ValsCalc (valscalc.com)
0.1: Fundamental Skills in Algebra
Online Textbook Practice Tests 1500 Calculus Problems Solved About
☰ Contents
< Previous
Next >

0.1: Fundamental Skills in Algebra

Algebra is the branch of mathematics that studies operations with numbers—namely, by generalizing arithmetic by using symbols to represent numbers and relationships. It underpins much of calculus and beyond, so any calculus student needs to master its basic operations. So this section covers the following topics:

Solving Equations and Inequalities

In a mathematical equation, two quantities have the same value. For example, \(3 = 2 + 1\) is a valid equation because both sides have the same value. But some equations have some unknown quantity, such as the equation \(2x + 5 = 7.\) What does \(x\) equal? We can manipulate equations by adding, subtracting, multiplying, or dividing a number on both sides. On both sides of \(2x + 5 = 7,\) subtracting \(5\) and then dividing by \(2\) yields \(2x = 2\) and then \(x = 1.\) The following guidelines generalize the strategy for solving similar equations. In general, the objective is to perform operations to isolate the variable.

SOLVING EQUATIONS WITH ONE VARIABLE
To solve for \(x\) in an equation, use the following steps:
  1. Condense On each side, simplify the expression by combining like terms.
  2. Separate By adding or subtracting, gather all the variables on one side of the equation and all the constants on the other side.
  3. Isolate Divide both sides by the coefficient.
  4. Check Substitute your value of \(x\) back into the equation to verify whether it is correct.
EXAMPLE 1
Solve for \(x \col\) \[3x + 6 + 10 = 7 - 8x + 2x \pd\]
  1. Condense On the left side, adding \(6\) and \(10\) gives \(16.\) On the right side, adding \(-8x\) to \(2x\) yields \(-6x.\) Thus, the condensed equation is \[3x + 16 = 7 - 6x \pd\]
  2. Separate We want to place all the \(x\)'s on the left and all the constants on the right. To clear the \(-6x\) on the right, we add \(6x\) to both sides; to clear the \(16\) on the left, we subtract \(16\) from both sides. Doing so yields \[9x = -9 \pd\]
  3. Isolate The coefficient is the number by which \(x\) is multiplied—namely, \(9.\) To cancel out the \(9\) and therefore isolate \(x,\) we divide both sides by \(9\) to get \[x = \boxed{-1}\]
  4. Check Substituting \(x = -1\) back into the given equation, we see \[ \ba 3(-1) + 6 + 10 &= 7 - 8(-1) + 2(-1) \nl -3 + 6 + 10 &= 7 + 8 - 2 \nl 13 &\equalsCheck 13 \pd \ea \] This step verifies that \(x = -1\) is the correct solution.
EXAMPLE 2
Solve for \(x \col\) \[4(x + 5) - 2x = 31 \pd\]
  1. Condense To expand \(4(x + 5),\) we distribute the \(4\)—that is, multiply each term inside the parentheses by \(4.\) Then the equation becomes \[ \ba 4x + 20 - 2x &= 31 \nl 2x + 20 &= 31 \pd \ea \]
  2. Separate We want all the \(x\)'s to be on the left side and all the constants to be on the right side. Subtracting \(20\) from both sides yields \[2x = 11 \pd\]
  3. Isolate To cancel out the coefficient of \(2,\) we divide both sides by \(2\) to attain \[x = \boxed{\tfrac{11}{2}}\] (It is also correct to write \(x = 5.5;\) in calculus we prefer to write fractions.)
  4. Check Substituting \(x = 11/2\) back into the given equation, we see \[ \ba 4 \par{\tfrac{11}{2} + 5} - 2 \par{\tfrac{11}{2}} &= 31 \nl (22 + 20) - 11 &= 31 \nl 31 &\equalsCheck 31 \pd \ea \] Hence, \(x = 11/2\) is the correct solution.

An inequality is a mathematical statement that compares two quantities; for example, \(5 \gt 4\) means \(5\) is greater than \(4.\) Likewise, the inequality \(x \leq 6\) means \(x\) can be any number that is less than or equal to \(6.\) We solve inequalities similarly to how we solve equations, as shown by the following examples.

