Chapter 1

  1. One possibility: $ 2(3 ) (3) (4) $ . (Left hand side = 24, Right hand side = 48.)

\[ \mathbf{7}.\ a)\ (1+2)^{2}\neq1^{2}+2^{2}\ v(LHS=9,RHS=5)\qquad b)\ \sqrt{9+16}\neq\sqrt{9}+\sqrt{16}.\ (LHS=5,RHS=7.) \]

  1. $ 3x^{2} + 4x^{2} = (3 + 4)x^{2} = 7x^{2} $ . Observe the “undistribution” in the first step! The more respectable name for undistribution is factoring, because its effect is to turn an algebraic expression into a product of two factors. You’ll learn more about it in the next section.

    1. Jacob is giving Esau the correct answer.

  1. Remember what $ x^{2} $ and $ x^{3} $ actually mean. These are shorthand expressions for xx and xxx, respectively. Thus, $ x{2}x{3} $ itself is shorthand for $ (xx)(xxx) $ , five copies of x multiplied together. But that, of course, is $ x^{5} $ .

  2. $ x^{6} $ , $ x^{5} $ , $ x^{10} $ , $ 6x^{6} $ , $ -30x^{5} $ . In the penultimate example, $ (2x{2})(3x{4}) $ equals $ 6x^{6} $ because $ (2x{2})(3x{4}) $ means $ (2xx)(3xxxx) $ ; those factors can be multiplied in any order, so we have 2 times 3 times six factors of x, which is, of course, $ 6x^{6} $ . A similar justification holds for the last example. $ ^{*} $

\[ 6x^{2}-2x-28\quad b)2x^{2}-x-6\quad c)2x^{6}-6x^{5}+x^{4}-6x^{2}+18x-3\quad d)-x^{4}+4x^{3}-14x^{2}+20x-21 \]

\[ \mathsf{e})-x^{6}-x^{5}+2x^{4}-x^{2}+x\qquad\mathsf{f})x^{3}+6x^{2}+11x+6\qquad\mathsf{g})x^{4}-1\qquad\mathsf{h})x^{6}-1\qquad\mathsf{i})x^{100}-1. \]

\[ \begin{aligned}16.\ a)\ (x+4)(x+2)\quad&b)\ (x-10)(x+10)\quad c)\ 5x(2x+1)\quad d)\ (x-5)(x-2)\quad e)\ 3(x-2)(x+1)\\ f)\ (x-5)(x+5)\quad&g)\ -4(x+4)^{2}\quad h)\ -15(x+3)(x-1)\quad i)\ (x-2)(x+2)(x^{2}+4)\quad j)\ (3x-2)(3x+2)\\ k)\ (3x^{2}-1)\ (3x^{2}+1)\ (9x^{4}+1)\quad&l)\ (2x+1)(x+2)\quad m)\ (3x+1)(x-3)\quad n)\ (2x+1)(2x-3)\end{aligned} \]

\[ \begin{aligned}17.\ a)x^{2}+25+10x\quad&b)x^{2}+25-10x\quad&c)x^{2}+121+22x\quad&d)x^{2}+144-24x\quad&e)4x^{2}+1+4x\\ f)9x^{2}+4-12x\quad&g)a^{2}+2+2\sqrt{2}a\quad&h)a^{2}+2-2\sqrt{2}a\quad&i)4x^{2}+\frac{1}{4}+2x\quad&j)4a^{2}+9b^{2}+12ab\\ k)x+1-2\sqrt{x}\end{aligned} \]

    1. $ (-8) (-4) $ asks the question, “-4 times what is -8?” The answer, of course, is 2.

  2. $ 10 (1/3) $ asks the question, “ $ (1/3) $ times what is 10?” The answer is clearly 30.

  3. No. For example, $ 6 $ asks, “0 times what is 6?” This question has no valid answer; 0 times anything is 0. For this reason, we say that division (of any nonzero number) by zero is undefined.

  4. Yes. For example, $ 0 $ asks, “6 times what is 0?” The answer is clearly 0. By the same reasoning, zero divided by any nonzero number is zero.

  5. This won’t work, for a peculiar reason. The question that 0/0 asks (“0 times what is 0?”) admits contradictory answers: 0 times 0 is 0, so it would seem that 0/0 = 0; then again, 0 times 1 is 0, so it would seem that 0/0 = 1. Indeed, 0/0 would seem to be equal to every number. But if that were true, then all numbers would be equal, which is absurd. Thus, to preserve the coherence of mathematics, we must leave 0/0 undefined.

  6. The first intuitive fraction rule implies that $ (2/3)(4/5)=2(1/3)4(1/5) $ . Since we can multiply numbers in any order without changing the result, we’ll reorder these as follows: $ (2)(4)(1/3)(1/5) $ . After multiplying the two whole numbers, this becomes $ 8(1/3)(1/5) $ . By the second intuitive fraction rule, this is $ 8(1/15) $ . Finally, we apply the first intuitive fraction one last time, whereby we find that our product equals 8/15, as claimed.

