Algebraic Simplification Techniques#

Simplifying Expressions Involving Fractions#

Adding or Subtracting Fractions

If the fractions have a common denominator, then they can be combined into a single fraction by adding or subtracting the numerators accordingly.

\[ \frac{A}{B} \pm \frac{C}{B} = \frac{A \pm C}{B} \]

e.g., \(\dfrac{2}{3} + \dfrac{5}{3} = \dfrac{7}{3}\)

If the fractions do not have a common denominator, then a common denominator can be formed by multiplying the denominators together.

\[ \frac{A}{B} \pm \frac{C}{D} = \frac{AD}{BD} \pm \frac{BC}{BD} = \frac{AD \pm BC}{BD} \]

e.g., \(\dfrac{2}{3} - \dfrac{5}{7} = \dfrac{2\cdot 7- 3\cdot5}{3\cdot 7} = -\dfrac{1}{21}\)

Multiplying Fractions

To multiply two fractions, multiply the numerators and multiply the denominators of the two fractions.

\[ \frac{A}{B} \times \frac{C}{D} = \frac{AC}{BD} \]

e.g., \(\dfrac{2}{3} \times \dfrac{5}{7} = \dfrac{2\cdot 5}{3\cdot 7} = \dfrac{10}{21}\)

Dividing Fractions

To divide two fractions, remember that division is the same as multiplication by the reciprocal. In other words, dividing by \(C/D\) is the same as multiplying by \(D/C\).

\[\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} = \frac{AD}{BC}\]

e.g., \(\dfrac{2}{3} \div \dfrac{5}{7} = \dfrac{2}{3} \times \dfrac{7}{5} = \dfrac{14}{15}\)

Example 1#

Subtracting fractions with the same denominator

Rewrite \(\dfrac{x+1}{x+4} - \dfrac{3x-2}{x+4}\) as a single ratio.

Step 1:  Subtract fractions with common denominator.

Since the ratios already have the same denominator, we need only apply the formula for subtracting fractions with a common denominator.

\[\begin{align*} \frac{A}{B} - \frac{C}{B} &= \frac{A-C}{B} \end{align*}\]
Step 2:  Apply the formula from Step 1.

Apply the formula from Step 1 with \(A = x+1\), \(B=x+4\), and \(C = 3x-2\).

\[\begin{align*} \dfrac{x+1}{x+4} - \dfrac{3x-2}{x+4} &= \frac{x+1 - (3x-2)}{x+4} && \text{subtract numerators}\\ &= \frac{x+1 - 3x+2}{x+4} && \text{distribute the minus sign}\\ &= \frac{3-2x}{x+4} && \text{combine like terms} \end{align*}\]

Example 2#

Adding fractions with different denominators

Rewrite \(\dfrac{3}{x} + \dfrac{4}{x-5}\) as a single ratio.

Step 1:   Apply formula for adding fractions with different denominators.

Since the ratios do not have the same denominator, we will apply the formula for adding fractions with different denominators.

\[\begin{align*} \frac{A}{B} + \frac{C}{D} &= \frac{AD}{BD} + \frac{BC}{BD} = \frac{AD+BC}{BD} \end{align*}\]
Step 2:   Apply the formula from Step 1.

Apply the formula from Step 1 with \(A = 3\), \(B=x\), \(C = 4\), and \(D = x-5\).

\[\begin{align*} \dfrac{3}{x} + \dfrac{4}{x-5} &= \frac{3(x-5)}{x(x-5)} + \frac{4x}{x(x-5)} && \text{get a common denominator}\\ &= \frac{3(x-5)+4x}{x(x-5)} && \text{add numerators}\\ &= \frac{3x - 15 + 4x}{x(x-5)} && \text{distribute the 3}\\ &= \frac{7x-15}{x(x-5)} && \text{combine like terms} \end{align*}\]