EXAMPLE 3
Solve the following inequalities.
  1. \(6x + 3 \gt 15\)
  2. \(-4x - 6 \leq 10\)
  3. \(\dfrac{1}{x} - 2 \gt 5 \sspace\) if \(\sspace x \gt 0\)
In each problem, our goal is to isolate \(x.\)
  1. On both sides, we subtract \(3\) and then divide by \(6 \col\) \[6x \gt 12 \implies \boxed{x \gt 2}\]
  2. On both sides, we add \(6\) and then divide by \(-4.\) But when we multiply or divide both sides by a negative number, the inequality sign is flipped. So we have \[-4x \leq 16 \implies \boxed{x \geq -4}\]
  3. Adding \(2\) to both sides gives \[\frac{1}{x} \gt 7 \pd\] We multiply both sides by \(x\) and then divide by \(7,\) as follows: \[1 \gt 7x \implies \tfrac{1}{7} \gt x \implies x \lt \tfrac{1}{7} \pd\] (Since \(x\) is given to be positive, we don't flip the inequality sign.) The problem gives \(x \gt 0,\) so we write the solution as \[\boxed{0 \lt x \lt \tfrac{1}{7}}\]

Recall that exponents represent repeated multiplication; for example, \(2^3\) \(= 2 \cdot 2 \cdot 2\) \(= 8.\) Let \(a\) and \(b\) be positive numbers, and let \(m\) and \(n\) be any numbers. Then the exponent laws are as follows: \begin{align} a^{m + n} &= a^m \, a^n \cma \label{eq:exp-law-add} \nl a^{m - n} &= \frac{a^m}{a^n} \cma \label{eq:exp-law-minus} \nl \par{a^{m}}^n &= a^{mn} \cma \label{eq:exp-law-mn} \nl \par{ab}^n &= a^n \, b^n \pd \label{eq:exp-law-a-b} \end{align} We won't prove these properties, but you can perform some testing to verify that they are true. (We will study these properties more in depth in Section 0.9.)

A polynomial is a sum or difference of variables with nonnegative integer (whole number) powers. A polynomial is classified by its highest-power term. For example, \(x^3 + 5x^2 - 8\) is a third-degree polynomial and \(x^6 + x - 2\) is a sixth-degree polynomial. A second-degree polynomial is called quadratic; we investigate quadratic expressions and functions more in Section 0.7. Polynomials are prevalent in calculus; we devote the remainder of this section to operations with polynomials.

EXAMPLE 4
Simplify \[5x^2 + 7x - 3 - 4 \par{x^2 - 2x + 3}\]
We distribute the \(-4\) to each term within the parentheses and add the result. Doing so and combining like terms, we get \[ \ba 5x^2 + 7x - 3 + (- 4x^2 + 8x - 12 ) &= 5x^2 + 7x - 3 - 4x^2 + 8x - 12 \nl &= \boxed{x^2 + 15x - 15} \ea \] Note that \(x^2\) and \(x\) are separate terms, so we cannot combine \(x^2\) and \(15x.\)
EXAMPLE 5
Expand the following quantities.
  1. \((2x + 4)(x - 3)\)
  2. \((5x + 1)^2\)
  1. Breaking up the first set of parentheses, we get \[2x(x - 3) + 4(x - 3) \pd\] Then distributing and combining like terms produce \[ \ba (2x^2 - 6x) + (4x - 12) &= 2x^2 - 6x + 4x - 12 \nl &=\boxed{2x^2 - 2x - 12} \ea \]
  2. It is wrong to write \[\wrongMath{(5x + 1)^2 = (5x)^2 + 1^2 = 25x^2 + 1 \pd}\] Instead, note that \((5x + 1)^2\) \(= (5x + 1)(5x + 1).\) Breaking up the first set of parentheses, we have \[5x(5x + 1) + 1(5x + 1) \pd\] Then we distribute and combine like terms, as follows: \[ \ba (25x^2 + 5x) + (5x + 1) &= 25x^2 + 5x + 5x + 1 \nl &= \boxed{25x^2 + 10x + 1} \ea \]

Factoring Polynomials

Factoring is the process of decomposing a number or expression into the product of two or more terms (factors). For example, the number \(10\) is factored as \(2 \times 5.\) There are several techniques to factor polynomials. The most important one is by extracting the greatest common factor (GCF), the largest factor common to all terms in the expression. This process is the first step of factoring a polynomial.

EXAMPLE 6
Factor the following expressions.
  1. \(4x + 8\)
  2. \(9ab + 3b\)
  1. The greatest common factor of \(4x\) and \(8\) is \(4,\) which we extract to get \[\boxed{4(x + 2)}\]
  2. The greatest common factor of the coefficients \(9\) and \(3\) is \(3.\) Both terms also contain a \(b,\) so extracting \(3b\) gives \[\boxed{3b(3a + 1)}\]

The following formulas enable us to factor differences of squares, sums of cubes, and differences of cubes.

