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MATH 00 Page Linear Equations Linear Equations Eercises 5 Linear Equations Answer to Eercises 6 Linear Inequalities 8 Equations Involving Absolute Value 0 Inequalities Involving Absolute Value Linear Inequalities and Absolute Value Eercises Linear Inequalities and Absolute Value Answer to Eercises Linear Functions 5 Linear Functions Eercises Linear Functions Answer to Eercises Eponents Eponents Eercises 6 Eponents Answer to Eercises 7 Polynomials 8 Polynomials Eercises 5 Polynomials Answer to Eercises 6 Factoring 7 Factoring Eercises Factoring Answer to Eercises
LINEAR EQUATIONS A linear equation can be defined as an equation in which the highest eponent of the equation variable is one. When graphed, the equation is shown as a single line. Eample: + = Linear equation, highest eponent of the variable is. A linear equation has only one solution. The solution of a linear equation is equal to the value of the unknown variable that makes the linear equation true. Eample: + = The value of the unknown variable that makes the equation = true is : + = = There are different forms of writing a linear equation. Each form has a different way of solving that makes the process easier. There is only one rule that applies to any form of linear equations, whatever you do on one side you need to do on the other side. That is, if you add on the left side, you need to add on the right side, if you multiply on the left side you need to multiply on the right side, and so on.. Equations of the form + a = b or a = b. Linear equations in this form are solved by adding or subtracting the same quantity to both sides with the idea of leaving the variable by itself. Eamples: 5 = 0 5 + 5 = 0 + 5 Add 5 to both sides to eliminate the number on =5 the left side of the equation and leave alone. y +. =. y +.. =.. Subtract. to both sides to eliminate the y = 0.9 number on the left side and leave y alone.. Equations of the form a = b. Linear equations in the form of multiplication are solved by dividing both sides of the equation by the number multiplying the variable. When a fraction is multiplying the variable, multiply both sides of the equation by the reciprocal of the fraction attached to the variable.
Eamples: = 8 8 = Divide both sides by to leave by itself. = 6 = 5 = Multiply both sides by the reciprocal of, which is. 5 8 = 5. Equations of the form a + b = c. This type of linear equation presents a combination of the first two forms of linear equations. First step in solving this form of linear equations is to eliminate the stand alone number on the variable side by adding or subtracting the same quantity to both sides of the equation. Net, eliminate the number attached to the variable by dividing or multiplying, depending on the case, to both sides of the equation. The idea is to leave variables on one side and constants on the other side. Eamples: y + 7 = 5 y + 7 7 = 5 7 Subtract 7 to both sides to remove it from the left side. y = 8 8 y = Divide both sides by to leave y alone and solve for y. y = 6 8 + = Follow the steps described above to solve for. 5 8 + = 5 8 = 5 5 8 5 = 8 5 8 5 = 8
. Equations of the form a + b = c + d. In this form of linear equations the main goal is to leave the variables on one side and the constants on the other side. To do this, perform all the steps required to solve the previous forms of linear equations step by step. Do not try to do all the steps at once, this can only result in confusion and mistakes. Eample: 6 5 = 8 + Start by moving the variable from the right side to the left side. 6 + 8 5 = 8 + 8 + Combine the two variables together. 5 = Add 5 to both sides to eliminate the 5 + 5 = + 5 constant on the left side of the equation. = 8 Divide both sides of the equation by. 8 = Solve for. = Distributive Property: a ( b + c) = ab + ac The distributive property is used to remove grouping symbols in linear equations. Grouping symbols are terms enclosed in parentheses. To ungroup, distribute the term outside the parentheses to each of the terms inside the parentheses by multiplication. Eample: ( ) + = 7 Multiply the term outside the parentheses to each of 6 + = 7 the terms inside the parentheses. 5 = 7 Eliminate the constant on the left side of the equation 5 + 5 = 7 + 5 by adding 5 to both sides of the equation. = = Divide both sides by and solve for. =
LINEAR EQUATIONS EXERCISES Solve for :. 8 6 =. 5 = + 7. ( 5 ) = ( 5). 5. 6. 7. 8. 9. 0. + = a b = b a + 5 = + 6 8 + + = + + = a b c = 0 + + = 5 6. ( ) + = 5 +. 5( 5) =. ( 5 + 7) = 0. 8 + ( + ) = 5( ) 5. 5( + ) ( + ) = 7 6. 7. 7 + = + 7 6 + = 5
LINEAR EQUATIONS ANSWERS TO EXERCISES. 8 6 =. 5 7 8 6+ 6= + 6 8 = 8 9 = = +. 5 ( ) = ( 5) = 7+ 5 = = 6 5+ 6 = 0 7 = 0 + + 5 = 6 =. + = = + 6 = 6+ 6 6= + 0 = = 5 ( ) ( ) 5. a b = b a a a = b b ( ) ( ) a a = b b ( ) a b = a b a b = a b a b = a b 6. + 5 = + 6 7. 8 + + = + 5( + ) = 5( + ) 6 8 + 5 + ( )( + ) + = ( + ) + + 5 ( + ) = 5 0( + ) ( + ) + ( ) = 8( ) 5 = 5 0 0 + + = 8 8 6 = 6 7 8= 8 8 = = 7 = 7 8. + = 9. a b c = 0 6
0. abc + = abc a b c bc + ac = ab bc + ac = ab c( a+ b) = ab 5 6 + 0 + = 0 5 6 6 + + 5 = 0 = 0 9 = 0 = + + =. ( ) ( ) ( ) + 6 + 5 0 = 0 7 = = 6 ( ) + = 5+ 6 + = 5+ 7 5= + = 6 =. 5( 5) =. ( 5 + 7) = 0 0 + 5 = 5 = 0 0 = 5 5+ 0= + 0 = 0 5 = 5 = = 5. 8 + ( + ) = 5( ) 5. 5( + ) ( + ) = 7 8+ + 6= 5 5 5= 5 8 6 5 + 0 = 7 = 9 = 7 0 + 9 = = 6. 7 + = + 7 ( + ) = ( + ) + + ( + ) = 8 6 = 8 6 = 7 = 6 7. 7 6 + = 7 ( ) + = ( ) 6 7+ = 6( ) = 8 + 8 = 9 = 7
