Review Sheet for Test 2

Math 101

7-27-2012

These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the absence of a topic does not imply that it won't appear on the test.

1. Simplify the expression, using only positive exponents in your answer. Assume that all variables represent positive quantities.

(a) $(5 a^2)^{-3} \cdot 2 (a^4)^2$ .

(b) $\dfrac{(2 b^3)^4}{8 (b^2)^3}$ .

(c) $(4 x^{-3} y^5)^3 \cdot 2 (x^{-1} y^6)^2$ .

(d) $\dfrac{(2x^{-2} y^5)^{-3}}{(x^8 y^{-11})^2}$ .

(e) $\left(\dfrac{(2x^3 y^{-7})^3}{56 (x^4 y^{-12})^2}\right)^{-2}$ .

2. Multiply the polynomials:

(a) $(2x - 3)(x + 4)$ .

(b) $(3x^2 - x - 1)(x - 2)$ .

(c) $(2x - 5)(2x + 5)$ .

(d) $(3w + 7)^2$ .

(e) $(1 - 6t)^2$ .

(f) $(2x + y)(x - 3y)$ .

(g) $(x^2 - y^4)^2$ .

(h) $(x + y + 1)(x + 2y)$ .

3. Factor completely:

(a) $x^2 - 64$ .

(b) $x^2 + 8x + 16$ .

(c) $4x^2 - 25$ .

(d) $x^2 - 5x + 6$ .

(e) $x^2 - 9x - 10$ .

(f) $4x^2 + 8x - 5$ .

(g) $a^4 - 16b^4$ .

(h) $x^3 - 8y^3$ .

(i) $x^3 + 5x^2 - 16x - 80$ .

(j) $x^2y - 3xy + xy^2 - 3y^2$ .

(k) $\dfrac{81}{a^2b^2} - 1$ .

(l) $\dfrac{4x^2}{y^2} + \dfrac{4x}{y} + 1$ .

(m) $a^2 - 3a - ab + 3b$ .

(n) $x^3 + x^2 - xy^2 - y^2$ .

(o) $x^2 + xy - 5x - 5y$ .

(p) $2x^3 + 7x^2 + 3x$ .

(q) $x^3 + 27y^3$ .

(r) $\dfrac{1}{8}a^3 - 125b^3$ .

(s) $64 - y^3$ .

(t) $ab + 6b + a^2 + 6a$ .

(u) $x^3 - 3x^2 - 4x + 12$ .

(v) $x^2y - 2xy^2 + x - 2y$ .

(w) $x^2 - 4$ .

(x) $4x^2 - 9y^2$ .

(y) $5x^3 - 125x$ .

(z) $\dfrac{x^2}{4} - 81$ .

*(aa) $x^4 - 1$ .

(bb) $x^3 + 8$ .

(cc) $m^3 - 27$ .

(dd) $8p^3 - q^3$ .

(ee) $5x^4 - 5x$ .

*(ff) $\dfrac{x^3}{64} + \dfrac{1}{27}$ .

(gg) $2x^3 + x^2 + 6x + 3$ .

(hh) $x^3 + 3x^2 - 4x - 12$ .

(ii) $a^2 - 3a - 2ab + 6b$ .

*(jj) $2x^4 - 2x^3 + 4x^2 - 4x$ .

4. Solve the equation by factoring:

(a) $x^2 - 3x - 4 = 0$ .

(b) $2x^3 - 10x^2 - 28x = 0$ .

(c) $7x^3 + 7x = 0$ .

(d) $x^2 + 9 = 10x$ .

*(e) $2x^3 + x^2 - 2x - 1 = 0$ .

5. (a) Divide $x^4 + x^3 + 1$ by $x^2 - 1$ using long division.

(b) Find the quotient and remainder when $2 x^3 + 3 x - 5$ is divided by $x + 3$ .

(c) Compute the quotient and remainder when $2x^4 + 3x^2 + 4x$ is divided by $2x^2$ .

(d) Compute the quotient and remainder when $2x^2 + 5xy - 3y^2$ is divided by $x + 3y$ .

(e) Factor $x^3 - 2x^2 - 11x + 12$ completely, given that $x - 4$ is one of the factors.

6. (a) Show that it is not valid to cancel x's in $\dfrac{x + 4}{x}$ to get $\dfrac{1 + 4}{1} = 5$ by giving a specific value of x for which $\dfrac{x + 4}{x}$ is not equal to 5.

(b) Show that $\dfrac{a}{\dfrac{b}{c}}$ is not always the same as $\dfrac{\dfrac{a}{b}}{c}$ by giving specific values of a, b, and c for which $\dfrac{a}{\dfrac{b}{c}}$ is not equal to $\dfrac{\dfrac{a}{b}}{c}$ .

