Inner terms of each multiplied, and Last term of each binomial multiplied.) First results in x^{2}, so the first terms must be (x)(x); Outer added to the Inner combines to 6x, and the Last is 5, so you need to find two numbers that add to produce +6 and multiply to produce +5. These two numbers are +1 and +5; (x + 1)(x + 5).
 a. Since the formula for the area of a rectangle is length × width, find the factors of the trinomial to get the dimensions. First check to see if there is a common factor in each of the terms or if it is the difference between two perfect squares, and it is neither of these. The next step would be to factor the trinomial into two binomials. To do this, you will be doing a method that resembles FOIL backwards (First terms of each binomial multiplied, Outer terms in each multiplied, Inner terms of each multiplied, and Last term of each binomial multiplied.) First results in x^{2}, so the first terms must be (x)(x); Outer added to the Inner combines to 1x, and the Last is –12, so you need to find two numbers that add to +1 and multiply to –12. These two numbers are –3 and +4; (x – 3)(x + 4). Thus, the dimensions are (x + 4) and (x – 3).
 b. Since the trinomial does not have a coefficient of one on its highest exponent term, the easiest way to find the answer to this problem is to use the distributive property. First, using the original trinomial, identify the sum and product by looking at the terms when the trinomial is in descending order (highest exponent first): 3x^{2} – 7x + 2. The sum is the middle term, in this case, –7x. The product is the product of the first and last terms, in this case, (3x^{2})(2) – 6x^{2}. Now, identify two quantities whose sum is –7x and product is 6x^{2}, namely –6x and –x. Rewrite the original trinomial using these two terms to replace the middle term in any order: 3x^{2} –6x – x + 2. Now factor by grouping by taking a common factor out of each pair of terms. The common factor of 3x^{2} and –6x is 3x and the common factor of –x and 2 is –1. Therefore, 3x^{2} – 6x – x + 2 becomes 3x(x – 2) – 1(x – 2). (Notice that if this expression were multiplied back out and simplified, it would correctly yield the original polynomial.) Now, this twoterm expression has a common factor of (x – 2), which can be factored out of each term using the distributive property: 3x(x – 2) – 1(x – 2) becomes (x – 2)(3x –1). The dimensions of the courtyard are (x – 2) and (3x – 1).
 d. In order to convert this number to standard notation, multiply 9.3 by the factor of 10^{7}. Since 10^{7} is equal to 10,000,000, 9.3 × 10,000,000 is equal to 93,000,000. As an equivalent solution, move the decimal point in 9.3 seven places to the right since the exponent on the 10 is positive 7.
 c. To convert to scientific notation, place a decimal point after the first nonzero digit to create a number between 1 and 10—in this case, between the 2 and the 4. Count the number of decimal places from that decimal to the place of the decimal in the original number. In this case, the number of places would be 5. This number, 5, becomes the exponent of 10 and is positive because the original number was greater than one. The answer then is 2.4 × 10^{5}.
 d. In order to convert this number to standard notation, multiply 5.3 by the factor of 10^{–6}. Since 10^{–6} is equal to 0.000001, 5.3 × 0.000001 is equal to 0.0000053. Equivalently, move the decimal point in 5.3 six places to the left since the exponent on the 10 is negative 6.
More practice problems on algebra word problems can be found at:
 Algebra Word Problems Practice Questions Set 1
 Algebra Word Problems Practice Questions Set 2
 Algebra Word Problems Practice Questions Set 3
 Algebra Word Problems Practice Questions Set 4
 Algebra Word Problems Practice Questions Set 5 (You are here)
 Algebra Word Problems Practice Questions Set 6
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