Brief Review of Algebra and Trigonometry for Calculus

Algebra

To keep the topic simple, in the following calculations, we neglect the question of limitations on the values of $x$ for which the statements are valid.

Exponents

Pay attention to how the given expressions are simplified using the laws of exponents.

$\frac{2 x^{5} + 3 x^{7}}{15 x^{6}} = \frac{2 x^{5}}{15 x^{6}} + \frac{3 x^{7}}{15 x^{6}} = \frac{2}{15 x} + \frac{x}{5}$
${(a^{- 7} b^{- 2} c^{3})}^{- 2} = a^{14} b^{4} c^{- 6} = \frac{a^{14} b^{4}}{c^{6}}$
${(\sqrt{a^{3}} \sqrt[3]{b^{2}})}^{- 1 ∕ 2} = \frac{1}{\sqrt{a^{3 ∕ 2} b^{2 ∕ 3}}}$ or

${(\sqrt{a^{3}} \sqrt[3]{b^{2}})}^{- 1 ∕ 2} = a^{- 3 ∕ 4} b^{- 1 ∕ 3} = \frac{1}{a^{3 ∕ 4} b^{1 ∕ 3}}$
$\frac{x^{2}}{y^{3}} (\sqrt{x} + \sqrt{y^{3}}) = \frac{x^{5 ∕ 2}}{y^{3}} + \frac{x^{2}}{y^{3 ∕ 2}}$

Operations of Algebraic Expressions

Study the operations and simplification of the algebraic expressions given below.

$(2 x + 3) (5 - 2 x) + x^{2} + 3 x (4 - \frac{1}{x})$

$= 10 x + 15 - 4 x^{2} - 6 x + x^{2} + 12 x - 3$

$= (- 4 x^{2} + x^{2}) + (10 x - 6 x + 12 x) + (15 - 3)$

$= - 3 x^{2} + 16 x + 12$
$(4 x^{2} + y^{2} z) (\frac{x}{z} + 3 z^{2}) - (x + y) (y^{2} - x z)$

$= \frac{4 x^{3}}{z} + 12 x^{2} z^{2} + x y^{2} + 3 y^{2} z^{3} - x y^{2} + x^{2} z - y^{3} + x y z$

$= \frac{4 x^{3}}{z} + 12 x^{2} z^{2} + 3 y^{2} z^{3} - y^{3} + x^{2} z + x y z$
$3 x {(4 x - z)}^{2} + \frac{8 x^{6} y - x^{5} y z + 3 x^{4} y z}{- x^{3} y}$

$= 48 x^{3} - 24 x^{2} z + 3 x z^{2} - 8 x^{3} + x^{2} z - 3 x z$

$= 40 x^{3} - 23 x^{2} z + 3 x z^{2} - 3 x z$
$\frac{5}{x + 3} + \frac{1}{x - 9} = \frac{5 (x - 9) + (x + 3)}{(x + 3) (x - 9)} = \frac{6 x - 42}{x^{2} - 6 x - 27}$

Factorization

We will be factoring algebraic expressions continuously in this course, so pay attention to the following examples.

Similar Terms

$\begin{array}{l} x^{3} - 5 x^{2} - 4 x + 20 + 10 x & do the operations of similar terms \\ = (x^{3} - 5 x^{2} + 6 x) + 20 & group similar terms \\ = x (x^{2} - 5 x + 6) + 20 & factor the common term with the lowest exponent \end{array}$

Conjugates

The binomial terms $(a + b)$ and $(a - b)$ are conjugates of each other. Observe that their product

$(a - b) (a + b) = a^{2} - b^{2}$

is a difference of squares. So, for example,

$\begin{array}{l} 4 x^{2} y^{4} - 6 z^{6} & = {(2 x y^{2})}^{2} - {(\sqrt{6} z^{3})}^{2} \\ = (2 x y^{2} + \sqrt{6} z^{3}) (2 x y^{2} - \sqrt{6} z^{3}) . \end{array}$

Quadratics

A quadratic is an expression of the form $a x^{2} + b x + c$ with $b$ and $c$ any real numbers and $a$ a nonzero real number.