7. Simplify, cancelling any common factors:

(a) $\dfrac{5x^2}{y^3} \cdot \dfrac{y^{11}}{10x^5}$ .

(b) $\dfrac{x^2 - x - 2}{x^2 + 3x + 2} \cdot \dfrac{x^2 - 5x + 6}{x^2 - 4x + 4}$ .

(c) $\dfrac{x^3 - 4x}{5x^2} \cdot \dfrac{x^2 - 5x + 6}{x^2 - 4x + 4}\cdot \dfrac{10x^3}{6x + 12}$ .

8. Simplify, cancelling any common factors:

(a) $\dfrac{\dfrac{x^2 + 5x}{2x^2 - 6x}}{\dfrac{x^2 - 25}{x^2 - 5x + 6}}$ .

(b) $\dfrac{x^2 - x - 2}{5x^4 - 15x^3} \div \dfrac{x^2 + 5x + 4}{x^3 + 4x^2}$ .

(c) $\dfrac{\dfrac{x^2 - 2xy}{x^3 + x^2y}}{\dfrac{x^2 - 4y^2}{x^2 - 4xy - 5y^2}}$ .

(d) $\dfrac{\dfrac{x^3 - 5x^2}{x^2 - 2x - 15}}{\dfrac{x^2 + 2x}{x^2 - 4}}$ .

9. Combine the fractions into a single fraction and simplify:

(a) $\dfrac{2}{x - 2} + \dfrac{3x}{x^2 - 2x} - \dfrac{2}{x}$ .

(b) $\dfrac{1}{x^2 - 5x + 6} - \dfrac{6}{x^2 - 9}$ .

(c) $2 - \dfrac{1}{x} + \dfrac{1}{x + 1}$ .

(d) $\dfrac{x}{x + y} + \dfrac{y}{x - y}$ .

(e) $\dfrac{x - 2y}{x^2 - xy} + \dfrac{2y}{x^2 - y^2}$ .

(f) $x - \dfrac{1}{x} + \dfrac{x + 1}{x - 1}$ .

Solutions to the Review Sheet for Test 2

1. Simplify the expression, using only positive exponents in your answer. Assume that all variables represent positive quantities.

(a) $(5 a^2)^{-3} \cdot 2 (a^4)^2$ .

$$(5 a^2)^{-3} \cdot 2 (a^4)^2 = 5^{-3} (a^2)^{-3} \cdot 2 (a^4)^2 = \dfrac{1}{125} a^{-6} \cdot 2 a^8 = \dfrac{2 a^2}{125}.\quad\halmos$$

(b) $\dfrac{(2 b^3)^4}{8 (b^2)^3}$ .

$$\dfrac{(2 b^3)^4}{8 (b^2)^3} = \dfrac{2^4 (b^3)^4}{8 (b^2)^3} = \dfrac{16 b^{12}}{8 b^6} = 2 b^6.\quad\halmos$$

(c) $(4 x^{-3} y^5)^3 \cdot 2 (x^{-1} y^6)^2$ .

$$(4 x^{-3} y^5)^3 \cdot 2 (x^{-1} y^6)^2 = 4^3 (x^{-3})^3 (y^5)^3 \cdot 2 (x^{-1})^2 (y^6)^2 = 64 x^{-9} y^{15} \cdot 2 x^{-2} y^{12} = 128 x^{-11} y^{27} = \dfrac{128 y^{27}}{x^{11}}.\quad\halmos$$

(d) $\dfrac{(2x^{-2} y^5)^{-3}}{(x^8 y^{-11})^2}$ .

$$\dfrac{(2x^{-2} y^5)^{-3}}{(x^8 y^{-11})^2} = \dfrac{2^{-3}(x^{-2})^{-3}(y^5)^{-3}}{(x^8)^2(y^{-11})^2} = \dfrac{\left(\dfrac{1}{2^3}\right) x^6 y^{-15}} {x^{16} y^{-22}} = \dfrac{1}{8} x^{-10} y^7 = \dfrac{y^7}{8 x^{10}}.\quad\halmos$$

(e) $\left(\dfrac{(2x^3 y^{-7})^3}{56 (x^4 y^{-12})^2}\right)^{-2}$ .