FACTORING SQUARES AND CUBES
\begin{alignat}{2} a^2 - b^2 &= (a + b)(a - b) \comment{\text{Difference of Squares}} \label{eq:diff-sq} \nl a^3 + b^3 &= (a + b)(a^2 - ab + b^2) \comment{\text{Sum of Cubes}} \label{eq:sum-cb} \nl a^3 - b^3 &= (a - b)(a^2 + ab + b^2) \comment{\text{Difference of Cubes}} \label{eq:diff-cb} \end{alignat}
EXAMPLE 7
Factor each expression.
  1. \(x^2 - 81\)
  2. \(y^3 + 27\)
  1. Observe that \(81\) \(= 9^2.\) Using \(\eqref{eq:diff-sq}\) with \(a = x\) and \(b = 9,\) we get \[\boxed{(x + 9)(x - 9)}\]
  2. Note that \(27\) \(= 3^3.\) Using \(\eqref{eq:sum-cb}\) with \(a = y\) and \(b = 3\) yields \[\boxed{(y + 3)(y^2 - 3y + 9)}\]
EXAMPLE 8
Factor \(x^4 - 16\) completely.
By exponent law \(\eqref{eq:exp-law-mn},\) \(x^4\) \(= \par{x^2}^2.\) Also note that \(16 = 4^2,\) so by \(\eqref{eq:diff-sq}\) we have \[(x^2 + 4)(x^2 - 4) \pd\] We again use \(\eqref{eq:diff-sq}\) to factor \((x^2 - 4),\) attaining \[\boxed{(x^2 + 4)(x + 2)(x - 2)}\] The factor \((x^2 + 4)\) can't be factored any further, so it is called prime.

Next we examine the strategy for factoring the family of quadratics \(x^2 + bx + c\) into two binomials. (A binomial is the sum or difference of two terms.) The objective is to find two numbers \(p\) and \(q\) whose product is \(c\) and sum is \(b;\) then \[x^2 + bx + c = (x + p)(x + q) \pd\] The following example demonstrates this procedure.

EXAMPLE 9
Factor \(x^2 - x - 20.\)
We identify \(b = -1\) and \(c = -20.\) What two numbers multiply to \(-20\) and add to \(-1 \ques\) By trial and error, the two numbers are \(p = -5\) and \(q = 4.\) Hence, the factored form is \[\boxed{(x - 5)(x + 4)}\]
EXAMPLE 10
Solve for \(x \col\) \[2x^2 = 12x - 16 \pd\]
To clear the right side, we subtract \(12x\) and add \(16\) to both sides; doing so gives \[2x^2 - 12x + 16 = 0 \pd\] The GCF of the left side is \(2,\) which we extract to get \[2(x^2 - 6x + 8) = 0 \pd\] Observe that \(p = -4\) and \(q = -2\) have a product of \(8\) and a sum of \(-6.\) Hence, factoring the expression in parentheses gives \[2(x - 4)(x - 2) = 0 \pd\] By the Zero Property of Multiplication, the left side equals \(0\) if at least one factor equals \(0.\) Equating each factor to \(0,\) we see \[ \baat{2} x - 4 &= 0 \qquad &x - 2 &= 0 \nl \implies x &= \boxed 4 &\qquad \implies x &= \boxed 2 \eaat \]

Let's now discuss factoring by grouping. Using this technique, we split a polynomial into pairs of terms, each of which we factor independently. Factoring by grouping works if each pair shares a common factor; then the polynomial's factored form contains the shared factor multiplied by the sum of the groups' GCFs, as shown by the following examples.

EXAMPLE 11
Factor \(x^3 + 5x^2 + 2x + 10.\)
We skip the preliminary step of extracting a GCF because none exists. To factor by grouping, we split the expression into two pairs and factor each independently by extracting GCFs: \[ \underbrace{x^3 + 5x^2}_{\orange{x^2} (x + 5)} + \underbrace{2x + 10}_{\orange{2}(x + 5)} \pd \] Each pair shares the common factor \((x + 5),\) and the sum of the GCFs is \((\orange{x^2} + \orange{2}).\) So the factored form is \[\boxed{(x + 5)(\orange{x^2} + \orange{2})}\] Note that \((\orange{x^2} + \orange{2})\) is prime; it can't be factored any further.

