## Exercises - Alternate Bases and Polynomials

1. For each, express the value in base 10.

1. $(73501)_8$

2. $(4213)_5$

3. $(110110001)_2$

4. $(3AF10E)_{16}$

5. $(D87BC)_{16}$

1. $30529$

2. $558$

3. $433$

4. $3862798$

5. $886716$

2. For each, convert the (base 10) value given to the indicated base.

1. $4582 = (\ \underline{\ ? \ }\ )_5$

2. $19921 = (\ \underline{\ ? \ }\ )_8$

3. $823 = (\ \underline{\ ? \ }\ )_2$

4. $1254 = (\ \underline{\ ? \ }\ )_{16}$

5. $262144 = (\ \underline{\ ? \ }\ )_{16}$

1. $(121312)_5$

2. $(46721)_8$

3. $(1100110111)_2$

4. $(4E6)_{16}$

5. $(40000)_{16}$

3. Find each sum without converting to another base first.

1. $(167)_8 + (635)_8$

2. $(243124)_5 + (3123)_5$

3. $(10111)_2 + (111011)_2$

4. $(A3598)_{16} + (25739)_{16}$

5. $(F1EAC)_{16} + (35B97)_{16}$

1. $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & & {\scriptsize 1} & {\scriptsize 1} &\\ & & 1 & 6 & 7\\ + & & 6 & 3 & 5 \\\hline & 1 & 0 & 2 & 4 \end{array} }$

2. $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & {\scriptsize 1} & {\scriptsize 1} & & {\scriptsize 1} & {\scriptsize 1} &\\ & 2 & 4 & 3 & 1 & 2 & 4\\ + & & & 3 & 1 & 2 & 3\\\hline & 3 & 0 & 1 & 3 & 0 & 2 \end{array} }$

3. $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & & {\scriptsize 1} & {\scriptsize 1} & {\scriptsize 1} & {\scriptsize 1} & {\scriptsize 1} &\\ & & & 1 & 0 & 1 & 1 & 1\\ + & & 1 & 1 & 1 & 0 & 1 & 1\\\hline & 1 & 0 & 1 & 0 & 0 & 1 & 0 \end{array} }$

4. $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & & & & {\scriptsize 1} &\\ & A & 3 & 5 & 9 & 8\\ + & 2 & 5 & 7 & 3 & 9\\\hline & C & 8 & C & D & 1 \end{array} }$

5. $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & {\scriptsize 1} & & {\scriptsize 1} & {\scriptsize 1} & {\scriptsize 1} &\\ & & F & 1 & E & A & C\\ + & & 3 & 5 & B & 9 & 7\\\hline & 1 & 2 & 7 & A & 4 & 3 \end{array} }$

4. Find each product without converting to another base first.

1. $(2314)_5 \cdot (412)_5$

2. $(3361)_7 \cdot (514)_7$

3. $(A3C4)_{16} \cdot (235)_{16}$

1.  $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & & \overset{\color{red}{2}}{\vphantom{0}} & \overset{\color{blue}{1},\color{red}{2}}{2} & \overset{\color{red}{1}}{3} & \overset{\color{blue}{1},\color{red}{3}}{1} & 4 \\ \times & & & & \color{red}{4} & 1 & \color{blue}{2} \\\hline & & \overset{\color{green}{1}}{1} & 0 & \overset{\color{green}{1}}{1} & 3 & 3 \\ & & 2 & 3 & 1 & 4 & \\ 2 & \smash{\overset{\color{green}{1}}{0}}\vphantom{0} & 3 & 2 & 1 & & \\\hline 2 & 1 & 2 & 0 & 4 & 2 & 3 \\ \end{array} }$ Note: multiplications by the blue $2$ and red $4$ create carries shown in the corresponding colors.Carries from adding the three rows are shown in green. Also, the following conversions were done to find some of the digits shown in the three rows added at left, and their sum. For example, as our first step in finding the product at left, we multiply the unit's digits, finding $2 \cdot 4 = 8$. Given the first calculation below, where $8 = (13)_5$, we write a $3$ in the unit's digit column in the first of what will ultimately be three rows added, and then carry a (blue) $1$ to the next column (to the left). Then, we multiply $2$ by $1$ and add the carry of $1$ to get $3$. This being less than the base requires no conversion. Then, we multiply $2$ by $3$ (with no carry) to get $6$. Appealing to the calculation $6=(11)_5$ below, we write a digit of $1$ and carry the other $1$, and so on.. $$\begin{array}{rclcl} 8 &=& 1 \cdot 5 + 3 &=& (13)_{5}\\ 6 &=& 1 \cdot 5 + 1 &=& (11)_{5}\\ 5 &=& 1 \cdot 5 + 0 &=& (10)_{5}\\ 16 &=& 3 \cdot 5 + 1 &=& (31)_{5}\\ 7 &=& 1 \cdot 5 + 2 &=& (12)_{5}\\ 13 &=& 2 \cdot 5 + 3 &=& (23)_{5}\\ 10 &=& 2 \cdot 5 + 0 &=& (20)_{5}\\ 7 &=& 1 \cdot 5 + 2 &=& (12)_{5}\\ \end{array}$$