$$\left(\dfrac{(2x^3 y^{-7})^3}{56 (x^4 y^{-12})^2}\right)^{-2} = \left(\dfrac{2^3 (x^3)^3 (y^{-7})^3} {56 (x^4)^2 (y^{-12})^2}\right)^{-2} = \left(\dfrac{8 x^9 y^{-21}}{56 x^8 y^{-24}}\right)^{-2} = \left(\dfrac{1}{7} x y^3\right)^{-2} = \left(\dfrac{1}{7}\right)^{-2} x^{-2} (y^3)^{-2} =$$

$$\left(\dfrac{1^{-2}}{7^{-2}}\right) x^{-2} y^{-6} = \dfrac{1}{\left(\dfrac{1}{7^2}\right)} x^{-2} y^{-6} = \dfrac{1}{\left(\dfrac{1}{49}\right)} x^{-2} y^{-6} = 49 x^{-2} y^{-6} = \dfrac{49}{x^2 y^6}.\quad\halmos$$

2. Multiply the polynomials:

(a) $(2x - 3)(x + 4)$ .

$$\matrix{& & 2x & - & 3 \cr & & x & + & 4 \cr \noalign{\vskip1pt}\noalign{\hrule}\noalign{\vskip1pt} & & 8x & - & 12 \cr 2x^2 & - & 3x & & \cr \noalign{\vskip1pt}\noalign{\hrule}\noalign{\vskip1pt} 2x^2 & + & 5x & - & 12 \cr}\quad\halmos$$

(b) $(3x^2 - x - 1)(x - 2)$ .

$$\matrix{& & 3x^2 & - & x & - & 1 \cr & & & & x & - & 2 \cr \noalign{\vskip1pt}\noalign{\hrule}\noalign{\vskip1pt} & & -6x^2 & + & 2x & + & 2 \cr 3x^3 & - & x^2 & - & x & & \cr \noalign{\vskip1pt}\noalign{\hrule}\noalign{\vskip1pt} 3x^3 & - & 7x^2 & + & x & + & 2 \cr}\quad\halmos$$

(c) $(2x - 5)(2x + 5)$ .

Using the rule $(a - b)(a + b) = a^2 - b^2$ ,

$$(2x - 5)(2x + 5) = 4x^2 - 25.\quad\halmos$$

(d) $(3w + 7)^2$ .

Using the rule $(a + b)^2 = a^2 - 2ab + b^2$ ,

$$(3w + 7)^2 = 9w^2 + 42w + 49.\quad\halmos$$

(e) $(1 - 6t)^2$ .

Using the rule $(a - b)^2 = a^2 - 2ab + b^2$ ,

$$(1 - 6t)^2 = 1 - 12t + 36t^2.\quad\halmos$$

(f) $(2x + y)(x - 3y)$ .

Using FOIL, I get

$$(2x + y)(x - 3y) = 2x^2 - 6xy + xy - 3y^2 = 2x^2 - 5xy - 3y^2. \quad\halmos$$

(g) $(x^2 - y^4)^2$ .

Using the form $(a - b)^2 = a^2 - 2ab + b^2$ , I get

$$(x^2 - y^4)^2 = x^4 - 2x^2y^4 + y^8.\quad\halmos$$

(h) $(x + y + 1)(x + 2y)$ .

$$(x + y + 1)(x + 2y) = (x + y + 1)(x) + (x + y + 1)(2y) = x^2 + xy + x + 2xy + 2y^2 + 2y = x^2 + x + 3xy + 2y^2 + 2y.\quad\halmos$$

3. Factor completely:

(a) $x^2 - 64 = (x - 8)(x + 8)$ .

(b) $x^2 + 8x + 16 = (x + 4)^2$ .

(c) $4x^2 - 25 = (2x - 5)(2x + 5)$ .

(d) $x^2 - 5x + 6 = (x - 2)(x - 3)$ .

(e) $x^2 - 9x - 10 = (x - 10)(x + 1)$ .

(f) $4x^2 + 8x - 5 = (2x - 1)(2x + 5)$ .

(g) $a^4 - 16b^4 = (a^2 - 4b^2)(a^2 + 4b^2) = (a - 2b)(a + 2b)(a^2 + 4b^2)$ .

(h) $x^3 - 8y^3 = (x - 2y)(x^2 + 2xy + 4y^2)$ .

(i) $x^3 + 5x^2 - 16x - 80$ .

I'll use factoring by grouping:

$$x^3 + 5x^2 - 16x - 80 = (x^3 + 5x^2) - (16x + 80) = x^2(x + 5) - 16(x + 5) = (x^2 - 16)(x + 5) = (x - 4)(x + 4)(x + 5).\quad\halmos$$

(j) $x^2y - 3xy + xy^2 - 3y^2$ .

