Tasks 01-03 - Core Algebra & Exponents

Building Your Algebraic Foundation

Problem 1: Order of Operations

Evaluate the following expressions. Show each step clearly.

\(5 + 3^2 \times 4 - 18 \div 3\)
\(2 \times [6 + 3 \times (4^2 - 14)]\)
\(\frac{3^3 - 2 \times 4}{5 - 2}\)
\((2 + 3)^3 - 5 \times (12 - 2 \times 4)\)
\(\frac{4^2 + 2 \times 3}{2} - 3 \times (5 - 3)^2\)

Solution

\(5 + 3^2 \times 4 - 18 \div 3\)
- \(= 5 + 9 \times 4 - 18 \div 3\)
- \(= 5 + 36 - 6\)
- \(= 35\)
\(2 \times [6 + 3 \times (4^2 - 14)]\)
- \(= 2 \times [6 + 3 \times (16 - 14)]\)
- \(= 2 \times [6 + 3 \times 2]\)
- \(= 2 \times [6 + 6]\)
- \(= 2 \times 12 = 24\)
\(\frac{3^3 - 2 \times 4}{5 - 2}\)
- \(= \frac{27 - 8}{3}\)
- \(= \frac{19}{3}\)
\((2 + 3)^3 - 5 \times (12 - 2 \times 4)\)
- \(= 5^3 - 5 \times (12 - 8)\)
- \(= 125 - 5 \times 4\)
- \(= 125 - 20 = 105\)
\(\frac{4^2 + 2 \times 3}{2} - 3 \times (5 - 3)^2\)
- \(= \frac{16 + 6}{2} - 3 \times 2^2\)
- \(= \frac{22}{2} - 3 \times 4\)
- \(= 11 - 12 = -1\)

Simplify the following expressions completely. Express answers with positive exponents only.

\((x^3)^2 \cdot x^{-4} \cdot x\)
\(\frac{(2a^3b^2)^3}{8a^5b^4}\)
\(\left(\frac{3x^2y}{z^2}\right)^2 \cdot \frac{z^3}{9xy^2}\)
\((5^3)^2 \cdot 5^{-4} \div 5^2\)
\(\frac{(m^2n^3)^2 \cdot m^{-3}n}{m^{-1}n^4}\)
\(\left(\frac{2x^{-2}y^3}{z^{-1}}\right)^{-2}\)

Solution

\((x^3)^2 \cdot x^{-4} \cdot x\)
- \(= x^6 \cdot x^{-4} \cdot x^1\)
- \(= x^{6-4+1} = x^3\)
\(\frac{(2a^3b^2)^3}{8a^5b^4}\)
- \(= \frac{8a^9b^6}{8a^5b^4}\)
- \(= a^{9-5}b^{6-4}\)
- \(= a^4b^2\)
\(\left(\frac{3x^2y}{z^2}\right)^2 \cdot \frac{z^3}{9xy^2}\)
- \(= \frac{9x^4y^2}{z^4} \cdot \frac{z^3}{9xy^2}\)
- \(= \frac{9x^4y^2 \cdot z^3}{z^4 \cdot 9xy^2}\)
- \(= \frac{x^3}{z}\)
\((5^3)^2 \cdot 5^{-4} \div 5^2\)
- \(= 5^6 \cdot 5^{-4} \cdot 5^{-2}\)
- \(= 5^{6-4-2} = 5^0 = 1\)
\(\frac{(m^2n^3)^2 \cdot m^{-3}n}{m^{-1}n^4}\)
- \(= \frac{m^4n^6 \cdot m^{-3}n}{m^{-1}n^4}\)
- \(= \frac{m^1n^7}{m^{-1}n^4}\)
- \(= m^{1-(-1)}n^{7-4} = m^2n^3\)
\(\left(\frac{2x^{-2}y^3}{z^{-1}}\right)^{-2}\)
- \(= \left(\frac{2y^3z}{x^2}\right)^{-2}\)
- \(= \frac{x^4}{4y^6z^2}\)