A quadratic is a perfect square if $b = \pm 2 \sqrt{a c}$ . If so, then

$a x^{2} + b x + c = {(\sqrt{a} x \pm \sqrt{c})}^{2} .$

The quadratic $4 x^{2} - 20 x + 25$ is a perfect square because

$a = 4, c = 25 and b = - 2 \sqrt{4 (25)} = - 2 \sqrt{100} = - 20 .$

Then,

$4 x^{2} - 20 x + 25 = {(2 x - 5)}^{2} .$

Similarly, $4 t^{2} - 12 t + 9$ is a perfect square, and

$4 t^{2} - 12 t + 9 = {(2 t - 3)}^{2} .$
If we have a quadratic expression of the form $x^{2} + b x + c$ , and we can find two numbers $k$ and $m$ such that $c = k m$ and $b = k + m$ , then

$x^{2} + b x + c = (x + k) (x + m) .$

In the quadratic equation $x^{2} - 2 x - 35$ , we see that $(- 7) (5) = - 35$ and $- 7 + 5 = - 2;$ hence, $x^{2} - 2 x - 35 = (x - 7) (x + 5) .$
The solutions of the quadratic equation $a x^{2} + b x + c = 0$ is given by the “quadratic formula”:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} .$

Using the quadratic formula, we find that the solutions of the equation $32 y^{2} + 4 y - 6 = 0$ are

$y = \frac{- 4 \pm \sqrt{16 + 768}}{64} = \frac{- 4 \pm 28}{64};$

that is, $y = - \frac{1}{2}$ and $y = \frac{3}{8}$ . Moreover,

$(y + \frac{1}{2}) (y - \frac{3}{8}) = y^{2} + \frac{y}{8} - \frac{3}{16}$

and we see that

$32 (y + \frac{1}{2}) (y - \frac{3}{8}) = 32 y^{2} + 4 y - 6 .$

Verify the following factorization:

$32 y^{2} + 4 y - 6 = 2 (2 (y + \frac{1}{2})) (8 (y - \frac{3}{8})) = 2 (2 y + 1) (8 y - 3) .$

Let us try another example: the solutions of $- 8 x^{2} + 10 x + 3 = 0$ are

$x = \frac{- 10 \pm \sqrt{100 + 96}}{- 16} = \frac{- 10 \pm 14}{- 16};$

hence, $x = \frac{3}{2}$ and $x = - \frac{1}{4}$ .

Since

$(x - \frac{3}{2}) (x + \frac{1}{4}) = x^{2} - \frac{5 x}{4} - \frac{3}{8},$

we have

$\begin{aligned} - 8 x^{2} + 10 x + 3 & = - 8 (x - \frac{3}{2}) (x + \frac{1}{4}) \\ = - (2 (x - \frac{3}{2}) 4 (x + \frac{1}{4})) \\ = - (2 x - 3) (4 x + 1) . \end{aligned}$

Other Factorizations

See the sections titled “Factoring Special Polynomials” and “Binomial Theorem” on page 1 of the “Reference Pages” at the beginning of the textbook. To apply these factorizations, identify the corresponding terms of $x$ and $y$ . For example, consider the following difference of cubes.

The expression $(8 a^{3} - 27 z^{6})$ is a difference of cubes where $x = 2 a$ and $y = 3 z^{2}$ ; hence,

$(8 a^{3} - 27 z^{6}) = (2 a - 3 z^{2}) (4 a^{2} + 6 a z^{2} + 9 z^{4}) .$

Rational Expressions

To simplify rational expressions, we factor and cancel equal terms.

$\frac{5 x + 35}{20 x} = \frac{5 (x + 7)}{5 (4 x)} = \frac{x + 7}{4 x}$
$\frac{x^{3} + x^{2}}{x + 1} = \frac{x^{2} (x + 1)}{x + 1} = x^{2}$
$\frac{x^{2} + 6 x + 5}{x^{2} - 25} = \frac{(x + 5) (x + 1)}{(x + 5) (x - 5)} = \frac{x + 1}{x - 5}$
$\frac{x^{3} + 5 x^{2} + 6 x}{x^{2} - x - 12} = \frac{x (x^{2} + 5 x + 6)}{(x - 4) (x + 3)} = \frac{x (x + 2) (x + 3)}{(x - 4) (x + 3)} = \frac{x (x + 2)}{x - 4}$

We can also simplify rational expressions by long division.

For example, to simplify $\frac{7 x - 9 - 4 x^{2} + 4 x^{3}}{2 x - 1},$ we divide:

\begin{array}{r} \require{enclose} 2x - 1 \enclose{longdiv}{4x^3 - 4x^2 + 7x - 9} \\[-3pt] \end{array}

arrange terms in the dividend and divisor in decreasing order of their exponents

\begin{array}{r} \require{enclose} {2x^2}\hspace{4.1em} \\[-3pt] 2x - 1 \enclose{longdiv}{4x^3 - 4x^2 + 7x - 9} \\[-3pt] \end{array}

divide ${\displaystyle\frac{4x^3}{2x}} = 2x^2$

\begin{array}{r} \require{enclose} {2x^2}\hspace{4.1em} \\[-3pt] 2x - 1 \enclose{longdiv}{4x^3 - 4x^2 + 7x - 9} \\[-3pt] {4x^3 - 2x^2}\hspace{4.01em} \\[-3pt] \end{array}