I'll use factoring by grouping:

$$x^2y - 3xy + xy^2 - 3y^2 = (x^2y - 3xy) + (xy^2 - 3y^2) = xy(x - 3) + y^2(x - 3) = (xy + y^2)(x - 3) = (x + y)(y)(x - 3). \quad\halmos$$

(k) $\dfrac{81}{a^2b^2} - 1$ .

$$\dfrac{81}{a^2b^2} - 1 = \left(\dfrac{9}{ab}\right)^2 - 1^2 = \left(\dfrac{9}{ab} - 1\right)\left(\dfrac{9}{ab} + 1\right). \quad\halmos$$

(l) $\dfrac{4x^2}{y^2} + \dfrac{4x}{y} + 1$ .

$$\dfrac{4x^2}{y^2} + \dfrac{4x}{y} + 1 = \left(\dfrac{2x}{y}\right)^2 + 2\cdot \dfrac{2x}{y} + 1^2 = \left(\dfrac{2x}{y} + 1\right)^2.\quad\halmos$$

(m) $a^2 - 3a - ab + 3b$ .

$$a^2 - 3a - ab + 3b = (a^2 - 3a) - (ab - 3b) = a(a - 3) - b(a - 3) = (a - b)(a - 3).\quad\halmos$$

(n) $x^3 + x^2 - xy^2 - y^2$ .

$$x^3 + x^2 - xy^2 - y^2 = (x^3 + x^2) - (xy^2 + y^2) = x^2(x + 1) - y^2(x + 1) = (x + 1)(x^2 - y^2) = (x + 1)(x - y)(x + y).\quad\halmos$$

(o) $x^2 + xy - 5x - 5y$ .

$$x^2 + xy - 5x - 5y = x(x + y) - 5(x + y) = (x - 5)(x + y).\quad\halmos$$

(p) $2x^3 + 7x^2 + 3x$ .

$$2x^3 + 7x^2 + 3x = x(2x^2 + 7x + 3) = x(2x + 1)(x + 3).\quad\halmos$$

(q) $x^3 + 27y^3$ .

$$x^3 + 27y^3 = (x + 3y)(x^2 - 3xy + 9y^2).\quad\halmos$$

(r) $\dfrac{1}{8}a^3 - 125b^3$ .

$$\dfrac{1}{8}a^3 - 125b^3 = \left(\dfrac{1}{2}a - 5b\right) \left(\dfrac{1}{4}a^2 + \dfrac{5}{2}ab + 25b^2\right).\quad\halmos$$

(s) $64 - y^3$ .

$$64 - y^3 = (4 - y)(16 + 4y + y^2).\quad\halmos$$

(t) $ab + 6b + a^2 + 6a$ .

$$ab + 6b + a^2 + 6a = (ab + 6b) + (a^2 + 6a) = b(a + 6) + a(a + 6) = (b + a)(a + 6).\quad\halmos$$

(u) $x^3 - 3x^2 - 4x + 12$ .

$$x^3 - 3x^2 - 4x + 12 = (x^3 - 3x^2) - (4x - 12) = x^2(x - 3) - 4(x - 3) = (x - 3)(x^2 - 4) = (x - 3)(x - 2)(x + 2).\quad\halmos$$

(v) $x^2y - 2xy^2 + x - 2y$ .

$$x^2y - 2xy^2 + x - 2y = (x^2y - 2xy^2) + (x - 2y) = xy(x - 2y) + (x - 2y) = (x - 2y)(xy + 1).\quad\halmos$$

(w) $x^2 - 4$ .

$$x^2 - 4 = x^2 - 2^2 = (x - 2)(x + 2).\quad\halmos$$

(x) $4x^2 - 9y^2$ .

$$4x^2 - 9y^2 = (2x)^2 - (3y)^2 = (2x - 3y)(2x + 3y).\quad\halmos$$

(y) $5x^3 - 125x$ .

$$5x^3 - 125x = 5x(x^2 - 25) = 5x(x^2 - 5^2) = 5x(x - 5)(x + 5).\quad\halmos$$

(z) $\dfrac{x^2}{4} - 81$ .

$$\dfrac{x^2}{4} - 81 = \left(\dfrac{x}{2}\right)^2 - 9^2 = \left(\dfrac{x}{2} - 9\right)\left(\dfrac{x}{2} + 9\right). \quad\halmos$$

*(aa) $x^4 - 1$ .

$$x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1).\quad\halmos$$

Reminder: The formulas you need for the next few problems are:

$$a^3 + b^3 = (a + b)(a^2 - ab + b^2).$$

$$a^3 - b^3 = (a - b)(a^2 + ab + b^2).$$

(bb) $x^3 + 8$ .

$$x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4).\quad\halmos$$

(cc) $m^3 - 27$ .

$$m^3 - 27 = m^3 - 3^3 = (m - 3)(m^2 + 3m + 9).\quad\halmos$$

(dd) $8p^3 - q^3$ .