LINEAR INEQUALITIES When we use the equal sign in an equation we are stating that both sides of the equation are equal to each other. In an inequality, we are stating that both sides of the equation are not equal to each other. It can also be seen as an order relation; that is, it tells us which one of the two epressions is smaller, or larger, than the other one. A linear inequality is an equation in which the highest variable eponent is one. Eamples: < less than, < + 5 less than or equal to, > greater than, > 6 greater than or equal to, The solution to an inequality is the value of the variable which makes the statement, or the inequality, true. Eamples: < This inequality is telling us that is less than. Therefore, any number less than three is a possible solution. Remember that on the number line, any number to the left is less than a given number, and any number to the right of that given number is greater. 0 Notice how is marked with an empty dot, this means that the answer does not include the number. + 5 This inequality is telling us that the equation + 5 is less than or equal to the equation. Solve for to find which number, or numbers, make this equation true. + 5 Solve for by leaving the variables on one side + 5 5 5 and the numbers on the other side, just as you 8 would solve a linear equation. 8 8 Notice how 8 is marked with a black dot, this means that the answer does include the number 8. 9 8 7 6 5 0 > This inequality is telling us that is greater than. Therefore, any number greater than negative two is a possible solution. 8
0 6 Solve for by leaving the variables on one side + 6 + and the numbers on the other side. 0 As you can see, solving linear inequalities is very similar to solving linear equations. There is only one thing you need to keep in mind when multiplying or dividing a negative number: the direction or sense of the inequality is reversed. For eample, if the sign is >, after multiplying or dividing by a negative number the sign changes to <. Eample: + Solve for as in the previous eamples. + 6 6 6 Multiply times the negative reciprocal to leave the alone and get an answer. 6 However, since we are multiplying by a negative the inequality sign must change from to. 9
EQUATIONS INVOLVING ABSOLUTE VALUE The absolute value of a number is always the positive value of that number. Since the absolute value is the distance of a number from the origin, and since distances are always positive, the absolute value is always a positive value. The same definition applies to equations involving absolute value; any equation inside the absolute value bars must always be equal to a positive number. Eample: = 8 This absolute value equation does not have a real answer. Any equation inside the absolute value bars must always be equal to a positive number. In this case, for to be equal to 8, must be : = 8 However, since the equation is inside the absolute value bars it will always yield a positive answer 8 = 8 and never a negative answer. When the equation involving absolute value is equal to a positive number, the value of the variable inside the absolute value bars can be negative or positive; therefore, the equation must be made equal to a positive and a negative answer. The answer will have two different possible values of, and both will make the statement valid. Eamples: + 6 = Make the equation equal to and and solve for in each case. + 6 = and + 6 = + 6 6 = 6 + 6 6 = 6 = 6 = 8 6 8 = = = = 6 The answer to + 6 = is = and = 6. 6 + 5 6 = 0 5 6 = Leave the absolute value epression by itself on the left side of the equation. Then solve for as in the previous eample. 5 6 = and 5 6 = 5 6 + 6 = + 6 5 6 + 6 = + 6 5 = 0 5 = = = 5 The answer is = and =. 5 0
INEQUALITIES INVOLVING ABSOLUTE VALUE Inequalities involving absolute value are solved in a similar form as equations involving absolute value. Since the variable inside the absolute value sign can be either positive or negative, both positive and negative values are used to solve for the variable. However, in the case of linear inequalities involving absolute value, the inequality is placed between the two values rather than having two different equations. Eamples: < To solve this linear inequality use both positive and negative values of and place the inequality in the middle without the absolute value sign. The inequality sign will be the same on both sides. < < This notation is read: is greater than but less than 0 + 0 Write both positive and negative values of on each side of the inequality and leave it in the middle. + 0 Solve for by moving each term to both sides of the inequality. 0 + 0 0 0 8 8 Remember that when dividing by a negative number the direction of the inequality sign changes. This notation is read: is greater than or eqal to but less than or equal to. 0 5
We eplained previously that equations that contain an absolute value cannot be equal to a negative number since an absolute value will always give you a positive number. In the case of linear inequalities, if an inequality is greater than or equal to a negative number, then there are infinite solutions. If the inequality is less than or equal to a negative number, then there is no real solution to the inequality. Eamples: + The absolute value will give you a positive number which will always be greater than any negative number. The inequality is true and has infinite solutions. + The absolute value will give you a positive number and no positive number will ever be less than or equal to a negative number. The inequality is false and has no real solution.