Factoring by grouping is useful for factoring quadratics in the form \(ax^2 + bx + c\) \((a \ne 0)\) when no GCF is present. The objective is to find two numbers \(p\) and \(q\) whose product is \(ac\) and sum is \(b,\) and then split \(bx\) into \(px + qx.\) The resulting expression can be factored by grouping, as shown by the next example.

EXAMPLE 12
Factor \(6x^2 + 5x + 1.\)
Identifying \(a = 6\) and \(c = 1,\) we have \(ac = 6.\) The numbers \(p = 2\) and \(q = 3\) multiply to \(6\) and add to \(5.\) We therefore split \(5x\) into \(2x + 3x\) and factor by grouping: \[\underbrace{6x^2 + 2x}_{\orange{2x} (3x + 1)} + \underbrace{3x + 1}_{\orange{1} (3x + 1)} \pd\] The shared factor is \((3x + 1),\) and the sum of the GCFs is \((\orange{2x} + \orange{1}).\) Hence, the factored form is \[\boxed{(3x + 1)(\orange{2x} + \orange{1})}\]

Operations with Fractions

Fractions represent the division of two quantities. Recall that the top part of a fraction is the numerator and the bottom part is the denominator. Multiplying fractions is easy—we simply multiply across; for example, \[\tfrac{2}{3} \times \tfrac{5}{6} = \tfrac{10}{18} = \tfrac{5}{9} \pd\] (In the last step we divided the numerator and denominator each by \(2.\) Multiplying or dividing the numerator and denominator by the same number does not alter the fraction's value.) In dividing two fractions, we flip the second fraction and perform multiplication. As an example, to divide \(\tfrac{4}{5}\) by \(\tfrac{8}{7},\) we have \[\tfrac{4}{5} \times \tfrac{7}{8} = \tfrac{28}{40} = \tfrac{7}{10} \pd\] The fractions \(\tfrac{8}{7}\) and \(\tfrac{7}{8}\) are reciprocals of each other. The same logic with numerical fractions applies to fractions with algebraic expressions, as in the next examples.

EXAMPLE 13
For \(x \ne -3,\) evaluate \[\frac{x - 2}{6} \cdot \frac{5x}{x + 3} \pd\]
Multiplying across and expanding, we get \[ \frac{(x - 2)(5x)}{6 (x + 3)} = \boxed{\frac{5x^2 - 10x}{6x + 18}} \]
EXAMPLE 14
If no division by \(0\) is present, then simplify \[2 \cdot \frac{5}{x + 6} \pd\]
Note that \(2\) \(= 2/1,\) so we have \[\frac{2}{1} \cdot \frac{5}{x + 6} = \boxed{\frac{10}{x + 6}}\] Thus, multiplying a fraction by a whole number is equivalent to multiplying just the numerator by that number.
EXAMPLE 15
Solve for \(x \col\) \[\frac{2x - 1}{6} = \frac{1}{x} \pd\]
To clear the denominator on the left, we multiply both sides by \(6.\) To clear the denominator on the right, we multiply both sides by \(x.\) Doing so, we get \[x(2x - 1) = 6 \or 2x^2 - x - 6 = 0 \pd\] Factoring gives \[(x - 2)(2x + 3) = 0 \pd\] Equating each factor to \(0,\) we have \[ \baat{2} x - 2 &= 0 \qquad &2x + 3 &= 0 \nl \implies x &= \boxed 2 &\qquad \implies x &= \boxed{-\tfrac{3}{2}} \eaat \] Neither solution yields division by \(0,\) so both answers are valid.
EXAMPLE 16
Given that no denominators equal \(0,\) simplify \[\frac{\dfrac{x^2 - 3x - 10}{x^3 + 1 \strut}}{\dfrac{\strut x^2 - 4x - 5}{x^2 + 2x + 1}} \pd\]
This expression is equivalent to the top fraction multiplied by the reciprocal of the bottom fraction—namely, \[ \frac{x^2 - 3x - 10}{x^3 + 1} \cdot \frac{x^2 + 2x + 1}{x^2 - 4x - 5} \pd \] Factoring each expression, we have \[ \frac{\cancel{(x - 5)}(x + 2)}{\cancel{(x + 1)}(x^2 - x + 1)} \cdot \frac{\cancel{(x + 1)}\cancel{(x + 1)}}{\cancel{(x - 5)}\cancel{(x + 1)}} = \boxed{\frac{x + 2}{x^2 - x + 1}} \]

