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${(8x)}^{3}$ means $(8x)(8x)(8x)$ since the parentheses indicate that the exponent 3 is directly connected to the factor $8x$ . Remember that the grouping symbols $\left(\right)$ indicate that the quantities inside are to be considered as one single number.
$34{(a+1)}^{2}$ means $34\cdot (a+1)(a+1)$ since the exponent 2 applies only to the factor $(a+1)$ .
Write each of the following without exponents.
Select a number to show that ${(2x)}^{2}$ is not always equal to $2{x}^{2}$ .
Suppose we choose $x$ to be 5. Consider both ${(2x)}^{2}$ and $2{x}^{2}$ .
Notice that ${(2x)}^{2}=2{x}^{2}$ only when $x=0$ .
Select a number to show that ${(5x)}^{2}$ is not always equal to $5{x}^{2}$ .
Select $x=3$ . Then ${(5\cdot 3)}^{2}={(15)}^{2}=225$ , but $5\cdot {3}^{2}=5\cdot 9=45$ . $225\ne 45$ .
In ${x}^{n}$ ,
Cube with
length
$=x$
width
$=x$
depth
$=x$
Volume
$=xxx={x}^{3}$
In Section [link] we were introduced to the order of operations. It was noted that we would insert another operation before multiplication and division. We can do that now.
Use the order of operations to simplify each of the following.
${2}^{2}+5=4+5=9$
${5}^{2}+{3}^{2}+10=25+9+10=44$
$\begin{array}{ll}{2}^{2}+(5)(8)-1\hfill & =4+(5)(8)-1\hfill \\ \hfill & =4+40-1\hfill \\ \hfill & =43\hfill \end{array}$
$\begin{array}{ll}7\cdot 6-{4}^{2}+{1}^{5}\hfill & =7\cdot 6-16+1\hfill \\ \hfill & =42-16+1\hfill \\ \hfill & =27\hfill \end{array}$
$\begin{array}{ll}{(2+3)}^{3}+{7}^{2}-3{(4+1)}^{2}\hfill & ={(5)}^{3}+{7}^{2}-3{(5)}^{2}\hfill \\ \hfill & =125+49-3(25)\hfill \\ \hfill & =125+49-75\hfill \\ \hfill & =99\hfill \end{array}$
$\begin{array}{ll}{[4{(6+2)}^{3}]}^{2}\hfill & ={[4{(8)}^{3}]}^{2}\hfill \\ \hfill & ={[4(512)]}^{2}\hfill \\ \hfill & ={[2048]}^{2}\hfill \\ \hfill & =4,194,304\hfill \end{array}$
$\begin{array}{ll}6({3}^{2}+{2}^{2})+{4}^{2}\hfill & =6(9+4)+{4}^{2}\hfill \\ \hfill & =6(13)+{4}^{2}\hfill \\ \hfill & =6(13)+16\hfill \\ \hfill & =78+16\hfill \\ \hfill & =94\hfill \end{array}$
$\begin{array}{ll}\frac{{6}^{2}+{2}^{2}}{{4}^{2}+6\cdot {2}^{2}}+\frac{{1}^{3}+{8}^{2}}{{10}^{2}-(19)(5)}\hfill & =\frac{36+4}{16+6\cdot 4}+\frac{1+64}{100-95}\hfill \\ \hfill & =\frac{36+4}{16+24}+\frac{1+64}{100-95}\hfill \\ \hfill & =\frac{40}{40}+\frac{65}{5}\hfill \\ \hfill & =1+13\hfill \\ \hfill & =14\hfill \end{array}$
Use the order of operations to simplify the following.
$\frac{{5}^{2}+{6}^{2}-10}{1+{4}^{2}}+\frac{{0}^{4}-{0}^{5}}{{7}^{2}-6\cdot {2}^{3}}$
3
For the following problems, write each of the quantities using exponential notation.