LINEAR INEQUALITIES AND ABSOLUTE VALUE EQUATIONS EXERCISES Solve for :. 7. 5 0. 6 +. 0 5. 8 + 6. ( + ) < 8 7. ( + ) > 8. + 0 9. + + 6 0. 7 5 =. =.. 6
LINEAR INEQUALITIES AND ABSOLUTE VALUE EQUATIONS ANSWER TO EXERCISES. 7. 5 0. 6 + 9 8 6 6 8. 0 5. 8 + 6. ( + ) < 8 8 6 + < 8 0 5 6 < 5 < 5 6 7. ( + ) > 8. + 0 9. + + 6 > 0 9 6+ + 8 9 > 5 8 9 7 8 5 < 7 6 9 6 7 0. 7 5 =. = 7 = + 5 = = 7 = 7 = = 6 7 = 7 7 = 7 = 6 or = 8 = 0 = or 0 = 8 =.. 6 6 6 6+ 6+ 9 9
LINEAR FUNCTIONS As previously described, a linear equation can be defined as an equation in which the highest eponent of the equation variable is one. A linear function is a function of the form f ( ) = a + b. The graph of a linear equation is a graphical view of the set of all points that make the equation true. The graph of any linear function is a straight line. A linear function can be represented in two ways, standard form and slope-intercept form. Standard form is a formal way of writing a linear equation, while slope-intercept form makes the equation easier to graph. Form Equation Note Standard A + By = C A and B are not 0. A > 0 Slope-intercept y = m + b m is the slope of the line and b is the y-intercept. To graph a linear function we must first identify the and y intercepts. The -intercept is the point where the graph crosses the -ais and the y-intercept is the point where the graph crosses the y-ais. To find the -intercept:. Set y = 0 in the equation.. Solve for. The value obtained is the -coordinate of the -intercept.. The -intercept is the point (, 0), with the value found in step. 5
To find the y-intercept:. Set = 0 in the equation.. Solve for y. The value obtained is the y-coordinate of the y-intercept.. The y-intercept is the point (0, y), with y the value found in step. Vertical Lines Equations of the form = a are vertical lines. The -coordinate of every point on the vertical line = a has the value "a," always, for any given value.. Horizontal Lines Equations of the form y = a are horizontal lines. The y-coordinate of every point on the horizontal line y = b has the value "b," always, for any given value. 6
Slope The slope of a line refers to the slant or inclination of the line. The slope is the ratio of the vertical change to the horizontal change between two points on the line. The slope can also be called the rise over run ratio because it tells you how many spaces to move up or down and how many spaces to move to the right. A positive sign will move the line up and a negative sign will move the line down. One important thing to remember is that the run will always be to the right, regardless of the sign. rise change in y m = = = run change in y The formula to find the slope of a line passing through the points ( ) and ( ),y,y is: y y m = Note: A horizontal line has slope of 0, while a vertical line has an undefined slope. Eample: Find the slope of the line: You can use any two points on a line to calculate its slope, your answer will be the same no matter which points you choose. y Choosing the points (, ) and (, 0): y m = y 0 = = = 5 5 6 7
Parallel Lines In the y-intercept form equation, y = m + b, m is the slope and b is the y-intercept. Two lines are parallel if their slopes are the same ( m = m ) and their y-intercepts are different ( b b ). Eample: ) y = + 9 m = b = 9 ) y = m = b = In the previous eample the slope of both equations is the same and their y-intercepts are different; therefore, the lines are parallel. 8 6 y 8 6 6 8 0 6 8 0 Perpendicular Lines Two lines are perpendicular if the slopes are the negative reciprocal of each other: m Eample: ) y = + 7 m = b = 7 ) y = + m = b = =. m The slope of equation is the negative reciprocal of the slope of equation ; therefore, the lines are perpendicular. 8 6 y 8 6 6 8 0 6 8 0 8
Finding the equation of a line To find the equation of a line when only points or a slope is given, use the point-slope form of a linear equation formula: y y = m where m is the slope of the line and ( ) ( ), y is a point on the line. Eamples: Find the equation of the line which passes through the point (, ) and whose slope is 5. Using the point-slope form equation we obtain: m( ) ( ( ) ) y y = y = 5 y = 5 + 0 y = 5 + Resultant Equation Find the equation of the line through the points (, 5) and ( 6, ). First find the slope between these two points using the slope equation: m = y y = 5 6 = = ( ) With the slope obtained and one of the two points given, use the point-slope form equation to find the equation of the line. y y = m ( ) y 5 = ( ( ) ) y 5 = + y = + Resultant Equation Eventually, you may be asked to find the equation of a line that is parallel or perpendicular to a given line. Just remember that the slopes of two parallel lines are eactly the same and that the slopes of two perpendicular lines are the negative reciprocals of each other. 9