Example 3#

Simplifying quotients of fractions

Simplify

\[ \cfrac{\left(\cfrac{3p}{\sqrt{180-6p}}\right)}{ \sqrt{180-6p}}. \]
Step 1:   Recall formula for the quotient of fractions.
\[ \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C} = \frac{AD}{BC} \]
Step 2:   Use formula from Step 1 to rewrite equation.
\[\begin{align*} \cfrac{\left(\cfrac{3p}{\sqrt{180-6p}}\right)}{ \sqrt{180-6p}} &= \frac{3p}{\sqrt{180-6p}} \div \sqrt{180-6p} \\ \\ &= \frac{3p}{\sqrt{180-6p}} \times \frac{1}{\sqrt{180-6p}} && \text{division is multiplication by reciprocal}\\ \\ &= \frac{3p}{\sqrt{180-6p}\times \sqrt{180-6p}} && \text{multiply numerators and denominators}\\ \\ &= \frac{3p}{180 - 6p} && \text{simplify} \end{align*}\]

Distributive Property of Multiplication#

Distributivity

The distributive property of multiplication can be used to rewrite a product (where at least one factor is a sum) as a sum.

\[ A(B+C) = AB + AC \]

e.g., \(7(3+2) = 7\cdot 3 + 7\cdot 2\)

Example 4#

Expanding a product of polynomials

Expand \(x^2(5x^3 + 7)\).

Step 1:   Use distributivity to expand given product.
\[\begin{align*} x^2(5x^3 + 7) &= 5x^2x^3 + 7x^2 && \text{distribute $x^2$}\\ &= 5x^5 + 7x^2 && \text{simplify} \end{align*}\]

The FOIL Method#

The FOIL method is a way to remember how to apply the distributive property of multiplication when expanding the product of two binomial expressions. FOIL is an acronym for

F-O-I-L

  • First (i.e., multiply the first terms from each binomial)

  • Outer (i.e., multiply the first term from the first factor and the second term from the second factor)

  • Inner (i.e., multiply the second term from the first factor and the first term from the second factor)

  • Last (i.e., multiply the second terms of each binomial)

\[ (a + b)(c + d) = \underbrace{ac}_{\text{(F)irst}} + \underbrace{ad}_{\text{(O)uter}} + \underbrace{bc}_{\text{(I)nner}} + \underbrace{bd}_{\text{(L)ast}} \]

Example 5#

Applying the FOIL method

Expand \((x+2)(3x-5)\) using the FOIL method.

Step 1:   Recall the formula for the FOIL method.
\[\begin{align*} (a + b)(c + d) &= ac + ad + bc + bd \end{align*}\]
Step 2:   Use the FOIL method to expand the expression.
\[\begin{align*} (x+2)(3x-5) &= x(3x) + x(-5) + 2(3x) + 2(-5) &&\text{FOIL}\\ &= 3x^2 - 5x + 6x - 10 && \text{simplify each term}\\ &= 3x^2 + x - 10 && \text{combine like terms} \end{align*}\]

Squaring a Binomial#

Applying the FOIL method to a binomial

When applying the FOIL method to the square of a binomial (i.e., \((a+b)^2\) or \((a-b)^2\)), we arrive at the following formulas:

\[ (a + b)^2 = a^2 + 2ab + b^2 \qquad\qquad (a - b)^2 = a^2 - 2ab + b^2 \]

Example 6#

Expanding a binomial

Expand \((3x-5)^2\) by squaring the binomial.

Step 1:   Apply binomial formula.

Apply \((a - b)^2 = a^2 - 2ab + b^2\) with \(a=3x\) and \(b=5\).

\[\begin{align*} (3x-5)^2 &= (3x)^2 - 2(3x)(5) + 5^2 && \text{square the binomial} \\ \\ &= 3^2x^2 - 30x + 25 && \text{simplify}\\ \\ &= 9x^2 - 30x + 25 \end{align*}\]

Example 7#

Expanding a product of a monomial and a binomial

Expand \(3x^5(4+x)^2\).

Step 1:   Apply binomial formula.

Apply \((a + b)^2 = a^2 + 2ab + b^2\) with \(a=4\) and \(b=x\).

\[\begin{align*} 3x^5(4+x)^2 &= 3x^5(4^2 + 2(4)(x) + x^2) && \text{square the binomial}\\ \\ &= 3x^5(16 + 8x + x^2) && \text{simplify} \end{align*}\]
Step 2:   Use the distributive property of multiplication.
\[\begin{align*} 3x^5(16 + 8x + x^2) &= 3x^5(16) + 3x^5(8x) + 3x^5(x^2) && \text{distribute $3x^5$} \\ \\ &= 48x^5 + 24x^6 + 3x^7 && \text{simplify} \end{align*}\]