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2.1 Algebraic expressions

From Förberedande kurs i matematik 1

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Theory

Exercises

Contents:

Distributive law
Expansion and factorisation
Difference of two squares
Rational expressions

Learning outcomes:

After this section you will have learned how to:

Simplify complicated algebraic expressions.
Factorise expressions, including perfect squares and the difference of two squares.
Expand expressions, including perfect squares and the difference of two squares.

Distributive Law

The distributive law specifies how to multiply a bracketed expression by a factor.

[Image]

Example 1

\displaystyle 4(x+y) = 4x + 4y
\displaystyle 2(a-b) = 2a -2b
\displaystyle x \left(\frac{1}{x} + \frac{1}{x^2} \right) = x\times \frac{1}{x} + x \times \frac{1}{x^2} = \frac{\not{x}}{\not{x}} + \frac{\not{x}}{x^{\not{2}}} = 1 + \frac{1}{x}
\displaystyle a(x+y+z) = ax + ay + az

Using the distributive law we can also see how to tackle a minus sign in front of a bracketed expression. The rule says that a minus sign in front of a bracket can be eliminated if all the terms inside the brackets switch signs.

Example 2

\displaystyle -(x+y) = (-1) \times (x+y) = (-1)x + (-1)y = -x-y
\displaystyle -(x^2-x) = (-1) \times (x^2-x) = (-1)x^2 -(-1)x = -x^2 +x
where we have in the final step used \displaystyle -(-1)x = (-1)(-1)x = 1\times x = x\,\mbox{.}
\displaystyle -(x+y-y^3) = (-1)\times (x+y-y^3) = (-1)\times x + (-1) \times y -(-1)\times y^3
\displaystyle \phantom{-(x+y-y^3)}{} = -x-y+y^3
\displaystyle x^2 - 2x -(3x+2) = x^2 -2x -3x-2 = x^2 -(2+3)x -2
\displaystyle \phantom{x^2-2x-(3x+2)}{} = x^2 -5x -2

Applying the distributive law this way round - converting a product of factors into a sum of terms - is called expanding. If the distributive law is applied in reverse we say we “factorise” the expression. We usually want to factorise as thoroughly as possible, by identifying the highet factor shared by all the terms.

Example 3

\displaystyle 3x +9y = 3x + 3\times 3y = 3(x+3y)
\displaystyle xy + y^2 = xy + y\times y = y(x+y)
\displaystyle 2x^2 -4x = 2x\times x - 2\times 2\times x = 2x(x-2)
\displaystyle \frac{y-x}{x-y} = \frac{-(x-y)}{x-y} = \frac{-1}{1} = -1

Squaring

On occasions the distributive law has to be used repeatedly to deal with larger expressions. If we consider

\displaystyle (a+b)(c+d)

and regard \displaystyle a+b as a factor that multiplies the bracketed expression \displaystyle (c+d) we get

\displaystyle \eqalign{

 \bbox[#AAEEFF,0pt]{\phantom{(a+b)}}\,(c+d)
   &= \bbox[#AAEEFF,0pt]{\phantom{(a+b)}}\,c
      + \bbox[#AAEEFF,0pt]{\phantom{(a+b)}}\,d\mbox{,}\cr
 (a+b)\,(c+d)
   &= (a+b)\,c + (a+b)\,d\mbox{.}}

Then the \displaystyle c and the \displaystyle d are multiplied into their respective brackets,

\displaystyle (a+b)c + (a+b)d = ac + bc + ad + bd \, \mbox{.}

A mnemonic for this formula is:

[Image]

Example 4

\displaystyle (x+1)(x-2) = x\times x + x \times (-2) + 1 \times x + 1 \times (-2) = x^2 -2x+x-2
\displaystyle \phantom{(x+1)(x-2)}{}=x^2 -x-2
\displaystyle 3(x-y)(2x+1) = 3(x\times 2x + x\times 1 - y \times 2x - y \times 1) = 3(2x^2 +x-2xy-y)
\displaystyle \phantom{3(x-y)(2x+1)}{}=6x^2 +3x-6xy-3y
\displaystyle (1-x)(2-x) = 1\times 2 + 1 \times (-x) -x\times 2 - x\times (-x) = 2-x-2x+x^2
\displaystyle \phantom{(1-x)(2-x)}{}=2-3x+x^2 where we have used \displaystyle -x\times (-x) = (-1)x \times (-1)x = (-1)^2 x^2 = 1\times x^2 = x^2.