Eamples: Find the equation of a line that is parallel to y = + and that passes through the point (, 7). y = + Since the equation is in slope-intercept form, the coefficient of is m = the slope of the line. Remember y=m+b (m is the slope!). Since the lines must be parallel, use the same slope and the given point to find the equation of the parallel line: m = m = ( ) ( ) y y = m y 7= y 7= 8 y = Parallel line to y = +. Find the equation of a line perpendicular to + y = 9and that passes through the point (, ). + y = 9 This equation is in standard form, we need to convert it to slope-intercept form: y=m+b. y = + 9 + 9 y = 9 y = + The slope is m =. Since the second equation must be perpendicular to the first equation, find the negative reciprocal. m = m = = m Using the slope m = and the point (, ): y y = m ( ) y ( ) = ( ( ) ) y+ = + y = Perpendicular line to + = 9. 0
LINEAR FUNCTIONS EXERCISES. Identify the false statement below a) The standard form of a linear equation is y = a + b b) The point (, 0) is on the graph of y = c) The graph of y = passes the vertical line test for functions d) The y-intercept of y = is (0, ). Identify the false statement below a) To find the -intercept of a graph, set y equal to zero and solve the equation for b) The graph of = is a horizontal line three units above the -ais c) Generally speaking, it is a good idea to plot three points when constructing the graph of a linear function d) To find the y-intercept of a graph, set equal to zero and solve the equation for y. The standard form of the equation 5y 6= 0 is 6 a) 6+ 5y = 0 b) y = + 6 5 c) 5y 6= 0 d) 6 5y = 0. Identify the true statement below a) All linear graphs are functions b) The graph of y = is a function c) To find the y-intercept of a function, let y equal to zero and solve for d) The graph of = is a function 5. Determine the slope and y-intercept of each equation: a) 6+ 7y = b) + 5= 0 c) y + 9= 0 6. Use the point slope form to find the equation of the line which passes through (, ) and whose slope is 7. 7. Write in standard form the equation of the line passing through the point (, ) with slope equal to m =. 8. Which of the following is the equation of a line in standard form passing through the point (, 0) and perpendicular to the line y =? a) + y = 6 b) + y = c) y = 6 d) y = 9. Find the equation of a line in standard form that passes through the points (, ) and (, ). 0. Determine whether the two lines are parallel, perpendicular or either: 5y = 6 and 6 0y = 7
LINEAR FUNCTIONS ANSWERS TO EXERCISES. A is the false statement. The standard form of a linear equation is A + By = C.. B is the false statement. The graph of = is a vertical line units to the right of the y-ais.. D: 6 5y = 0,A> 0. B: y = is an eample of a linear function. 6 5. a) m= and b= 7 b) m = undefinded and b does not eist c) m= 0 and b= 6. y = 7 7 7. y = 8. B: + y = is a perpendicular line to y = 9. + y = 7 0. The lines are parallel.
EXPONENTS Eponents are used to write long multiplications in a short way. The eponent will tell you how many times the number or variable needs to be multiplied. In this case the variable is a. Eamples: ( )( ) 6 6 = 6 a a a a = a In eponential notation, the number or variable being multiplied several times is called the base. The eponent, or number that tells you how many times you need to multiply, is called the power. is the base, and is the power Eponents are mostly used when dealing with variables, or letters, since it is easier and simpler to write than. Also, letters are used for they represent an unknown number, and so, the term: variable. There are a few rules used for simplifying eponents: Zero Eponent Rule Any number or letter raised to the zero power is always equal to. Eample: 0 = a 0 = Product Rule When multiplying the same base, the eponents are added together. Eample: Same base,, add the eponents, + = 7 = 7 Quotient Rule When dividing the same base, subtract the eponents. 5 Eample: Same base,, subtract the eponents, 5 = 5 =
Power Rule When the operation contains parentheses, multiply the eponent in the inside with the eponent on the outside. 6 Eample: ( ) y Multiply the eponents, 6 = 6 ( y ) = y When there is a fraction inside the parentheses, the eponent multiplies on the current power of the numerator and the denominator. However, this rule does not apply if you have a sum or difference within the parentheses; in that case a different rule will apply. Eamples: ( ) = = 9 = 6 y ( y ) = y 8!Be careful: a + b c + d a + b is not the same as! c + d In fact: a + b a + ab + b = c + d c + cd + d Negative Signs If the negative sign is outside the parentheses, perform the operations inside the parentheses and carry out the negative sign to the final answer. Eample: ( ) = ()()() = (7) = 7 However, if the negative sign is inside the parentheses, the negative sign will be affected by the eponent. Eample: ( ) = ( )( )( ) = 7 If the negative sign is inside the parentheses and the eponent is an even number, the answer will be positive. If the eponent is an odd number, then the answer will be negative. Eamples: ( ) = 8 Since the negative sign is outside the parentheses, carry it out to the final answer. ( 5) = ( 5)( 5) = 5 Since the negative sign is inside the parentheses, it needs to be carried out through the operation. ( ) = ( )( ) = 6 Even number of eponents, positive answer. ( ) = ( )( )( ) = 6 Odd number of eponents, negative answer.
