Powers of sums and products

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Powers and roots can be distributed over products and quotients.  They may not be distributed over sums or differences, no matter how tempting it may be. Sums or differences raised to a power must be used as a factor the indicated number of times, then multiplied using the distributive property.

An exponent applies only to the factor immediately below it unless parentheses have been used to indicate otherwise.

“Like terms” have the same variables, to the same powers… and only “like” terms may be combined by adding their coefficients.

The answer shown below is wrong.  Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead?

(f-g^2)^2-fg^2-(-fg)^2-f^2

=f^2-g^4-fg^2-(-fg)^2-f^2

=f^2-g^4-f^2g^2-(-fg)^2-f^2

=f^2-g^4-f^2g^2+fg^2-f^2

=-g^4

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Hint 1: The correct answer to the problem on the first line above is: g^4-3fg^2-f^2g^2

Hint 2: The work shown above contains sign errors, simplification errors, and exponentiation errors. Re-read the text at the top of the posting if you have not found them all…

The Distributive Property

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The “distributive property” of multiplication and division.  When do you distribute, and when don’t you?  That is the question.

The answer shown below is wrong.  Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead?

\dfrac{(4)(3)(2k+6)}{6}

=\dfrac{(12)(8k+24)}{6}

=\dfrac{(12)(8k+24)}{(2)(3)}

=\dfrac{(6)(4k+12)}{3}

=(2)(4k+4)

=8k+4

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Hint 1: The correct answer to the original problem is: 4k+12

Hint 2: The complete description of the distributive property of multiplication is: “the distributive property of multiplication over…”?

Hint 3: Several mistakes have been made in the work above – don’t just seek the correct answer – look for mistakes between every pair of lines.

Negative Exponents

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Negative exponents are another source of occasional confusion. No matter where you find a negative exponent, you can turn it into a positive exponent by taking the reciprocal of the expression it applies to.

The answer shown below is wrong.  Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead?

\dfrac{c^{-2}d^3}{c^3d^{-4}}

=\dfrac{d^3d^4}{c^{-2}c^3}

=\dfrac{d^7}{c^{-6}}

=d

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Hint 1: The correct answer to the original problem is: \dfrac{d^7}{c^5}

Hint 2: Several mistakes have been made in the work above – don’t just seek the correct answer – look for the mistakes between every pair of lines.

Exponents

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The “laws of exponents” are a frequent source of errors.  The rules that apply when simplifying expressions that involve exponents can be figured out quickly on your own if you (in your mind’s eye) expand integral exponents into repeated multiplication. I encourage students to master being able to explain why each of these rules is as it is instead of memorizing them, as memorized versions are more likely to get jumbled together in your thinking when working problems.

The answer shown below is wrong.  Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead?

\dfrac{(ab^2)^3(5a^2)}{a^2b}

=\dfrac{a^5b^525a^2}{a^2b}

=\dfrac{25a^{10}b^5}{a^2b}

=25a^5b^5.

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Hint 1: The correct answer to the original problem is: 5a^3 b^5

Hint 2: Several mistakes have been made in the work above – don’t just seek the correct answer – look for the mistakes between every pair of lines.

Negative Signs

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Negative signs bother many students, particularly when they are followed by fractions. In such situations, it is important to remember that the vinculum (the horizontal line between numerator and denominator) serves as a grouping symbol – like parentheses would. I recommend that students always put a numerator with more than one term in parentheses before bringing a leading negative sign into the numerator.

The answer shown below is wrong.  Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead?

\dfrac{2-w}{3}-\dfrac{2w-5}{2}

=\dfrac{2-w}{3}\cdot\dfrac{2}{2}-\dfrac{2w-5}{2}

=\dfrac{2(2-w)}{6}+\dfrac{-2w-5}{6}

=\dfrac{4-w-2w-5}{6}

=\dfrac{-1-3w}{6}

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Hint 1: the correct answer to the original problem is: \dfrac{-8w+19}{6}

Hint 2: several mistakes have been made in the work above – they are not all sign or distribution errors

Negative Signs

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The distributive property is a source of many mistakes in algebra. When a negative sign, multiplication, or division is next to a set of parentheses, everything inside the parentheses will be affected.

Negative signs in particular bother many students. The most reliable approach is probably to rewrite subtraction as the addition of a negative, then distribute the negative sign to all terms inside the parentheses.

The answer shown below is wrong.  Try working your way through the problem backwards, from the answer up. As you find each mistake, try to identify the thinking behind it… what perspective led to the mistake? What should have been done instead?

(1-d+e)-(2d-3e+4)-(-4e+3d-2)

=(1-d+e)-(-2d+3e-4)+(4e-3d-2)$

=1-d+e+2d-3e+4+4e-3d-2

=-d+2d-3d+e-3e+4e+1+4-2

=-6d+2e+3

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Hint 1: the correct answer to the original problem is: 8e-6d-1

Hint 2: a number of mistakes have been made in the work above – they are not all sign errors. Identify all of the errors made at each step and the likely reasons they were made before solving the problem correctly yourself.