Part A: Conversions

Convert to scientific notation:

Convert to standard form:

Part B: Calculations

Perform the following calculations and express answers in scientific notation:

Solution

Part A:

Part B:

\((2.5 \times 10^6) \times (3.2 \times 10^{-3})\)
- \(= 8.0 \times 10^3\)
\(\frac{9.6 \times 10^8}{3.2 \times 10^5}\)
- \(= 3.0 \times 10^3\)
\((4.0 \times 10^4)^2\)
- \(= 16 \times 10^8 = 1.6 \times 10^9\)
\((3.6 \times 10^9) + (2.4 \times 10^8)\)
- \(= (36 \times 10^8) + (2.4 \times 10^8)\)
- \(= 38.4 \times 10^8 = 3.84 \times 10^9\)

Solve the following equations and inequalities. Express solutions using set notation or interval notation as appropriate.

Solution

\(|3x - 6| = 9\)
- Case 1: \(3x - 6 = 9 \Rightarrow x = 5\)
- Case 2: \(3x - 6 = -9 \Rightarrow x = -1\)
- Solution: \(x \in \{-1, 5\}\)
\(|x - 4| < 3\)
- \(-3 < x - 4 < 3\)
- \(1 < x < 7\)
- Solution: \((1, 7)\)
\(|2x + 8| \geq 6\)
- \(2x + 8 \leq -6\) OR \(2x + 8 \geq 6\)
- \(x \leq -7\) OR \(x \geq -1\)
- Solution: \((-\infty, -7] \cup [-1, \infty)\)
\(3|x - 2| - 4 = 5\)
- \(3|x - 2| = 9\)
- \(|x - 2| = 3\)
- \(x - 2 = 3\) OR \(x - 2 = -3\)
- Solution: \(x \in \{-1, 5\}\)

Factor the following expressions completely:

\(15x^3 - 10x^2 + 5x\)
\(x^2 - 144\)
\(49x^2 - 64y^2\)
\(x^2 + 16x + 64\)
\(9x^2 - 12x + 4\)
\(8x^3 - 32x\)
\(3x^2 - 75\)
\(x^2 - 18x + 81\)

Solution

\(15x^3 - 10x^2 + 5x\)
- \(= 5x(3x^2 - 2x + 1)\)
\(x^2 - 144\)
- \(= (x + 12)(x - 12)\)
\(49x^2 - 64y^2\)
- \(= (7x + 8y)(7x - 8y)\)
\(x^2 + 16x + 64\)
- \(= (x + 8)^2\)
\(9x^2 - 12x + 4\)
- \(= (3x - 2)^2\)
\(8x^3 - 32x\)
- \(= 8x(x^2 - 4)\)
- \(= 8x(x + 2)(x - 2)\)
\(3x^2 - 75\)
- \(= 3(x^2 - 25)\)
- \(= 3(x + 5)(x - 5)\)
\(x^2 - 18x + 81\)
- \(= (x - 9)^2\)

Solve or simplify as indicated:

Simplify: \(\frac{(3x^2)^3 \cdot x^{-4}}{27x^2}\)
Factor: \(25x^2 - 100\)
Express \(0.0000567\) in scientific notation
Evaluate: \(\frac{2^3 + 3 \times 2^2}{14 - 2 \times 3}\)
Simplify: \((a^{-2}b^3)^{-2} \cdot (ab^{-1})^3\)

Solution

\(\frac{(3x^2)^3 \cdot x^{-4}}{27x^2} = \frac{27x^6 \cdot x^{-4}}{27x^2} = \frac{27x^2}{27x^2} = 1\)
\(25x^2 - 100 = 25(x^2 - 4) = 25(x + 2)(x - 2)\)
\(5.67 \times 10^{-5}\)
\(\frac{8 + 12}{8} = \frac{20}{8} = \frac{5}{2}\)
\(a^4b^{-6} \cdot a^3b^{-3} = a^7b^{-9} = \frac{a^7}{b^9}\)