2.  $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & & \overset{\color{blue}{2},\color{red}{2}}{\phantom{0}} & \overset{\color{blue}{2},\color{red}{2}}{3} & \overset{\color{blue}{3},\color{red}{4}}{3} & 6 & 1 \\ \times & & & & \color{red}{5} & 1 & \color{blue}{4} \\\hline & & 2 & \overset{\color{green}{1}}{0} & 1 & 3 & 4 \\ & & 3 & 3 & 6 & 1 & \\ 2 & \smash{\overset{\color{green}{1}}{3}}\vphantom{3} & 5 & 2 & 5 & & \\\hline 2 & 4 & 3 & 6 & 5 & 4 & 4 \\ \end{array} }$ Note: multiplications by the blue $4$ and red $5$ create carries shown in the corresponding colors.Carries from adding the three rows are shown in green. Also, the following conversions were done to find some of the digits shown in the three rows added at left, and their sum: $$\begin{array}{rclcl} 24 &=& 3 \cdot 7 + 3 &=& (33)_{7}\\ 15 &=& 2 \cdot 7 + 1 &=& (21)_{7}\\ 14 &=& 2 \cdot 7 + 0 &=& (20)_{7}\\ 30 &=& 4 \cdot 7 + 2 &=& (42)_{7}\\ 19 &=& 2 \cdot 7 + 5 &=& (25)_{7}\\ 17 &=& 2 \cdot 7 + 3 &=& (23)_{7}\\ \end{array}$$

3.  $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & & \overset{\color{blue}{3},\color{purple}{3},\color{red}{1}}{\phantom{0}} & \overset{\color{blue}{1},\color{purple}{2}}{A} & \overset{\color{blue}{3},\color{red}{1}}{3} & \overset{\color{blue}{1}}{C} & 4\\ & & & & \color{red}{2} & \color{purple}{3} & \color{blue}{5}\\\hline & & \overset{\color{green}{1}}{3} & 3 & \overset{\color{green}{1}}{2} & D & 4\\ & \smash{\overset{\color{green}{1}}{1}}\vphantom{1} & E & B & 4 & C & \\ 1 & 4 & 7 & 8 & 8 & & \\\hline 1 & 6 & 9 & 6 & F & 9 & 4\\ \end{array} }$ Note: multiplications by the blue $5$, purple $3$, and red $2$ create carries shown in the corresponding colors.Carries from adding the three rows are shown in green. Also, remembering that in base $16$: $A=10$, $B=11$, $C=12$, $D=13$, $E=14$, and $F=15$; note that the following conversions were done to find some of the digits shown in the three rows added at left, and their sum: $$\begin{array}{rclcl} 20 &=& 1 \cdot 16 + 4 &=& (14)_{16}\\ 61 &=& 3 \cdot 16 + 13 &=& (3D)_{16}\\ 18 &=& 1 \cdot 16 + 2 &=& (12)_{16}\\ 51 &=& 3 \cdot 16 + 3 &=& (13)_{16}\\ 36 &=& 2 \cdot 16 + 4 &=& (14)_{16}\\ 30 &=& 1 \cdot 16 + 14 &=& (1E)_{16}\\ 24 &=& 1 \cdot 16 + 8 &=& (18)_{16}\\ 20 &=& 1 \cdot 16 + 4 &=& (14)_{16}\\ 22 &=& 1 \cdot 16 + 6 &=& (16)_{16}\\ 25 &=& 1 \cdot 16 + 9 &=& (19)_{16}\\ \end{array}$$