To add or subtract two fractions with the same denominator, we simply add or subtract the numerators; a simple example is \(\tfrac{3}{5} + \tfrac{1}{5}\) \(= \tfrac{4}{5}.\) But if no common denominator is present, then we multiply both parts of one fraction (or both fractions) to force the existence of one. For example, \[\tfrac{1}{2} + \tfrac{1}{3} = \tfrac{3}{6} + \tfrac{2}{6} = \tfrac{5}{6} \pd\] The same idea applies to rational functions.

EXAMPLE 17
Given that no denominators equal \(0,\) simplify \[\frac{5x - 1}{x + 3} + \frac{6x}{x + 7} \pd\]
To force the existence of a common denominator, we multiply the numerator and denominator of the first fraction by \((x + 7),\) and we multiply the numerator and denominator of the second fraction by \((x + 3).\) Doing so gives \[ \ba \frac{(5x - 1)(x + 7)}{(x + 3)(x + 7)} + \frac{6x (x + 3)}{(x + 3)(x + 7)} &= \frac{(5x - 1)(x + 7) + 6x(x + 3)}{(x + 3)(x + 7)} \nl &= \boxed{\frac{11x^2 + 52x - 7}{x^2 + 10x + 21}} \ea \]

To simplify a quotient of polynomials that may share no common factors, we use polynomial division, which is analogous to standard long division for numbers. The following steps describe how to perform this operation.

HOW TO PERFORM POLYNOMIAL DIVISION
To divide a polynomial \(P\) by a polynomial \(Q\) where \(P\) and \(Q\) are polynomials such that the degree of \(Q\) is less than the degree of \(P,\) use the following steps:
  1. Write out the terms of \(P\) in order of descending powers. To the left of the expression, draw a right parenthesis \()\); above the expression, draw a horizontal line. To the left of the parenthesis, write out the terms of \(Q\) in order of descending powers.
  2. Divide the first term of \(P\) by the first term of \(Q;\) place that answer above the horizontal line. Multiply \(Q\) by that answer and write the result in a new row at the bottom.
  3. Subtract the result of Step 2 from \(P\) to generate a new polynomial.
  4. Divide the first term of the new polynomial by the first term of \(Q.\) Multiply \(Q\) by that result and subtract it from the new polynomial. Repeat this process until the resulting polynomial has a degree smaller than the degree of \(Q.\)
EXAMPLE 18
Perform polynomial division on \[\frac{2x^3 + x^2 + x - 4}{x - 1} \pd\]
We separate the expressions \(2x^3 + x^2 + x - 4\) and \(x - 1\) using a right parenthesis \()\) and draw a horizontal line above the first polynomial. Next we divide the first terms—that is, \(2x^3\) divided by \(x\)—to get \(2x^2,\) which we write at the top. Then \(2x^2 (x - 1)\) \(= 2x^3 - 2x^2,\) which we subtract from \(2x^3 + x^2\) \(+ \, x - 4\) to generate a new polynomial, \(3x^2 + x - 4.\) We repeat this process, as shown by the following setup. \[ \require{enclose} \begin{array}{rll} 2x^2 + 3x + 4 \\[-3pt] x - 1 \enclose{longdiv}{2x^3 + \phantom{2} x^2 + x - 4}\kern-.2ex \nl \begin{alignedat}{3} 2x^3 \hspace{0.32em} &- &&\, \, 2x^2 & \nl \hline & &&3x^2 &+ \phantom{3} x \phantom{-4} \nl & &&3x^2 &- 3x \phantom{+0} \nl \hline & && &\phantom{+} 4x - 4 \nl & && &\phantom{+} 4x - 4 \nl \hline & && & \hspace{0.4em} 0 \end{alignedat} \end{array} \] Thus, \[\frac{2x^3 + x^2 + x - 4}{x - 1} = \boxed{2x^2 + 3x + 4}\]
EXAMPLE 19
Use polynomial division to simplify \[\frac{x^4 + 3x^3 - x}{x^2 + 1} \pd\]
Let's explicitly write \(+ 0x^2\) and \(+ 0\) to maintain alignment and organization. We therefore have the following setup. \[ \require{enclose} \begin{array}{rll} x^2 + 3x - 1 \\[-3pt] x^2 + 1 \enclose{longdiv}{x^4 + 3x^3 + 0x^2 - x + 0}\kern-.2ex \nl \begin{alignedat}{3} x^4 \hspace{0.48em} &\phantom{+} \phantom{3x^3} &&+ \hspace{0.05em} x^2 &\phantom{-} \phantom{x + 0} \hspace{0.3em} \nl \hline &\phantom{+} 3x^3 &&-x^2 &- x \phantom{+ 0} \hspace{1em} \nl &\phantom{+} 3x^3 &&\phantom{-x^2} &+ 3x \hspace{1.803em} \nl \hline &\phantom{+} &&-x^2 &-4x \hspace{1.85em} \nl &\phantom{+} &&-x^2 &\phantom{-4x} - 1 \nl \hline & && &-4x + 1 \nl \end{alignedat} \end{array} \] The remainder \((-4x + 1)\) has a lower degree than \((x^2 + 1),\) so no more division can occur. Therefore, we have \[\frac{x^4 + 3x^3 - x}{x^2 + 1} = \boxed{x^2 + 3x - 1 + \frac{-4x + 1}{x^2 + 1}}\]