Negative Eponents Whenever the problem or the answer to the problem contains negative eponents, they need to be changed to positive. An answer with negative eponents will most likely be counted wrong. To change negative eponents into positive eponents, get the reciprocal fraction. In simpler words, if the negative eponent is on the top, move it down; if the negative eponent is on the bottom, move it up. Eamples: Get the reciprocal, or move the negative eponent down. = = 5y Get the reciprocal of only the base with the negative eponent, the number stays in its place. 5y 5 5y = = y a b Get the reciprocal of the base with the negative eponent, the base with the positive eponent stays in its place. a b = a b = a b 5
EXPONENTS EXERCISES Simplify: 5 7.. 0 (for 0). ( ) 0 a + b (for a + b 0). 5 5. 6. ( ) 7. 5 8. 5 9. ( ) 0. y y. ( ). 9 9. ( a ). ( a ) 5. 7 c c c c c c 5 6. ( a b c ) ( ab) 7. ( a ) 8. ( a b) ( a b ) 9. 0. k. ( ) a ( a ). ( a b ). ( ) 5. ( ) 9 5. 8 7 6
EXPONENTS ANSWERS TO EXERCISES 5 7. =. 0 = 0 a =. 5 = = 5 5 = = 9. ( + b) 5. = = = 6. ( ) 7. 5 = = = 5 5 5 8. 5 5 5 = = 6 ( )( ) 9. ( ) = = 0. y y ( ) ( ) 0 y y = = =. ( ) = = = 8. 9 9 = 9 8. ( a ) = ( a )( a )( a )( a ) = a. ( ) ( )( )( )( ) 5. 7 c c c c c c 5 c c 0 0 a = a a a a = a 8 0 0 6 = = 6. ( ) ( ) a b c ab = a b c 7. ( a ) = a 9. 9 = 6 = 6. ( a ) = a = 6 a 0 8. ( ) ( ) 0. a b a b = a b = = k k k 6 ( a ) a b. = = 8 6 ( a b ) a b a. ( ) 5 =. ( ) 9 5. 8 8 8 9 = = = = 7 7 7 = 7
POLYNOMIALS Polynomials can be defined as the sum or difference of terms or epressions. Each term can be either a constant or variable, have one or more terms, and be composed of like terms or different terms. The epressions that can eist in a polynomial are defined as: Integer- a negative or positive whole number including zero. Eample: -5, -.-,0,, 5,6,7 etc Constants A single number in the equation that does not contain any variable. Eample:, 6 Coefficient The numerical part of a monomial. Eample: 7 is the coefficient of Degree The highest power to which a variable is raised. 7 y Eamples: Identify each term in the following polynomial: Constant: the single number in the equation Coefficient: the number in front of the variable Degree: the highest power of the variable 6 Give the degree of the following polynomial: 5 + 8 Degree: 6 It is a sith degree polynomial because the highest eponent of is 6. There are also different types of polynomials: Monomial A constant, or the product of a constant, and one or more variables raised to an integer. Eample: yz Polynomial Any finite sum (or difference) of terms. Eample: y z + 9 y z Binomial A polynomial consisting of eactly two terms. Eample: 7 Trinomial A polynomial consisting of eactly three terms. Eample: + There are special binomial rules that can be followed that can make them easier to solve. 8
( a + b) = a + ab + b ( a + b) = a + a b + ab + b ( a b) = a ab + b ( a b) = a a b + ab b ( a + b)( a b) = a b Eamples: ( + ) Solve using the binomial properties ( a + b) = a + ab + b ( + ) = + ( )( ) + ( + ) = + + ( y ) Solve using the binomial properties ( a b) = a a b + ab b ( y ) = y ( y) ( ) + ( y)( ) ( y ) = y 9y + 7y 7 Combining like Terms Polynomials can be short epression or really long ones. Probably the most common thing you will be doing with polynomials is combining like terms in order to simplify long polynomial epressions. Combining like terms is the process by which we combine eact same terms containing the same variable with the same eponent together to shorten the epression. Eamples: + Both terms contain the same variable. Therefore, they can be combined. + = 7 9 + + + Combine like terms together and then simplify. + + + 9 ( ) + ( + ) + ( + 9) + + 5 9
Evaluating Polynomials To evaluate polynomials substitute the variables for the given value and solve the equation. Eamples: Evaluate + at =. ( ) ( ) ( ) + Plug in for ; be careful with ( ) ( ) 7 9 + + 5 9 + 6+ 9 If P( ) = 5 + 7, find P ( ) ( ) ( ) ( ) = 5 ( ) 8+ 7 ( ) = 9 P = 5 + 7 P parentheses and negative signs.. This is the same thing as plugging a for Addition and Subtraction of Polynomials The addition and subtraction of polynomials consist of combining like terms by grouping together the same variable terms with the same degrees. In the case of subtraction, if the subtraction sign (or negative sign) is outside parentheses, the first thing to do is to distribute the negative sign to each of the terms inside the parentheses. Eamples: Simplify: ( + + 5) + ( + ) ( 5 ) ( ) + + + + Combine same degree terms together. + + + + 5 ( + ) + ( ) + ( + ) + ( 5 ) = + + Simplify: ( + 5y) + ( y) 0
