The whole of algebra in one set of exercises: collecting like terms, fractional exponents, removing brackets, logarithms both ways, algebraic fractions, polynomial division and factoring.

This lesson brings no new theory: it is a set of exercises on everything from lessons 20-24. It is worth working through all of them, because every later topic in this course - equations, functions, derivatives - rests on these manipulations. If one type of exercise gives you trouble, go back to the lesson it comes from now rather than meeting the same gap three chapters later.

Algebra - review exercises

Simplify each of the following expressions.

\begin{aligned}\text{a)}\quad 2ab - 4ac + ba - 2cb + 3ba &= 6ab - 4ac - 2cb = \\&= 2(3ab - 2ac - cb)\end{aligned}

\begin{aligned}\text{b)}\quad 3x^{2}yz - zx^{2}y + 4yxz^{2} - 2x^{2}zy + 3z^{2}yx - 3zy^{2} &= \\= 0\cdot x^{2}yz + 7z^{2}yx - 3zy^{2} &= 7z^{2}yx - 3zy^{2} = \\&= yz(7xz - 3y)\end{aligned}

\text{c)}\quad c^{p}\cdot c^{-q} : c^{-2} = c^{\,p + (-q) - (-2)} = c^{\,p - q + 2}

\text{d)}\quad \frac{\left(x^{\frac{1}{2}}\right)^{-\frac{2}{3}} : \left(y^{\frac{3}{4}}\right)^{2}\cdot\left(x^{\frac{3}{5}}\right)^{-\frac{5}{3}}}{\left(x^{\frac{1}{4}}\right)^{-1}\cdot\left(y^{\frac{1}{3}}\right)^{6}} = \frac{x^{-\frac{1}{3}} : y^{\frac{3}{2}}\cdot x^{-1}}{x^{-\frac{1}{4}}\cdot y^{2}} = x^{-\frac{13}{12}}\cdot y^{-\frac{7}{2}}

-\tfrac{1}{3} + (-1) - \left(-\tfrac{1}{4}\right) = -\tfrac{4}{3} + \tfrac{1}{4} = -\tfrac{16}{12} + \tfrac{3}{12} = -\tfrac{13}{12}

-\tfrac{3}{2} - 2 = -\tfrac{7}{2}

Remove the brackets from each of the following expressions.

\begin{aligned}\text{a)}\quad 2f(3g - 4h)(g + 2h) &= 2f(3g^{2} + 6gh - 4gh - 8h^{2}) = \\&= 2f(3g^{2} + 2gh - 8h^{2}) = \\&= 6fg^{2} + 4fgh - 16fh^{2}\end{aligned}

\begin{aligned}\text{b)}\quad (5x - 6y)(2x + 6y)(5x - y) &= (10x^{2} + 30xy - 12xy - 36y^{2})(5x - y) = \\&= (10x^{2} + 18xy - 36y^{2})(5x - y) = \\&= 50x^{3} - 10x^{2}y + 90x^{2}y - 18xy^{2} - 180xy^{2} + 36y^{3} = \\&= 50x^{3} + 80x^{2}y - 198xy^{2} + 36y^{3}\end{aligned}

\begin{aligned}\text{c)}\quad 4\{3[p - 2(q - 3)] - 3(r - 2)\} &= 4\{3[p - 2q + 6] - 3r + 6\} = \\&= 4\{3p - 6q + 18 - 3r + 6\} = \\&= 4(3p - 6q + 24 - 3r) = \\&= 12p - 24q - 12r + 96\end{aligned}

Evaluate using a calculator or by applying the change-of-base formula where necessary.

\text{a)}\quad \log 0.0270 \approx -1.569

\text{b)}\quad \ln 47.89 \approx 3.869

\text{c)}\quad \log_{7}126.4 = \log_{7}10\cdot\log_{10}126.4 = \frac{1}{\log_{10}7}\cdot\log_{10}126.4 \approx \frac{2.102}{0.845} \approx 2.487

Write the following expressions in logarithmic form.

\begin{aligned}\text{a)}\quad V &= \frac{\pi h}{4}(D - h)(D + h) \\\log V &= \log \pi + \log h - \log 4 + \log(D - h) + \log(D + h)\end{aligned}

\begin{aligned}\text{b)}\quad P &= \tfrac{1}{16}(2d - 1)^{2}N\sqrt{S} = 2^{-4}(2d - 1)^{2}N\sqrt{S} \\\log P &= \log 2^{-4} + \log(2d - 1)^{2} + \log N + \log\sqrt{S} \\\log P &= -4\log 2 + 2\log(2d - 1) + \log N + \tfrac{1}{2}\log S\end{aligned}

Rewrite the following expressions without using logarithms.

\begin{aligned}\text{a)}\quad \log x &= \log P + 2\log Q - \log k - 3 \\3 &= \log P + \log Q^{2} - \log k - \log x = \log\frac{PQ^{2}}{xk} \\\frac{PQ^{2}}{xk} &= 10^{3} \quad\Rightarrow\quad PQ^{2} = 1000xk \quad\Rightarrow\quad x = \frac{PQ^{2}}{1000k}\end{aligned}

\begin{aligned}\text{b)}\quad \log R &= 1 + \tfrac{1}{3}\log M + 3\log S \\1 &= \log R - \log\sqrt[3]{M} - \log S^{3} = \log\frac{R}{S^{3}\sqrt[3]{M}} \\\frac{R}{S^{3}\sqrt[3]{M}} &= 10 \quad\Rightarrow\quad R = 10S^{3}\sqrt[3]{M}\end{aligned}

\begin{aligned}\text{c)}\quad \ln P &= \tfrac{1}{2}\ln(Q + 1) - 3\ln R + 2 \\2 &= \ln P - \ln\sqrt{Q + 1} + \ln R^{3} = \ln\frac{PR^{3}}{\sqrt{Q + 1}} \\\frac{PR^{3}}{\sqrt{Q + 1}} &= e^{2} \quad\Rightarrow\quad PR^{3} = e^{2}\sqrt{Q + 1} \quad\Rightarrow\quad P = \frac{e^{2}\sqrt{Q + 1}}{R^{3}}\end{aligned}