  7. Write a as a/1, then multiply.

    1. Yes, because $ 4a{2}b{2}-18a{5}b{3}=2a{2}b{2}(2-9a^{3}b) $ .

  1. k is a term of an algebraic expression if the expression can be written in the form $ k + () $ .

  2. No. Counterexample: $ $ . (LHS = 2, RHS = 4)

  3. $ $ , $ $ , $ $ , $ c+d $ .

  1. a, d, g, h are false. 32. 1/9900 33. False. 34. b) $ y^{2}/x $ , $ 42/(2x+y) $ , ab, $ 1/(bc) $ .

  2. Multiply top and bottom by -1. [Downstairs, use the useful fact that $ -(a-b)=b-a $ ]

    1. No real simplification is possible. (You can factor the top, but that doesn’t really simplify it.)

\[ b)\frac{1}{x+4}\quad c)\frac{x}{yz}\quad d)\frac{a+b}{cd}\quad e)\frac{a+b}{c^{2}d}\quad f)\frac{1}{acd}\quad g)6\quad h)\frac{1}{3a-2b}\quad i)-1\quad j)\frac{a^{8}}{d^{6}}\quad k)2(x-3) \]

37. a) 3b b) $ 4a - b + c $ c) 6a - 6b d) b - 3a e) -y - 2z f) 0

  1. $ -3x^{2} + 4x + 2 $ 40. a) 1807/360 b) -52/105 c) 215/152

  2. The result would be the same. (There would just be some extra cancellation to take care of at the end.)

    1. a/5 b) $ $ c) 2/15 d) $ $ e) $ $ f) $ $ g) $ $
  1. $ $ i) $ $ j) $ $ k) $ $ l) $ $ [Were you close? Recall exercise 35.]

\[ \mathsf{m})\frac{-h}{x(x+h)}\qquad\mathsf{n})\frac{x^{3}-5x^{2}+2x-2}{x^{2}(x-1)^{2}} \]

    1. $ $ b) $ $ c) $ b^{2} - c^{2} $ d) $ $
    1. False (LHS = -9) b) False (If x < 0, then -x > 0.) c) False (LHS = 9) d) True

  1. True (Cancelling a common factor from top and bottom)

  2. False $ ((2x + 1). $ is not a factor of the numerator. g) True

\[ 45.a)\frac{2}{x-2}\quad b)\frac{-1}{x(x+h)}\quad c)h+2x\quad d)\frac{a^{2}+1}{a^{2}}\quad e)\frac{2}{x}\quad f)\frac{x}{x-1}\quad g)\frac{x-1}{x}\quad h)-\frac{1}{x}\quad i)\frac{12}{a^{2}-b^{2}}\quad j)b-a \]

Chapter 2

\[ \mathbf{1.}\;x^{87}\quad\mathbf{2.}\;x^{6}\quad\mathbf{3.}\;x^{20}\quad\mathbf{4.}\;x^{30}\quad\mathbf{5.}\;x^{7}\quad\mathbf{6.}\;16x^{24}\quad\mathbf{7.}\;6000x^{18}\quad\mathbf{8.}\;p^{2}q^{9}r^{13}\quad\mathbf{9.}\;x^{62}\quad\mathbf{10.}\;1\quad\mathbf{11.}\;a^{2}b^{3}\quad\mathbf{12.}\;1 \]

\[ \mathbf{13}.\ a)\ 3/x^{3}\quad b)\ 1/27x^{3}\quad c)\ a^{3}+b^{3}\quad d)\ 8\quad e)\frac{x^{5}}{y^{11}}\quad f)\ 3\quad g)\ 12x^{n+2}y^{2n}\quad h)\ 4a^{11}b^{15}\quad i)\ 2+\frac{x}{y}+\frac{y}{x}\quad j)\ 2/a \]

14.5/2

    1. is false; LHS = $ (ab^{5} + 1)/(2b^{5}c) $ . (c) is false; LHS = $ (a^{2} + ab)/(1 + ad) $ .

  2. If $ x^{-n} $ is a factor of the denominator, then $ = = = a() = $ .

  3. No. This would be equal to 1/0, and division by zero is undefined.

  4. b, d, h, j, o are false. 21. b, d, e are false.

  5. We can take “5 apples, 4 times” (i.e. 5 apples, then 5 more, then 5 more, then 5 more), or “4 apples 5 times” (i.e. 4 apples, then 4 more, then 4 more, then 4 more, then 4 more), but “5 apples times 4 apples” is nonsense, so trying to base another argument on it is nonsense squared, so to speak.

    1. $ 10 $ b) $ 77 - 3 $ c) 120 d) 1/3

  7. Not if a < 0. In fact, the general rule is that $ = |a| $ , where $ |a| $ denotes a’s absolute value.

    1. 6 b) −1 c) 100 d) 7 e) 2 f) 6 g) 711 h) cannot be simplified any further i) −2/5 j) 1/2 k) −1 l) 0 m) 4 n) 16 o) 5