5. Find each difference without converting to another base first.

1. $(432)_8 - (176)_8$

2. $(204)_5 - (143)_5$

3. $(10000)_2 - (1101)_2$

1. $\require{cancel}$
 $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & \overset{3}{\cancel{4}} & \overset{{}^1 2}{\cancel{3}} & {}^1 2\\ - & 1 & 7 & 6\\\hline & 2 & 3 & 4\\ \end{array}}$ Don't forget that the subtractions done in each column need to be done in base $8$. So here, we find $$\begin{array}{rcl} (12)_8 - (6)_8 &=& (4)_8\\ (12)_8 - (7)_8 &=& (3)_8\\ (3)_8 - (1)_8 &=& (2)_8 \end{array}$$

2.  $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & \overset{1}{\cancel{2}} & {}^1 0 & 4\\ - & 1 & 4 & 3\\\hline & & 1 & 1\\ \end{array}}$ Don't forget that the subtractions done in each column need to be done in base $5$. So here, we find $$\begin{array}{rcl} (4)_5 - (3)_5 &=& (1)_5\\ (10)_5 - (4)_5 &=& (3)_1\\ (1)_5 - (1)_5 &=& (0)_5 \end{array}$$

3.  $\displaystyle{ \begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & \cancel{1} & \overset{1}{\cancel{0}} & \overset{1}{\cancel{0}} & \overset{1}{\cancel{0}} & {}^1 0\\ - & & 1 & 1 & 0 & 1\\\hline & & & & 1 & 1\\ \end{array}}$ The "borrowing" here is handled a bit differently, given the string of $0$'s in the top number. Recall what happens in similar circumstnaces in base 10 as demonstrated by the example below, where we "borrow" $1$ from $1000$, replacing it with $999$: $$\begin{array}[t]{c@{\,},c@{\,},c@{\,},c@{\,},c@{\,}c} & \cancel{1} & \overset{9}{\cancel{0}} & \overset{9}{\cancel{0}} & \overset{9}{\cancel{0}} & {}^1 0\\ - & & 1 & 4 & 8 & 7\\\hline & & 8 & 5 & 1 & 3\\ \end{array}$$ At left, recall in base $2$ the digit $1$ is the greatest digit, and consequently $1$ in base $2$ "plays the same role" as $9$ in base $10$. We need to "borrow" a $1$ for the subtraction in the unit's column, so the $1000$ to the left of the $0$ in that column becomes $111$ accordingly. Also, don't forget that the subtractions done in each column need to be done in base $2$. So here, $(10)_2 - (1)_2 = (1)_2$.

6. Write two different representations of each value given. For all, keep the same base for the overall number, but use base $10$ for the individual digits. Include at least one "negative digit" and at least one "digit" greater than or equal to the base in the representations you find. (Don't forget to use commas to separate the "digits" to avoid any confusion!)

1. $3076$

2. $(2304)_5$

3. $(1122)_7$

1. $3076 = (3,-1,17,6)_{10} = (30,7,6)_{10}$

2. $(2304)_5 = (2,3,-1,9)_5 = (2,3,-3,19)_{5}$

3. $(1122)_7 = (-1,15,2,2)_{7} = (1,-9,72,2)_{7}$

7. Express as a single polynomial each sum, difference, or product of polynomials (or mixture of these) given. Do this using general associativity, general commutativity, inverses, and the distributive property -- identifying these as you use them.