multiply $2x^2(2x - 1) = 4x^3 - 2x^2$

\begin{array}{r} \require{enclose} {2x^2}\hspace{4.1em} \\[-3pt] 2x - 1 \enclose{longdiv}{\phantom{-}4x^3 - 4x^2 + 7x - 9} \\[-3pt] \underline{- 4x^3 + 2x^2}\hspace{4em} \\[-3pt] {\phantom{- 4x^3 } - 2x^2 + 7x - 9} \\[-3pt] \end{array}

subtract $4x^3 - 4x^2 - (4x^3 - 2x^2)$
$= 4x^3 - 4x^2 - 4x^3 + 2x^2$
$= -2x^2$;
bring down $7x - 9$;
the new dividend is $-2x^2 + 7x -9$

\begin{array}{r} \require{enclose} 2x^2 - \phantom{0}x\hspace{1.7em} \\[-3pt] 2x - 1 \enclose{longdiv}{\phantom{-}4x^3 - 4x^2 + 7x - 9} \\[-3pt] \underline{- 4x^3 + 2x^2}\hspace{4em}\\[-3pt] {\phantom{- 4x^3 } - 2x^2 + 7x - 9}\\[-3pt] {\phantom{- 4x^3 } - 2x^2 + \phantom{7}x \phantom{- 9}}\hspace{0.4em} \\ \end{array}

we repeat the process:
divide ${\displaystyle\frac{-2x^2}{2x}} = -x$
then multiply

\begin{array}{r} \require{enclose} 2x^2 - \phantom{0}x\hspace{1.7em} \\[-3pt] 2x - 1 \enclose{longdiv}{\phantom{-}4x^3 - 4x^2 + 7x - 9} \\[-3pt] \underline{- 4x^3 + 2x^2}\hspace{4em} \\[-3pt] {\phantom{- 4x^3 } - 2x^2 + 7x - 9} \\[-3pt] {\phantom{4x^3} \underline{ \phantom{{}-{}} 2x^2 - \phantom{7}x \phantom{{}-{}} \phantom{9} } } \\[-3pt] {\phantom{4x^3 - 4x^2 + } 6x - 9} \\ \end{array}

subtract: note that
$-(-2x^2+x)=2x^2-x$;
the new dividend is $6x-9$

\begin{array}{r} \require{enclose} 2x^2 - \phantom{0}x + 3 \\[-3pt] 2x - 1 \enclose{longdiv}{\phantom{-}4x^3 - 4x^2 + 7x - 9} \\[-3pt] \underline{- 4x^3 + 2x^2}\hspace{4em} \\[-3pt] {\phantom{- 4x^3 } - 2x^2 + 7x - 9} \\[-3pt] {\phantom{4x^3} \underline{ \phantom{{}-{}} 2x^2 - \phantom{7}x \phantom{{}-{}} \phantom{9} } } \\[-3pt] {\phantom{4x^3 - 4x^2 + } 6x - 9} \\[-3pt] {\phantom{4x^3 - 4x^2 } \underline{ - 6x + 3}} \\[-3pt] {\phantom{4x^3 - 4x^2 - 6x } - 6} \\ \end{array}

repeating the process

The remainder is $- 6$ ; hence,

$\frac{7 x - 9 - 4 x^{2} + 4 x^{3}}{2 x - 1} = 2 x^{2} - x + 3 - \frac{6}{2 x - 1} .$

Rationalization

To change quotient expressions with square roots, we rationalize; that is, we multiply by the conjugate to change the expression into a different, more manageable one.

For example, to simplify

$\frac{x^{2} - 81}{\sqrt{x} - 3},$

it is helpful to realize that the conjugate of $\sqrt{x} - 3$ is $\sqrt{x} + 3$ , and $(\sqrt{x} - 3) (\sqrt{x} + 3) = x - 9$ . So,

\begin{align*}\frac{x^2 - 81}{\sqrt{x} - 3} & = \frac{(x - 9)(x + 9)(\sqrt{x} + 3)}{x - 9}\\ & = (x + 9)(\sqrt{x} + 3) \end{align*}

Similarly,

\begin{align*} \frac{\sqrt{h + 1} - 1}{h} & = \frac{(\sqrt{h + 1} - 1)(\sqrt{h + 1} + 1)}{h(\sqrt{h + 1} + 1)}\\ & = \frac{1 + h -1}{h(\sqrt{1 + h} + 1)}\\ & = \frac{1}{\sqrt{1 + h} + 1} \end{align*}