$$8p^3 - q^3 = (2p)^3 - q^3 = (2p - q)(4p^2 + 2pq + q^2).\quad\halmos$$

(ee) $5x^4 - 5x$ .

$$5x^3 - 5x = 5x(x^3 - 1) = 5x(x^3 - 1^3) = 5x(x - 1)(x^2 + x + 1).\quad\halmos$$

*(ff) $\dfrac{x^3}{64} + \dfrac{1}{27}$ .

$$\dfrac{x^3}{64} + \dfrac{1}{27} = \left(\dfrac{x}{4}\right)^3 + \left(\dfrac{1}{3}\right)^3 = \left(\dfrac{x}{4} + \dfrac{1}{3}\right) \left(\dfrac{x^2}{16} + \dfrac{x}{12} + \dfrac{1}{9}\right). \quad\halmos$$

In the next few problems, I'll use factoring by grouping.

(gg) $2x^3 + x^2 + 6x + 3$ .

$$2x^3 + x^2 + 6x + 3 = (2x^3 + x^2) + (6x + 3) = x^2(2x + 1) + 3(2x + 1) = (2x + 1)(x^2 + 3).\quad\halmos$$

(hh) $x^3 + 3x^2 - 4x - 12$ .

$$x^3 + 3x^2 - 4x - 12 = (x^3 + 3x^2) - (4x + 12) = x^2(x + 3) - 4(x + 3) = (x + 3)(x^2 - 4) = (x + 3)(x - 2)(x + 2).\quad\halmos$$

(ii) $a^2 - 3a - 2ab + 6b$ .

$$a^2 - 3a - 2ab + 6b = (a^2 - 3a) - (2ab - 6b) = a(a - 3) - 2b(a - 3) = (a - 3)(a - 2b).\quad\halmos$$

*(jj) $2x^4 - 2x^3 + 4x^2 - 4x$ .

$$2x^4 - 2x^3 + 4x^2 - 4x = 2x(x^3 - x^2 + 2x - 2) = 2x[(x^3 - x^2) + (2x - 2)] = 2x[x^2(x - 1) + 2(x - 1)] =$$

$$2x(x - 1)(x^2 + 2).\quad\halmos$$

4. Solve the equation by factoring:

(a) $x^2 - 3x - 4 = 0$ .

$$\eqalign{ x^2 - 3x - 4 & = 0 \cr (x - 4)(x + 1) & = 0 \cr}$$

Therefore, $x = 4$ or $x = -1$ .

(b) $2x^3 - 10x^2 - 28x = 0$ .

$$\eqalign{ 2x^3 - 10x^2 - 28x & = 0 \cr 2x(x^2 - 5x - 14) & = 0 \cr 2x(x - 7)(x + 2) & = 0 \cr}$$

$2x = 0$ gives $x = 0$ , and the other factors give $x = 7$ and $x = -2$ . Therefore $x = 0$ , $x = 7$ , or $x = -2$ .

(c) $7x^3 + 7x = 0$ .

$$\eqalign{ 7x^3 + 7x & = 0 \cr 7x(x^2 + 1) & = 0 \cr}$$

$7x = 0$ gives $x = 0$ .

If $x^2 + 1 = 0$ , then $x^2 = -1$ , which has no real solutions.

The only solution is $x = 0$ .

(d) $x^2 + 9 = 10x$ .

You need to get 0 on one side of the equation to use factoring to solve.

$$\eqalign{ x^2 + 9 & = 10x \cr x^2 - 10x + 9 & = 0 \cr (x - 1)(x - 9) & = 0 \cr}$$

Therefore, $x = 1$ or $x = 9$ .

*(e) $2x^3 + x^2 - 2x - 1 = 0$ .

$$\eqalign{ 2x^3 + x^2 - 2x - 1 & = 0 \cr (2x^3 + x^2) - (2x + 1) & = 0 \cr x^2(2x + 1) - 1 \cdot (2x + 1) & = 0 \cr (2x + 1)(x^2 - 1) & = 0 \cr (2x + 1)(x - 1)(x + 1) & = 0 \cr}$$

$2x + 1 = 0$ gives $2x = -1$ , so $x = -\dfrac{1}{2}$ . The other two factors give $x = 1$ and $x = -1$ .

Therefore, $x = -\dfrac{1}{2}$ , $x = 1$ , or $x = -1$ .