1. $(x^3 + 2x^2 - 5x + 4) + (2x^3 - 7x^2 - 3x + 1)$

2. $(9y^4 - 2y + 8) + (y^3 - 5y^2 - 3y - 6)$

3. $(5t^3 - t^2 + 7) + (-11t^3 + 4t^2 - 7)$

4. $(x^3 + 2x^2 - 5x + 4) - (2x^3 - 7x^2 - 3x + 1)$

5. $(9p^4 - 2p + 8) - (p^3 - 5p^2 - 3p - 6)$

6. $(5x^3 - x^2 + 7) - (-11x^3 + 4x^2 - 7)$

7. $(2x + 3)(5x - 7)$

8. $(3q^2 + 8q + 2)(7q - 5)$

9. $(3x^3 -x + 6)(4x^2 + 2x + 5)$

10. $(2a+1)^2 + (a-1)^3$

11. $(5x+3)(5x-3) - (4x+3)(4x-3)$

12. $(x-1)(x^2+x+1) - x^3$

13. $(x+1)^4$

14. $(4m^2-m+8)(m^3+2m^2+3m+4)$

15. $(x+1)^3+(x+1)^2+(x+1)+1$

The resulting polynomial for each will be

1. $3x^3-5x^2 -8x + 5$

2. $9y^4 + y^3 - 5y^2 - 5y + 2$

3. $-6t^3 + 3t^2$

4. $-x^3 + 9x^2 - 2x + 3$

5. $9p^4 - p^3 + 5p^2 + p + 14$

6. $16x^3 -5x^2 + 14$

7. $10x^2 + x - 21$

8. $21q^3 + 41q^2 - 26q - 10$

9. $12x^5 + 6x^4 + 11x^3 + 22x^2 + 7x + 30$

10. $a^3 + a^2 + 7a$

11. $9x^2$

12. $-1$

13. $x^4 + 4x^3 + 6x^2 + 4x + 1$

14. $4m^5 + 7m^4 + 18m^3 + 29m^2 + 20m + 32$

15. $x^3 + 4x^2 + 6x + 4$

8. Express as a single polynomial each sum, difference, or product of polynomials by doing a "(base $x$)" arithmetic calculation without carries and possibly involving integer "digits" that are either negative or greater than or equal to $x$, in addition to the "normal" digits available in base $x$ (i.e., $0$ to one less than $x$).

1. $(x^3 + 2x^2 - 5x + 4) + (2x^3 - 7x^2 - 3x + 1)$

2. $(9y^4 - 2y + 8) + (y^3 - 5y^2 - 3y - 6)$

3. $(9p^4 - 2p + 8) - (p^3 - 5p^2 - 3p - 6)$

4. $(5x^3 - x^2 + 7) - (-11x^3 + 4x^2 - 7)$

5. $(2x + 3)(5x - 7)$

6. $(3q^2 + 8q + 2)(7q - 5)$

7. $(2x+5)(2x-5)$

8. $(x^2 + 2x + 1)(x^2 - x + 1)$

1. $\begin{array}{ccccc} & 1 & 2 & -5 & 4 \\ + & 2 & -7 & -3 & 1 \\\hline & 3 & -5 & -8 & 5 \end{array} \quad \Longrightarrow \quad 3x^3 - 5x^2 - 8x + 5$

2. $\begin{array}{ccccc} & 9 & 0 & 0 & -2 & 8 \\ + & & 1 & -5 & -3 & -6 \\\hline & 9 & 1 & -5 & -5 & 2 \end{array} \quad \Longrightarrow \quad 9y^4 + y^3 - 5y^2 - 5y + 2$

3. $\begin{array}{ccccc} & 9 & 0 & 0 & -2 & 8 \\ - & & 1 & -5 & -3 & -6 \\\hline & 9 & -1 & 5 & 1 & 14 \end{array} \quad \Longrightarrow \quad 9p^4 - p^3 + 5p^2 + p + 14$

4. $\begin{array}{ccccc} & 5 & -1 & 0 & 7 \\ - & -11 & 4 & 0 & -7 \\\hline & 16 & -5 & 0 & 14 \end{array} \quad \Longrightarrow \quad 16x^3 - 5x^2 + 14$