Two important special cases of the above formula are when \displaystyle a+b and \displaystyle c+d are the same expression

\displaystyle (a+b)^2 = a^2 +2ab + b^2

\displaystyle (a-b)^2 = a^2 -2ab + b^2

Example 5

\displaystyle (x+2)^2 = x^2 + 2\times 2x+ 2^2 = x^2 +4x +4
\displaystyle (-x+3)^2 = (-x)^2 + 2\times 3(-x) + 3^2 = x^2 -6x +9

where \displaystyle (-x)^2 = ((-1)x)^2 = (-1)^2 x^2 = 1 \times x^2 = x^2\,\mbox{.}
\displaystyle (x^2 -4)^2 = (x^2)^2 - 2 \times 4x^2 + 4^2 = x^4 -8x^2 +16
\displaystyle (x+1)^2 - (x-1)^2 = (x^2 +2x +1)- (x^2-2x+1)
\displaystyle \phantom{(x+1)^2-(x-1)^2}{}= x^2 +2x +1 -x^2 + 2x-1
\displaystyle \phantom{(x+1)^2-(x-1)^2}{} = 2x+2x = 4x
\displaystyle (2x+4)(x+2) = 2(x+2)(x+2) = 2(x+2)^2 = 2(x^2 + 4x+ 4)
\displaystyle \phantom{(2x+4)(x+2)}{}=2x^2 + 8x + 8
\displaystyle (x-2)^3 = (x-2)(x-2)^2 = (x-2)(x^2-4x+4)
\displaystyle \phantom{(x-2)^3}{}=x \times x^2 + x\times (-4x) + x\times 4 - 2\times x^2 - 2 \times (-4x)-2 \times 4
\displaystyle \phantom{(x-2)^3}{}=x^3 -4x^2 + 4x-2x^2 +8x -8 = x^3-6x^2 + 12x -8

These rules are also used in the reverse direction to factorise expressions.

Example 6

\displaystyle x^2 + 2x+ 1 = (x+1)^2
\displaystyle x^6-4x^3 +4 = (x^3)^2 - 2\times 2x^3 +2^2 = (x^3-2)^2
\displaystyle x^2 +x + \frac{1}{4} = x^2 + 2\times\frac{1}{2}x + \bigl(\frac{1}{2}\bigr)^2 = \bigl(x+\frac{1}{2}\bigr)^2

Difference of two squares

A third special case of the first formula in the last section is the difference of two squares rule.

Difference of two squares:

\displaystyle (a+b)(a-b) = a^2 -b^2

This formula can be obtained directly by expanding the left hand side

\displaystyle (a+b)(a-b)

 = a \cdot a + a\cdot (-b) + b\cdot a + b \cdot (-b)
 = a^2 -ab+ab-b^2
 = a^2 -b^2\mbox{.}

Example 7

\displaystyle (x-4y)(x+4y) = x^2 -(4y)^2 = x^2 -16y^2
\displaystyle (x^2+2x)(x^2-2x)= (x^2)^2 - (2x)^2 = x^4 -4x^2
\displaystyle (y+3)(3-y)= (3+y)(3-y) = 3^2 -y^2 = 9-y^2
\displaystyle x^4 -16 = (x^2)^2 -4^2 = (x^2+4)(x^2-4) = (x^2+4)(x^2-2^2)
\displaystyle \phantom{x^4-16}{}=(x^2+4)(x+2)(x-2)

Rational expressions

Working with fractions containing algebraic expressions is very similar to carrying out ordinary calculations with fractions.