Multiply and simplify the resulting expressions.

\begin{aligned}\text{a)}\quad (3x - 1)(x^{2} - x - 1) &= 3x^{3} - 3x^{2} - 3x - x^{2} + x + 1 = \\&= 3x^{3} - 4x^{2} - 2x + 1\end{aligned}

\begin{aligned}\text{b)}\quad (a^{2} + 2a + 2)(3a^{2} + 4a + 4) &= 3a^{4} + 4a^{3} + 4a^{2} + 6a^{3} + 8a^{2} + 8a + 6a^{2} + 8a + 8 = \\&= 3a^{4} + 10a^{3} + 18a^{2} + 16a + 8\end{aligned}

Simplify each of the following expressions.

\text{a)}\quad \frac{x^{3}}{y^{2}} : \frac{x}{y^{3}} = \frac{x^{3}}{y^{2}}\cdot\frac{y^{3}}{x} = \frac{x^{3}y^{3}}{xy^{2}} = x^{2}y

\text{b)}\quad \frac{ab}{c} : \frac{ac}{b}\cdot\frac{bc}{a} = \frac{ab}{c}\cdot\frac{b}{ac}\cdot\frac{bc}{a} = \frac{ab^{3}c}{a^{2}c^{2}} = \frac{b^{3}}{ac}

Perform the division.

(x^{2} + 2x - 3) : (x - 1) = x + 3

$x^{2}$	$- x$
0	$3x$	$- 3$
	$3x$	$- 3$
		$0$

(x - 1)(x + 3) = x^{2} + 2x - 3

(n^{3} + 0n^{2} + 0n + 27) : (n + 3) = n^{2} - 3n + 9

$n^{3}$	$+ 3n^{2}$
0	$- 3n^{2}$	$+ 0n$
	$- 3n^{2}$	$- 9n$
	0	$9n$	$+ 27$
		$9n$	$+ 27$
			$0$

(n + 3)(n^{2} - 3n + 9) = n^{3} + 0n^{2} + 0n + 27

(3a^{3} + 2a^{2} + 0a + 1) : (a + 1) = 3a^{2} - a + 1

$3a^{3}$	$+ 3a^{2}$
0	$- a^{2}$	$+ 0a$
	$- a^{2}$	$- a$
	0	$a$	$+ 1$
		$a$	$+ 1$
			$0$

(a + 1)(3a^{2} - a + 1) = 3a^{3} + 2a^{2} + 0a + 1

Factor the following expressions.

\text{a)}\quad 36x^{3}y^{2} - 8x^{2}y = 4x^{2}y(9xy - 2)

\begin{aligned}\text{b)}\quad x^{3} + 3x^{2}y + 2xy^{2} + 6y^{3} &= x^{2}(x + 3y) + 2y^{2}(x + 3y) = \\&= (x + 3y)(x^{2} + 2y^{2})\end{aligned}

\text{c)}\quad 4x^{2} + 12x + 9 = (2x + 3)^{2}

\begin{aligned}\text{d)}\quad (3x + 4y)^{2} - (2x - y)^{2} &= [3x + 4y - (2x - y)][3x + 4y + (2x - y)] = \\&= (3x + 4y - 2x + y)(3x + 4y + 2x - y) = \\&= (x + 5y)(5x + 3y)\end{aligned}

\text{e)}\quad x^{2} + 10x + 24 = (x + 6)(x + 4)

\text{f)}\quad x^{2} - 10x + 16 = (x - 8)(x - 2)

\text{g)}\quad x^{2} - 5x - 36 = (x + 4)(x - 9)

\begin{aligned}\text{h)}\quad 6x^{2} + 5x - 6 &= 6x^{2} + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = \\&= (2x + 3)(3x - 2)\end{aligned}

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Frequently asked questions

How do you multiply and divide powers with the same base?

Add the exponents when multiplying and subtract them when dividing, whether they are whole, negative or fractional. Hence c^p · c^(−q) : c^(−2) = c^(p − q + 2).

How do you evaluate a logarithm whose base is neither 10 nor e?

With the change-of-base formula: log₇126,4 = log126,4 / log7 = 2.102 / 0.845 ≈ 2.487. The base of the fraction can be anything, as long as it is the same above and below.

What does it mean to rewrite an expression without logarithms?

Collapse the right-hand side into a single logarithm and then drop it on both sides. From log x = log P + 2log Q − log k − 3 you get log(PQ²/(xk)) = 3, so PQ² = 1000xk, and hence x = PQ²/(1000k).

How do you check a polynomial division?

Multiply the divisor by the quotient - the dividend must come back. For (n³ + 27) : (n + 3) = n² − 3n + 9 we have (n + 3)(n² − 3n + 9) = n³ + 27, so the division is right.

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Factoring algebraic expressions

Evaluating expressions and rearranging formulae