5. (a) Divide $x^4 + x^3 + 1$ by $x^2 - 1$ using long division.

$$\hbox{\epsfysize=1.5in \epsffile{rev2-3a.eps}}$$

$$\dfrac{x^4 + x^3 + 1}{x^2 - 1} = (x^2 + x + 1) + \dfrac{x + 2}{x^2 - 1}.\quad\halmos$$

(b) Find the quotient and remainder when $2 x^3 + 3 x - 5$ is divided by $x + 3$ .

$$\hbox{\epsfysize=1.5in \epsffile{rev2-3b.eps}}$$

(Since $2 x^3 + 3 x - 5$ is missing an "$x^2$ " term, I put in "$0 \cdot x^2$ " as a place holder.)

$$\dfrac{2 x^3 + 3 x - 5}{x + 3} = (2 x^2 - 6 x + 21) + \dfrac{-68}{x + 3}.\quad\halmos$$

(c) Compute the quotient and remainder when $2x^4 + 3x^2 + 4x$ is divided by $2x^2$ .

$$\dfrac{2x^4 + 3x^2 + 4x}{2x^2} = \dfrac{2x^4}{2x^2} + \dfrac{3x^2}{2x^2} + \dfrac{4x}{2x^2} = x^2 + \dfrac{3}{2} + \dfrac{2}{x}.$$

Since there's only one term on the bottom, it's easier to break the fraction up into pieces than to do the long division.

(d) Compute the quotient and remainder when $2x^2 + 5xy - 3y^2$ is divided by $x + 3y$ .

$$\hbox{\epsfysize=1.25in \epsffile{rev2-3c.eps}}$$

$$\dfrac{2x^2 + 5xy - 3y^2}{x + 3y} = 2x - y.\quad\halmos$$

(e) Factor $x^3 - 2x^2 - 11x + 12$ completely, given that $x - 4$ is one of the factors.

Divide $x^3 - 2x^2 - 11x + 12$ by $x - 4$ :

$$\hbox{\epsfysize=1.5in \epsffile{rev2-3d.eps}}$$

Thus,

$$x^3 - 2x^2 - 11x + 12 = (x - 4)(x^2 + 2x - 3) = (x - 4)(x + 3)(x - 1).\quad\halmos$$

6. (a) Show that it is not valid to cancel x's in $\dfrac{x + 4}{x}$ to get $\dfrac{1 + 4}{1} = 5$ by giving a specific value of x for which $\dfrac{x + 4}{x}$ is not equal to 5.

For $x = 2$ , $\dfrac{x + 4}{x} = \dfrac{2 + 4}{2} = \dfrac{6}{2} = 3 \ne 5$ . Thus, you can't cancel x's in $\dfrac{x + 4}{x}$ to get 5.

(b) Show that $\dfrac{a}{\dfrac{b}{c}}$ is not always the same as $\dfrac{\dfrac{a}{b}}{c}$ by giving specific values of a, b, and c for which $\dfrac{a}{\dfrac{b}{c}}$ is not equal to $\dfrac{\dfrac{a}{b}}{c}$ .

If $a = 1$ , $b = 1$ , and $c = 2$ , then

$$\dfrac{a}{\dfrac{b}{c}} = \dfrac{1}{\dfrac{1}{2}} = 2, \quad\hbox{but}\quad \dfrac{\dfrac{a}{b}}{c} = \dfrac{\dfrac{1}{1}}{2} = \dfrac{1}{2}.$$

Thus, $\dfrac{a}{\dfrac{b}{c}}$ is not in general equal to $\dfrac{\dfrac{a}{b}}{c}$ .

7. Simplify, cancelling any common factors:

(a) $\dfrac{5x^2}{y^3} \cdot \dfrac{y^{11}}{10x^5}$ .

$$\dfrac{5x^2}{y^3} \cdot \dfrac{y^{11}}{10x^5} = \dfrac{y^8}{2x^3}.\quad\halmos$$

(b) $\dfrac{x^2 - x - 2}{x^2 + 3x + 2}\cdot \dfrac{x^2 - 5x + 6}{x^2 - 4x + 4}$ .

$$\dfrac{x^2 - x - 2}{x^2 + 3x + 2}\cdot \dfrac{x^2 - 5x + 6}{x^2 - 4x + 4} = \dfrac{(x - 2)(x + 1)}{(x + 1)(x + 2)}\cdot \dfrac{(x - 2)(x - 3)}{(x - 2)^2} = \dfrac{x - 3}{x + 2}.\quad\halmos$$

(c) $\dfrac{x^3 - 4x}{5x^2}\cdot \dfrac{x^2 - 5x + 6}{x^2 - 4x + 4}\cdot \dfrac{10x^3}{6x + 12}$ .