Solving Equations and Inequalities Algebra is the branch of mathematics concerned with generalizing operations with numbers. Solving equations and inequalities is the heart of algebra. The following steps enable us to solve an equation for some variable \(x \col\)

  1. Condense On each side, simplify the expression by combining like terms.
  2. Separate By adding or subtracting, gather all the variables on one side of the equation and all the constants on the other side.
  3. Isolate Divide both sides by the coefficient.
  4. Check Substitute your value of \(x\) back into the equation to verify whether it is correct.

A similar process enables us to solve inequalities for \(x.\) Remember that the inequality sign is flipped when both sides are multiplied or divided by a negative number. If \(a\) and \(b\) are positive numbers, and \(m\) and \(n\) are any numbers, then the exponent laws are as follows: \begin{align} a^{m + n} &= a^m \, a^n \cma \eqlabel{eq:exp-law-add} \nl a^{m - n} &= \frac{a^m}{a^n} \cma \eqlabel{eq:exp-law-minus} \nl \par{a^{m}}^n &= a^{mn} \cma \eqlabel{eq:exp-law-mn} \nl \par{ab}^n &= a^n \, b^n \pd \eqlabel{eq:exp-law-a-b} \end{align}

Factoring Polynomials Factoring is the process by which a number or expression is decomposed into the product of two or more terms (factors). A polynomial is a sum or difference of variables with nonnegative integer (whole number) powers; a polynomial is classified by its highest power. The first step to factoring a polynomial is to extract the greatest common factor (GCF), the largest number or expression that goes into each term in the expression. Subsequent methods of factoring are as follows.

To factor \(ax^2 + bx + c\) \((a \ne 0),\) we find two numbers \(p\) and \(q\) whose product is \(ac\) and sum is \(b.\) Then we split \(bx\) into \(px + qx\) and factor the resulting polynomial by grouping.

Operations with Fractions The procedures for adding, subtracting, multiplying, and dividing numerical fractions are identical to those for fractions containing algebraic expressions. Multiplying fractions is the simplest operation—simply multiply the numerators together and multiply the denominators together. Dividing two fractions is equivalent to multiplying the first fraction by the reciprocal of the second fraction. To add or subtract two fractions, attain a common denominator; if one does not exist, then multiply the numerator and denominator of one fraction (or both fractions) to force the existence of a common denominator. Polynomial division enables us to decompose a quotient of two polynomials; it is analogous to long division for numbers. The steps for evaluating \(P/Q,\) where \(P\) and \(Q\) are polynomials such that the degree of \(Q\) is less than the degree of \(P,\) are as follows:

  1. Write out the terms of \(P\) in order of descending powers. To the left of the expression, draw a right parenthesis \()\); above the expression, draw a horizontal line. To the left of the parenthesis, write out the terms of \(Q\) in order of descending powers.
  2. Divide the first term of \(P\) by the first term of \(Q;\) place that answer above the horizontal line. Multiply \(Q\) by that answer and write the result in a new row at the bottom.
  3. Subtract the result of Step 2 from \(P\) to generate a new polynomial.
  4. Divide the first term of the new polynomial by the first term of \(Q.\) Multiply \(Q\) by that result and subtract it from the new polynomial. Repeat this process until the resulting polynomial has a degree smaller than the degree of \(Q.\)