Multiplication and division of algebraic fractions follow the same rules that apply to ordinary fractions,

\displaystyle \frac{a}{b} \cdot \frac{c}{d}

 = \frac{a\, c}{b\, d}
 \quad \mbox{and} \quad
 \frac{\displaystyle\ \frac{a}{b}\ }{\displaystyle\frac{c}{d}}
 = \frac{a\, d}{b\, c} \; \mbox{.}

Example 8

\displaystyle \frac{3x}{x-y} \times \frac{4x}{2x+y} = \frac{3x\times 4x}{(x-y)\times(2x+y)} = \frac{12x^2}{(x-y)(2x+y)}
\displaystyle \frac{\displaystyle \frac{a}{x}}{\displaystyle \frac{x+1}{a}} = \frac{a^2}{x(x+1)}
\displaystyle \frac{\displaystyle \frac{x}{(x+1)^2}}{\displaystyle \frac{x-2}{x-1}} = \frac{x(x-1)}{(x-2)(x+1)^2}

A fractional expression can have its numerator and denominator multiplied by the same factor

\displaystyle \frac{x+2}{x+1}

 = \frac{(x+2)(x+3)}{(x+1)(x+3)}
 = \frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4)}
 = \dots

We can cancel factors that the numerator and denominator have in common

\displaystyle \frac{(x+2)(x+3)(x+4)}{(x+1)(x+3)(x+4) }

 = \frac{(x+2)(x+4)}{(x+1)(x+4)}
 = \frac{x+2}{x+1} \mbox{.}

Example 9

\displaystyle \frac{x}{x+1} = \frac{x}{x+1} \cdot \frac{x+2}{x+2} = \frac{x(x+2)}{(x+1)(x+2)}
\displaystyle \frac{x^2 -1}{x(x^2-1)}= \frac{1}{x}
\displaystyle \frac{(x^2-y^2)(x-2)}{(x^2-4)(x+y)} = \left\{\,\text{Difference of two squares}\,\right\} = \frac{(x+y)(x-y)(x-2)}{(x+2)(x-2)(x+y)} = \frac{x-y}{x+2}

When fractional expressions are added or subtracted they may need to be placed over a common denominator.

\displaystyle \frac{1}{x} - \frac{1}{x-1}

 = \frac{1}{x} \times\frac{x-1}{x-1} - \frac{1}{x-1} \times\frac{x}{x}
 = \frac{x-1}{x(x-1)} - \frac{x}{x(x-1)}
 = \frac{x-1-x}{x(x-1)}
 = \frac{-1}{x(x-1)} \; \mbox{.}

As with ordinary fractions, it's possible to find a common denominator by simply multiplying the two denominators, but it is better to find the smallest, simplest expression that both denominators go into: the lowest common denominator or LCD.

Example 10

\displaystyle \frac{1}{x+1} + \frac{1}{x+2}\quad has \displaystyle \ \text{LCD} = (x+1)(x+2)

Convert the first term using \displaystyle (x+2) and the second term using \displaystyle (x+1)

\displaystyle \begin{align*}

   \frac{1}{x+1} + \frac{1}{x+2}
     &= \frac{x+2}{(x+1)(x+2)} + \frac{x+1}{(x+2)(x+1)}\\[4pt]
     &= \frac{x+2+x+1}{(x+1)(x+2)}
      = \frac{2x+3}{(x+1)(x+2)}\:\mbox{.}
   \end{align*}

\displaystyle \frac{1}{x} + \frac{1}{x^2}\quad has \displaystyle \ \text{LCD} = x^2

We only need to convert the first term to get a common denominator
\displaystyle \frac{1}{x} + \frac{1}{x^2}
= \frac{x}{x^2} + \frac{1}{x^2} = \frac{x+1}{x^2}\,\mbox{.}

\displaystyle \frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}\quad has \displaystyle \ \text{LCD}= x^2(x+1)^2(x+2)

The first term is converted using \displaystyle x(x+2) while the other term is converted using \displaystyle (x+1)^2

\displaystyle \begin{align*}

   \frac{1}{x(x+1)^2} - \frac{1}{x^2(x+2)}
     &= \frac{x(x+2)}{x^2(x+1)^2(x+2)}
        - \frac{(x+1)^2}{x^2(x+1)^2(x+2)}\\[4pt]
     &= \frac{x^2+2x}{x^2(x+1)^2(x+2)} - \frac{x^2+2x+1}{x^2(x+1)^2(x+2)}\\[4pt]
     &= \frac{x^2+2x-(x^2+2x+1)}{x^2(x+1)^2(x+2)}\\[4pt]
     &= \frac{x^2+2x-x^2-2x-1}{x^2(x+1)^2(x+2)}\\[4pt]
     &= \frac{-1}{x^2(x+1)^2(x+2)}\,\mbox{.}
   \end{align*}