+ 5y+ y Combine s with s and y s with y s. ( + ) + ( 5y y) 5 + y Simplify: ( + + 5 ) ( 8 5+ 6) ( + + 5 ) ( 8 5 + 6 ) + + 5 + 8 + 5 6 ( ) + ( + 8 ) + ( 5+ 5) + ( 6) = + + 0 0 Multiplication of Polynomials To multiply polynomials the same rules apply as with eponents since we are dealing with variables with different eponents. Remember that when multiplying, the eponents add together. Eample: ( )( 0 ) Multiply the constant terms together first and then the variables. ( 0)( ) = Add eponents together. 5 = 0 Net is the one-term polynomial times a multi-term polynomial. Distribute the one-term polynomial outside the parentheses to each term inside the polynomial. Eample: ( 0) + Distribute the to all the terms inside the parentheses. ( ) ( ) ( 0) = = + 0 Multiplication of two two-term polynomials is a little more comple. To make it easier we use the FOIL method: a method that simplifies the process by following a pattern. FOIL stands for First, Outer, Inner and Last. Start with the first terms in each parenthesis, followed by the outer terms, then the inner terms, and finally the last terms. This method makes it easier to remember which terms to multiply and reduces the chance of forgetting to multiply some terms. Eample: Use the FOIL method to simplify ( + 5)( + ) "first": ( )( ) = ( + 5)( + )
"outer": ( )( ) = 6 + 6+ 5+ 5 "inner": ( 5)( ) = 5 = + + 5 "last": ( 5)( ) = 5 In order to multiply one multi-term polynomial by another multi-term polynomial, break the smaller polynomial apart and multiply each individual term by the longer polynomial. Eample: ( )( 7) + + + Take the smaller polynomial and multiply each of its individual terms by the largest polynomial. Division of Polynomials ( + + 7)( ) + ( + + 7)( ) ( ) + ( ) + ( ) 7( ) + ( ) + ( ) + ( ) 7( ) + + 7 + + 6 + 8 + ( + ) + ( + 6 ) + ( 7+ 8) = + 5 + 0 9 When dealing with polynomials divided by a single term the division can be treated as a simplification problem and the action is just to reduce its terms to the lowest possible. Eamples: Simplify: + In this case, there is a common factor in the numerator (top) and denominator (bottom), so it is easy to reduce this fraction. There are two ways of proceeding. Split the division into two fractions, each with only one term on top, and then reduce: + = + = + Or factor out the common factor from the top and bottom and then cancel: ( + ) + = = + Either way, the answer is the same: + Simplify: 8 6 9 +
Method : 8 + 6 9 8 = 9 6 + 9 = + Method : If you recognize a common factor that can be taken out of the parentheses and used to cancel the term in the denominator use it. 9 ( + ) 9 = + The answer is the same by any method: + Long Division If you come about a more complicated polynomial division you can use the long division method to do the operation. Long division of polynomial works just as a regular long division, with the eception that in this case variables are included. Eample: Divide 6 by +. Step Set up the division. + 6 Step Let s look at the first term inside,. On the outside we have an, so to get to we need to multiply. Write this term on top. + 6 Step Multiply the term on the top by the term outside and write it at the bottom, ( + ) = +. However, to be able to eliminate the terms we need to change the sign,. + 6 Step Perform the required operations.
+ 6 8 Step 5 Bring down the net term inside the division, in this case. + 6 8 Step 6 Repeat step through to eliminate the remaining terms. + 8 6 8 8 + 6 The solution to the long division is 8. Divide 5 0 + by + : + + + 5 0 ( ) + = 6 + 0 Subtract polynomials and carry down the net term, which is 0 + = +, to subtract, ( ) 6 + ( ) 6 6 = 6 + Subtract polynomials and carry down the net term, + = + =, to subtract, ( ) ( + ) = +, to subtract, ( ) 7 Since the division did not come out even, the answer is the polynomial on top of the division plus the 7 remainder as a fraction with the divisor as the denominator: + + +.