$a$ squared
$(-3)$ cubed
3 squared times $y$ to the fifth
$a$ cubed minus $(b+7)$ squared
${a}^{3}-{\left(b+7\right)}^{2}$
$(21-x)$ cubed plus $(x+5)$ to the seventh
$(8)(8)xxxx$
$2\cdot 3\cdot 3\cdot 3\cdot 3xxyyyyy$
$2\left({3}^{4}\right){x}^{2}{y}^{5}$
$2\cdot 2\cdot 5\cdot 6\cdot 6\cdot 6xyyzzzwwww$
$10xyy(c+5)(c+5)(c+5)$
$4x4x4x4x4x$
${\left(4x\right)}^{5}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}{4}^{5}{x}^{5}$
$(9a)(9a)(9a)(9a)$
$(-7)(-7)(-7)aabbba(-7)baab$
${\left(-7\right)}^{4}{a}^{5}{b}^{5}$
$(a-10)(a-10)(a+10)$
$(z+w)(z+w)(z+w)(z-w)(z-w)$
${\left(z+w\right)}^{3}{\left(z-w\right)}^{2}$
$(2y)(2y)2y2y$
$3xyxxy-(x+1)(x+1)(x+1)$
$3{x}^{3}{y}^{2}-{\left(x+1\right)}^{3}$
For the following problems, expand the quantities so that no exponents appear.
${4}^{3}$
${6}^{2}$
$6\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}6$
${7}^{3}{y}^{2}$
$8{x}^{3}{y}^{2}$
$8\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}y\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}y$
${(18{x}^{2}{y}^{4})}^{2}$
${(9{a}^{3}{b}^{2})}^{3}$
$\left(9aaabb\right)\left(9aaabb\right)\left(9aaabb\right)\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}9\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}9\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}9aaaaaaaaabbbbbb$
$5{x}^{2}{(2{y}^{3})}^{3}$
$10{a}^{3}{b}^{2}{(3c)}^{2}$
$10aaabb\left(3c\right)\left(3c\right)\text{\hspace{0.17em}or}\text{\hspace{0.17em}}10\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}3aaabbcc$
${(a+10)}^{2}{({a}^{2}+10)}^{2}$
$({x}^{2}-{y}^{2})({x}^{2}+{y}^{2})$
$\left(xx-yy\right)\left(xx+yy\right)$
For the following problems, select a number (or numbers) to show that
${(5x)}^{2}$ is not generally equal to $5{x}^{2}$ .
${(7x)}^{2}$ is not generally equal to $7{x}^{2}$ .
Select $x=2.$ Then, $196\ne 28.$
${(a+b)}^{2}$ is not generally equal to ${a}^{2}+{b}^{2}$ .
For what real number is ${(6a)}^{2}$ equal to $6{a}^{2}$ ?
zero
For what real numbers, $a$ and $b$ , is ${(a+b)}^{2}$ equal to ${a}^{2}+{b}^{2}$ ?
Use the order of operations to simplify the quantities for the following problems.
${4}^{3}-18$
${8}^{2}+3+5(2+7)$
${3}^{4}+{2}^{4}{(1+5)}^{3}$
${2}^{2}(10-{2}^{3})$
${(4+3)}^{2}+1\xf7(2\cdot 5)$
${1}^{6}+{0}^{8}+{5}^{2}{(2+8)}^{3}$
$\frac{{2}^{3}-7}{{5}^{2}}$
$\frac{{6}^{2}-1}{5}+\frac{{4}^{3}+(2)(3)}{10}$
$\frac{5[{8}^{2}-9(6)]}{{2}^{5}-7}+\frac{{7}^{2}-{4}^{2}}{{2}^{4}-5}$
5
$\frac{{(2+1)}^{3}+{2}^{3}+{1}^{3}}{{6}^{2}}-\frac{{15}^{2}-{[2(5)]}^{2}}{5\cdot {5}^{2}}$
$\frac{{6}^{3}-2\cdot {10}^{2}}{{2}^{2}}+\frac{18({2}^{3}+{7}^{2})}{2(19)-{3}^{3}}$
$\frac{1070}{11}\text{\hspace{0.17em}}\text{or}\text{\hspace{0.17em}}97.\overline{27}$
( [link] ) Use algebraic notation to write the statement "a number divided by eight, plus five, is equal to ten."
( [link] ) Draw a number line that extends from $-5$ to 5 and place points at all real numbers that are strictly greater than $-3$ but less than or equal to 2.
( [link] ) Is every integer a whole number?
( [link] ) Use the commutative property of multiplication to write a number equal to the number $yx$ .
$xy$
( [link] ) Use the distributive property to expand $3(x+6)$ .
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