$$\dfrac{x^3 - 4x}{5x^2}\cdot \dfrac{x^2 - 5x + 6}{x^2 - 4x + 4}\cdot \dfrac{10x^3}{6x + 12} = \dfrac{x(x^2 - 4)}{5x^2}\cdot \dfrac{(x - 2)(x - 3)}{(x - 2)^2}\cdot \dfrac{10x^3}{6(x + 2)} =$$

$$\dfrac{x(x - 2)(x + 2)}{5x^2}\cdot \dfrac{(x - 2)(x - 3)}{(x - 2)^2} \cdot \dfrac{10x^3}{6(x + 2)} = \dfrac{x^2(x - 3)}{3}.\quad\halmos$$

8. Simplify, cancelling any common factors:

(a) $\dfrac{\dfrac{x^2 + 5x}{2x^2 - 6x}}{\dfrac{x^2 - 25}{x^2 - 5x + 6}}$ .

$$\dfrac{\dfrac{x^2 + 5x}{2x^2 - 6x}}{\dfrac{x^2 - 25}{x^2 - 5x + 6}} = \dfrac{x^2 + 5x}{2x^2 - 6x} \cdot \dfrac{x^2 - 5x + 6}{x^2 - 25} = \dfrac{x(x + 5)}{2x(x - 3)} \cdot \dfrac{(x - 2)(x - 3)}{(x - 5)(x + 5)} = \dfrac{x - 2}{2(x - 5)}.\quad\halmos$$

(b) $\dfrac{x^2 - x - 2}{5x^4 - 15x^3}\div \dfrac{x^2 + 5x + 4}{x^3 + 4x^2}$ .

$$\dfrac{x^2 - x - 2}{5x^4 - 15x^3}\div \dfrac{x^2 + 5x + 4}{x^3 + 4x^2} = \dfrac{\dfrac{x^2 - x - 2}{5x^4 - 15x^3}} {\dfrac{x^2 + 5x + 4}{x^3 + 4x^2}} = \dfrac{x^2 - x - 2}{5x^4 - 15x^3}\cdot \dfrac{x^3 + 4x^2}{x^2 + 5x + 4} =$$

$$\dfrac{(x - 2)(x + 1)}{5x^3(x - 3)}\cdot \dfrac{x^2(x + 4)}{(x + 1)(x + 4)} = \dfrac{x - 2}{5x(x - 3)}.\quad\halmos$$

(c) $\dfrac{\dfrac{x^2 - 2xy}{x^3 + x^2y}}{\dfrac{x^2 - 4y^2}{x^2 - 4xy - 5y^2}}$ .

$$\dfrac{\dfrac{x^2 - 2xy}{x^3 + x^2y}} {\dfrac{x^2 - 4y^2}{x^2 - 4xy - 5y^2}} = \dfrac{x^2 - 2xy}{x^3 + x^2y}\cdot \dfrac{x^2 - 4xy - 5y^2}{x^2 - 4y^2} = \dfrac{x(x - 2y)}{x^2(x + y)}\cdot \dfrac{(x - 5y)(x + y)}{(x - 2y)(x + 2y)} = \dfrac{x - 5y}{x(x + 2y)}.\quad\halmos$$

(d) $\dfrac{\dfrac{x^3 - 5x^2}{x^2 - 2x - 15}}{\dfrac{x^2 + 2x}{x^2 - 4}}$ .

$$\dfrac{\dfrac{x^3 - 5x^2}{x^2 - 2x - 15}}{\dfrac{x^2 + 2x}{x^2 - 4}} = \dfrac{x^3 - 5x^2}{x^2 - 2x - 15}\cdot \dfrac{x^2 - 4}{x^2 + 2x} = \dfrac{x^2(x - 5)}{(x - 5)(x + 3)}\cdot \dfrac{(x - 2)(x + 2)}{x(x + 2)} = \dfrac{x(x - 2)}{x + 3}.\quad\halmos$$

9. Combine the fractions into a single fraction and simplify:

(a) $\dfrac{2}{x - 2} + \dfrac{3x}{x^2 - 2x} - \dfrac{2}{x}$ .

$$\dfrac{2}{x - 2} + \dfrac{3x}{x^2 - 2x} - \dfrac{2}{x} = \dfrac{2}{x - 2} + \dfrac{3x}{x(x - 2)} - \dfrac{2}{x} = \dfrac{x}{x}\cdot \dfrac{2}{x - 2} + \dfrac{3x}{x(x - 2)} - \dfrac{x - 2}{x - 2}\cdot \dfrac{2}{x} =$$

$$\dfrac{2x + 3x - 2(x - 2)}{x(x - 2} = \dfrac{3x + 4}{x(x - 2)}\quad\halmos$$

(b) $\dfrac{1}{x^2 - 5x + 6} - \dfrac{6}{x^2 - 9}$ .