POLYNOMIALS EXERCISES Perform the indicated operations:. ( 7+ ) + ( 8 ). ( + 6+ ) + ( 7). ( + ) + ( + + 0). ( ) + ( + ) + ( + ) 5. ( + 6+ 9) + ( + 7) 6. ( + 08) ( + + 6) 7. ( 8 + 7) ( + 7 ) 8. ( 5+ ) ( 7 ) 9. ( + 6+ ) + ( 5 ) ( + 5+ 5) 0. ( + ) ( + ) ( + 7). ( + )( + ). ( + )( + + ). ( )( + 7 ). ( y)( + y + y ) 5. ( + 6y)( y y ) Solve by long division: 6. ( 8+ ) ( ) 7. ( 8 + 6 + 0+ ) ( ) 8. ( 7 + 8) ( + ) 9. ( 5y 6y ) ( + y) 0. ( 8 5) ( + 5) 5
POLYNOMIALS ANSWER TO EXERCISES 7+ + 8 = 5+. ( ) ( ). ( + 6+ ) + ( 7) = 5 +. ( + ) + ( + + ) = + +. ( ) + ( + ) + ( + ) = + + 6 5. ( ) ( ) 6. ( 0+ 8) ( + + 6) = + 7. ( 8 + 7) ( + 7 ) = + 6 6 8. ( 5+ ) ( 7 ) = + + 5 9. ( ) ( ) ( ) 0. ( ) ( ) ( ) 0 + 6+ 9 + + 7 = + 8+ 6 + 6 + + 5 + 5 + 5 = 6 6 + + + 7 = 5 8 + + = + 7+. ( )( ). ( )( ). ( )( ). ( )( ) 5. ( )( ) + + + = + 9 + + + 7 = + 5 5+ y + y + y = y + y 6y + 6y y y = + y 8y y 6. ( ) ( ) 8+ = + 7. ( ) ( ) 8 + 6 + 0+ = + + 5 + 7 7 + 8 + = 7 9+ + 5y 6y + y = 6y 8. ( ) ( ) 9. ( ) ( ) 0. ( ) ( ) 8 5 + 5 = 6
FACTORING Factoring is similar to breaking up a number into its multiples. For eample, 0=5*. The multiples are 5 and. In a polynomial it is the same way, however, the procedure is somewhat more complicated since variables, not just numbers, are involved. There are different ways of factoring an equation depending on the compleity of the polynomial. Factoring out the greatest common factor The first thing to do when factoring is to look at all terms and break up each term into its multiples: Eamples: Factor: 8 + + 0 ( 8 + + 0) In this polynomial the only variable in common to all is. (++5) The two is also common to all terms. Therefore, this is as far as the polynomial can be factored. Factoring by grouping Factor: 6 + 8 + 6 6 + 8 + 6 The common variable is, and the smallest eponent is. The common multiple is. + + 8 Therefore, the greatest common factor is. ( ) You can also look at it this way: ( )( ) = 6 ( )( ) = 8 ( )( 8) = 6 7
There will be occasions when there are no common factors for all terms, but there will be terms that have a variable in common while other terms have a different variable in common. In this case, we can factor by grouping together these common terms. Eamples: Factor: a + ay + b + by a + ay + b + by Notice how there are two variables, and y. Group a + b + ay + by together the s and the y s and then a + b + y a + b factor out the common factor in each group. ( ) ( ) Factor: + y y y by grouping two terms at a time. ( ) + y ( ) Notice how the two polynomials inside the parentheses are similar. To make them the same change the sign in the second polynomial. + y There are two variables present, and y. Start ( ) y ( ) Now that they are the same, factor out the common term. y Now it is completely factored out. ( )( ) Factoring a difference of squares For a difference of squares, a b, the factors will be ( a b)( a + b). Eamples: y = ( y)( + y) ( ) ( ) ( )( ) ( )( )( ) = = + = + + Factoring the sum and difference of two cubes For the addition or subtraction of two cubes, the following formulas apply: ( )( ) ( )( ) a + b = a + b a ab + b a b = a b a + a + bb Eamples: + 8 ( ) ( ) ( ) ( )( ) = + Manipulate to be in a + b form. ( ) = + + follow formula. 8
( )( ) = + + 8y 7 Manipulate to be in a b = y follow formula. ( ) ( ) ( y ) ( y) ( y)( ) ( ) = + + ( y )( y 6y 9) = + + form. Factoring trinomials Factoring trinomials is based on finding the two integers whose sum and product meet the given requirements. Eample: Find two integers whose sum is and whose product is 0. 9 Step The product is positive, the sum is negative, therefore, both integers must be =+ and + = negative. ( )( ) ( ) ( ) Step Possible pairs of factors that will give a positive answer:. ( )( 0) = 0. ( )( ). ( )( 5) = 0. ( )( ) 0 = 0 5 6 = 0 Step The pair whose sum is is 5 and 6. Therefore, the critical integers are 5 5 6 =. and 6, ( ) Factoring trinomials of the form + b + c To factor trinomials of the form + b + c follow the same principle described above. Find two integers whose sum equals the middle term and whose product equals the last term. Eamples: Factor: 6 Step Find two integers whose sum is equal to and whose product is equal to 6. The product is negative; therefore, we need integers that have different sign. The sum is negative, so the bigger integer is negative. Step Possible pairs of factors that will give us 6:. ( )( ) = 6. ( )( ) = 6. ( )( 6) = 6. ( )( 6) = 6 Step The pair whose sum is is and, ( + = ) The critical integers are and..