\displaystyle \frac{x}{x+1} - \frac{1}{x(x-1)} -1 \quad has \displaystyle \ \text{LCD}=x(x-1)(x+1)

We must convert all the terms so that they have the common denominator \displaystyle x(x-1)(x+1)

\displaystyle \begin{align*}

   \frac{x}{x+1} - \frac{1}{x(x-1)} -1
     &= \frac{x^2(x-1)}{x(x-1)(x+1)} - \frac{x+1}{x(x-1)(x+1)}
        - \frac{x(x-1)(x+1)}{x(x-1)(x+1)}\\[4pt]
     &= \frac{x^3-x^2}{x(x-1)(x+1)} - \frac{x+1}{x(x-1)(x+1)}
        - \frac{x^3 -x}{x(x-1)(x+1)}\\[4pt]
     &= \frac{x^3-x^2 -(x+1) -(x^3-x)}{x(x-1)(x+1)}\\[4pt]
     &= \frac{x^3-x^2 -x-1 -x^3+x}{x(x-1)(x+1)}\\[4pt]
     &= \frac{-x^2-1}{x(x-1)(x+1)}\,\mbox{.}
   \end{align*}

To simplify large expressions it is often necessary both to cancel factors and to multiply numerators and denominators by factors. Where possible, therefore, we should keep expressions in a factorised form, and not expand expressions we will later need to factorise again.

Example 11

\displaystyle \frac{1}{x-2} - \frac{4}{x^2-4} = \frac{1}{x-2} - \frac{4}{(x+2)(x-2)} = \left\{\,\mbox{MGN} = (x+2)(x-2)\,\right\}

\displaystyle \phantom{\frac{1}{x-2} - \frac{4}{x^2-4}}{} = \frac{x+2}{(x+2)(x-2)} - \frac{4}{(x+2)(x-2)}

\displaystyle \phantom{\frac{1}{x-2} - \frac{4}{x^2-4}}{} = \frac{x+2 -4}{(x+2)(x-2)} = \frac{x-2}{(x+2)(x-2)} = \frac{1}{x+2}
\displaystyle \frac{x + \displaystyle \frac{1}{x}}{x^2+1} = \frac{\displaystyle \frac{x^2}{x} + \frac{1}{x}}{x^2+1} = \frac{\displaystyle \frac{x^2+1}{x}}{x^2+1} = \frac{x^2+1}{x(x^2+1)} = \frac{1}{x}
\displaystyle \frac{\displaystyle \frac{1}{x^2} - \frac{1}{y^2}}{x+y} = \frac{\displaystyle \frac{y^2}{x^2y^2} - \frac{x^2}{x^2y^2}}{x+y} = \frac{\displaystyle \frac{y^2-x^2}{x^2y^2}}{x+y} = \frac{y^2-x^2}{x^2y^2(x+y)}

\displaystyle \phantom{\smash{\frac{\displaystyle \frac{1}{x^2} - \frac{1}{y^2}}{x+y}}}{} = \frac{(y+x)(y-x)}{x^2y^2(x+y)} = \frac{y-x}{x^2y^2}

Exercises

Study advice

The basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.

Keep in mind that...

If you make a mistake somewhere the rest of the calculation will be wrong, so be careful!

Use many intermediate steps. If you are unsure of a calculation do it in many small steps rather than one big step.

Do not expand unnecessarily. You later may be forced to factorise what you earlier expanded.

Reviews

Learn more about algebra in the English Wikipedia

Understanding Algebra - English text on the Web

Useful web sites

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/2.1_Algebraic_expressions"

3. Roots and logarithms

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2.1 Algebraic expressions

From Förberedande kurs i matematik 1

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Distributive Law

Squaring

Difference of two squares

Rational expressions