5. $\begin{array}{ccccc} & 2 & 3\\ \times & 5 & -7\\\hline & -14 & -21\\ 10 & 15 & \\\hline 10 & 1 & -21 \end{array} \quad \Longrightarrow \quad 10x^2 + x - 21$

6. $\begin{array}{ccccc} & 3 & 8 & 2\\ \times & & 7 & -5\\\hline & -15 & -40 & -10 \\ 21 & 56 & 14 & \\\hline 21 & 41 & -26 & -10 \end{array} \quad \Longrightarrow \quad 21q^3 + 41q^2 - 26q - 10$

7. $\begin{array}{ccccc} & 2 & 5\\ \times & 2 & -5\\\hline & -10 & -25\\ 4 & 10 & \\\hline 4 & 0 & -25 \end{array} \quad \Longrightarrow \quad 4x^2 - 25$

8. $\begin{array}{ccccc} \times & & 1 & 2 & 1\\ & & 1 & -1 & 1\\\hline & & 1 & 2 & 1 \\ & -1 & -2 & -1 & \\ 1 & 2 & 1 & & \\\hline 1 & 1 & 0 & 1 & 1 \end{array} \quad \Longrightarrow \quad x^4 + x^3 + x + 1$

9. Express the following combinations of values written in two unknown bases as a polynomial of multiple variables.

1. $(2,5,3)_x (1,-4,-6)_y$

2. $(7,-3,0)_a (1,0,-2,9)_b$

3. $(1,-3)_p^2 - (1,-1,2)_q^2$

1. $(2x^2 + 5x + 3)(y^2 - 4y - 6)$
$= 2x^2 y^2 -8x^2 y -12x^2 + 5xy^2 - 20xy - 30x +3y^2 - 12y - 18$

2. $(7a^2 - 3a)(b^3-2b+9) = 7a^2b^3 -14a^2b + 63a^2 -3ab^3 +6ab - 27a$

3. $(p-3)^2 - (q^2-q+2)^2$
$= (p^2-6p+9) - (q^4 -2q^3 + 5q^2 -4q + 4)$
$= p^2 - 6p - q^4 + 2q^3 - 5q^2 + 4q + 13$

10. Express the following combination of polynomials (many of multiple variables) as a single polynomial of multiple variables.

1. $\displaystyle{(x^3-3x^2y+4xy^2+y^3)+(7x^3+x^2y-9xy^2+y^3)}$

2. $\displaystyle{(5xy^4 - 7xy^2 + 4x^2 - 3) \, - \, (-3xy^4 + 2xy^2 - 2y + 4)}$

3. $\displaystyle{(2x^3-5x^2y+xy^2-y^3)-(8x^3-x^2y-3xy^2+y^3)}$

4. $\displaystyle{(3x+y)(7x-3y)}$

5. $\displaystyle{(2x^3 - 5y)^2}$

6. $\displaystyle{(x+t)(x^2-xt+t^2)}$

7. $\displaystyle{(5a + 4b)^3}$

8. $\displaystyle{3x(4xy^3-7x^2y-3y)}$

9. $\displaystyle{(2a-3b)(3a+4ab+b)}$

10. $\displaystyle{(2-x+3y)(2-x-3y)(4x^3-3+2x) + (5x-3)^3}$

1. ${\displaystyle{8x^3-2x^2y-5xy^2+2y^3}}$

2. ${\displaystyle{8xy^4 - 9xy^2 + 4x^2 + 2y - 7}}$

3. ${\displaystyle{-6x^3-4x^2y+4xy^2-2y^3}}$

4. ${\displaystyle{21x^2-2xy-3y^2}}$

5. ${\displaystyle{4x^6 - 20x^3 y + 25y^2}}$

6. ${\displaystyle{x^3 + t^3}}$

7. ${\displaystyle{125a^3 + 300a^2b + 240ab^2 + 64b^3}}$

8. ${\displaystyle{12x^2y^3-21x^3y-9xy}}$

9. ${\displaystyle{6a^2+8a^2b-7ab-12ab^2-3b^2}}$

10. ${\displaystyle{4 x^5-16 x^4-36 x^3 y^2+143 x^3-236 x^2-18 x y^2+155 x+27 y^2-39}}$