Step The factors for 6 are ( )( + ). Factor: + 6 + 60 Step Find two integers whose sum is 6 and whose product is 60. Both integers must be positive since their sum and product are positive. Step Possible pairs of factors:. ( )( 60) = 60. ( )( 0) = 60 5. ( )( ). ( )( 0) = 60. ( )( 5) = 60 6. ( )( ) 5 = 60 6 0 = 60 Step The pair whose sum is 6 is (6, 0). Therefore, the critical integers are 6, 0. Step The factors for + 6+ 60 are ( + 6)( + 0 ). Factoring trinomials of the form a + b + c In this type of trinomials the coefficient of is not, therefore we must consider the product of Eamples: Factor: 7 Step Find two integers whose sum is 7 and whose product is ( )( ) = 8. The a c. integers must be of different signs since their product is a negative number. The larger integer is negative since their sum is negative. Step Possible pairs of factors: = 8. ( )( 8) = 8. ( )( ). ( )( 8) = 8. ( )( ) = 8 Step The pair whose sum is 7 is and 8. The critical integers are and 8. Step Use the critical integers to break the first degree term into two parts: 7 = 8 + Step 5 Factor by grouping the first two and last two terms: 8 + = ( ) + ( ) 0
There are now two terms in the epression. The binomial factor in each of these two terms should be the same; otherwise, there is an error. In this case, the common binomial factor is( ). Step 6 Factor out the common binomial factor: ( ) + ( ) = ( )( + ) Factor: 6 + 0 Step Find two integers whose sum is and whose product is 6(0) = 0. Since the product is positive and the sum is negative, the two integers are negative. Step Possible pairs of factors: 8 5 = 0. ( )( 0) = 0. ( )( 0) = 0 7. ( )( ). ( )( 60) = 0 5. ( 5)( ) = 0 8. ( )( ). ( )( 0) = 0 6. ( 6)( 0) = 60 0 = 0 Step The sum of 8 and 5 is. Therefore, the critical integers are 8 and 5. Step Use the critical integers to break the first degree term into two parts: 6 + 0 = 6 5 8 + 0 Step 5 Factor, separately, the first two and last two terms: 6 5 8 + 0 = ( 5) ( 5) Step 6 Factor out the common binomial factor. ( 5) ( 5) = ( 5)( )
FACTORING EXERCISES. 0 8. ( ) 5( ) ( ). + 9 8. 5. 6. 6 0 + 5 y y a 8 6 7. ( + 5) + ( + 5) + 8. 9. 6 t + t 8 + 7 0. 00 00 600. 8 5. + y + y. ( + ) + ( + ) + 5( + ). + 5. y + y + + 6 6. 8a 50 7. + 8 + 5 8. 6 0 50 9. 9 + + 0 0. +
FACTORING ANSWER TO EXERCISES. ( ) 5( ) ( ). 0 8 + 8 ( )( 5 ) ( ) ( ) ( )( ) + ( )( 6 + ) + ( )(( ) + ( )) ( )( + )( ) = ( ) ( + ). + 9 8. ( + ) 9( + ) ( 9)( + ) 0 + 5 y ( 0 + 5) y ( 5) y 9 = ( + )( ) Difference of squares: ( a b ) = ( a b)( a + b) ( )( + )( + ) where ( ) ( 5 y)( 5 + y) a = 5 and b= y 5. 6 6 y 6. ( ) ( ) a 8 6 6 y = y Difference of cubes: Difference of squares: ( a b ) = ( a b)( a + b) where ( )( ) y y y 6 6 a b = ( a b)( a + a + bb ) a b y = and = a a = and b = = + + a = a a + + = + a a ( a) Sum & Difference of Cubes: ( )( ) ( )( ) a + b = a + b a ab + b a b = a b a + a + bb = ( + y)( y + y )( y)( + y + y ) = ( + y)( y)( + y + y )( y + y )
7. ( + 5) + ( + 5) + = ( + 7)( + 7) = ( + 7) 8. 6 ( )( ) ( ) ( ) t + t = t t = t t + t+ 9. ( ) ( ) 8 + 7 = 6+ 9 = 0. 00 00 600 = 00( )( + ). 8 5 = ( + 5)( ). + y + y = ( + y)( + y). ( + ) + ( + ) + 5( + ) = ( + )( + )( + 5). + = ( + )( ) 5. y + y+ + 6= ( + )( y+ ) 6. 8a 50 = ( a+ 5)( a 5) 7. + 8+ 5 = ( + 5)( + ) 8. 6 0 50= ( 5)( + 5) 9. 9 + + 0 = ( + )( + 5) 0. + = ( )( + )