$$\dfrac{1}{x^2 - 5x + 6} - \dfrac{6}{x^2 - 9} = \dfrac{1}{(x - 2)(x - 3)} - \dfrac{6}{(x - 3)(x + 3)} = \dfrac{1}{(x - 2)(x - 3)}\cdot \dfrac{x + 3}{x + 3} - \dfrac{6}{(x - 3)(x + 3)}\cdot \dfrac{x - 2}{x - 2} =$$

$$\dfrac{x + 3}{(x - 2)(x - 3)(x + 3)} - \dfrac{6(x - 2)}{(x - 2)(x - 3)(x + 3)} = \dfrac{x + 3 - 6(x - 2)}{(x - 2)(x - 3)(x + 3)} = \dfrac{x + 3 - 6x + 12}{(x - 2)(x - 3)(x + 3)} =$$

$$\dfrac{-5x + 15}{(x - 2)(x - 3)(x + 3)} = \dfrac{-5(x - 3)}{(x - 2)(x - 3)(x + 3)} = \dfrac{-5}{(x - 2)(x + 3)}.\quad\halmos$$

(c) $2 - \dfrac{1}{x} + \dfrac{1}{x + 1}$ .

$$2 - \dfrac{1}{x} + \dfrac{1}{x + 1} = 2\cdot \dfrac{x(x + 1)}{x(x + 1)} - \dfrac{1}{x}\cdot \dfrac{x + 1}{x + 1} + \dfrac{1}{x + 1}\cdot \dfrac{x}{x} = \dfrac{2x(x + 1)}{x(x + 1)} - \dfrac{x + 1}{x(x + 1)} + \dfrac{x}{x(x + 1)} =$$

$$\dfrac{2x(x + 1) - (x + 1) + x}{x(x + 1)} = \dfrac{2x^2 + 2x - x - 1 + x}{x(x + 1)} = \dfrac{2x^2 + 2x - 1}{x(x + 1)}.\quad\halmos$$

(d) $\dfrac{x}{x + y} + \dfrac{y}{x - y}$ .

$$\dfrac{x}{x + y} + \dfrac{y}{x - y} = \dfrac{x}{x + y}\cdot \dfrac{x - y}{x - y} + \dfrac{y}{x - y}\cdot \dfrac{x + y}{x + y} = \dfrac{x^2 - xy}{(x - y)(x + y)} + \dfrac{yx + y^2}{(x - y)(x + y)} =$$

$$\dfrac{x^2 - xy + yx + y^2}{(x - y)(x + y)} = \dfrac{x^2 + y^2}{(x - y)(x + y)}.\quad\halmos$$

(e) $\dfrac{x - 2y}{x^2 - xy} + \dfrac{2y}{x^2 - y^2}$ .

$$\dfrac{x - 2y}{x^2 - xy} + \dfrac{2y}{x^2 - y^2} = \dfrac{x - 2y}{x(x - y)} + \dfrac{2y}{(x - y)(x + y)} = \dfrac{x - 2y}{x(x - y)}\cdot \dfrac{x + y}{x + y} + \dfrac{2y}{(x - y)(x + y)}\cdot \dfrac{x}{x} =$$

$$\dfrac{x^2 - xy - 2y^2}{x(x - y)(x + y)} + \dfrac{2xy}{x(x - y)(x + y)} = \dfrac{x^2 - xy - 2y^2 + 2xy}{x(x - y)(x + y)} = \dfrac{x^2 + xy - 2y^2}{x(x - y)(x + y)} =$$

$$\dfrac{(x + 2y)(x - y)}{x(x - y)(x + y)} = \dfrac{x + 2y}{x(x + y)}.\quad\halmos$$

(f) $x - \dfrac{1}{x} + \dfrac{x + 1}{x - 1}$ .

$$x - \dfrac{1}{x} + \dfrac{x + 1}{x - 1} = x \cdot \dfrac{x(x - 1)}{x(x - 1)} - \dfrac{1}{x} \cdot \dfrac{x - 1}{x - 1} + \dfrac{x + 1}{x - 1} \cdot \dfrac{x}{x} = \dfrac{x^2(x - 1)}{x(x - 1)} - \dfrac{x - 1}{x(x - 1)} + \dfrac{x(x + 1)}{x(x - 1)} =$$

$$\dfrac{x^2(x - 1) - (x - 1) + x(x + 1)}{x(x - 1)} = \dfrac{x^3 - x^2 - x + 1 + x^2 + x}{x(x - 1)} = \dfrac{x^3 + 1}{x(x - 1)} = \dfrac{(x + 1)(x^2 - x + 1)}{x(x - 1)}.\quad\halmos$$

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