# We start with the basic operations on polynomials, that is adding, subtracting, and multiplying.

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1 R. Polnomials In this section we want to review all that we know about polnomials. We start with the basic operations on polnomials, that is adding, subtracting, and multipling. Recall, to add subtract polnomials we simpl combine like terms. To multipl polnomials we have to multipl each term in the first polnomial b each term in the second polnomial and combine like terms. When multipling two binomials we can use the FOIL method (First-Outer- Inner-Last). Eample : Perfm the following operations. a. ( ) ( ) b. ( ) ( ) c. ( ) d. ( ) ( )( ) a. Since the parentheses are not needed in this epression, we can simpl combine like terms. ( ) ( ) 7 9 b. This time we need to start b distributing the negative. Then we combine like terms. We get ( ) ( ) c. To multipl these two binomials we will simpl multipl each term in the first binomial b each term in the second binomial and combine like terms. We get ( )( ) d. Lastl, we again multipl each term in the first polnomial b each term in the second polnomial and combine like terms. We get ( )( ) 7 6 Net we want to review basic facting of polnomials. There are four tpes of facting we want to review: facting the GCF, fact b grouping, fact b trial facts, facting b fmula. The technique we use f facting is as follows.

2 General Strateg f Facting. Fact out the GCF.. Count the number of terms in the remaining polnomial. If it has a. Four terms- we fact b grouping b. Three terms- we fact b trial facts c. Two terms- we use one of the following facting fmulas a b a b a b, called the difference of squares i. ( )( ) ii. a b ( a a ab b ) iii. a b ( a b)( a ab b ), called the difference of cubes, called the sum of cubes. Check each fact to see if ou can fact it further. If so, then we fact again. Eample : Fact completel. a. 8 6 b. b 8 c. 8 a. We start b facting out the GCF, which is clearl. We get 8 6 ( ) We see that we are left with a three term polnomial (a trinomial). So we fact b trial facts. That is, we know that ever trinomial that facts, will fact as two binomials. So we simpl guess at the wa it facts and check b multipling it back out. Some of the ke ideas are the first terms in each binomial will have to have a product that is the leading term of the trinomial and the second terms in each binomial will have to have a product that is the last term of the trinomial. So b trial facts we get ( ) ( )( ) b. Since we can see there is clearl no GCF here we can begin b counting the terms. Since we have two terms we are going to have to fact b a fmula. Clearl our onl choice is the difference of squares. So we will make each term into a square as follows. ( b ) b 8 9 Now we use the fmula to fact. This gives ( b ) 9 ( b 9)( b 9) Now we need to check each fact to see if we can do me facting. We see that the first binomial is again a difference of squares. Therefe we must fact it again. We get ( b 9)( b 9) ( b )( b )( b 9) We cannot fact a sum of squares, therefe the epression is completel facted. c. Lastl, we see no GCF so we can start b facting b grouping. That is, we group together the first two terms and the last two terms and fact what we can out of those pairs. Then we see if there is a common binomial fact. If not we tr grouping another wa. We proceed as follows 8 ( ) 8( ) ( )( 8) Now we check to see if there is me facting to be done. Clearl the first binomial is the difference of square and the second binomial is the difference of cubes. So we fact with the fmulas given above. We get

3 ( )( 8) ( )( )( )( ) ( ) ( )( ) Since the trinomial does not fact, the epression is completel facted. Facting polnomials leads us directl to solving polnomial equations. To do this we use the following propert. Zero Product Propert If a b, then a b. So to solve a polnomial equation, we get one side of the equation to be zero, fact completel, set each fact to zero and continue solving. Eample : Solve. a. b. c. ( ) ( ) a. We start b getting a zero on one side and facting completel. This gives ( )( ) Now, accding to the zero product propert we can set each fact to zero. We can then finish solving. We get, So the solution set is { }. b. Again we start b getting a zero on one side and facting. We then set each fact to zero and solve. We have So the solution set is { 0, }. ( ) c. This time we need to start b multipling out the binomials, then we can proceed as usual. We get

4 9 So the solution set is { 8, 9}. ( )( ) 0 ( 9)( 8) Lastl, we want to do some wd problems which require us to use polnomials. Eample : A certain graphing calculat is rectangular in shape. The length of the rectangle is cm me than twice the width. If the area of the calculat is cm, what are the length and width of the calculat? First we should draw a picture to represent the situation. Since the length of the rectangle is cm me than twice the width we have w Lw Since we know that the area of a rectangle is A l w ( w )w A l w we have Now that we have an equation, we simpl need to solve it. We get w w w ( w w 7) ( w 9)( w 8) ( w ) w w 9 w 8 w 9 w 8 Since w is the width of a rectangle, it cannot be negative. Thus, w 8. Now we can simpl find l. Since l w, l (8) 8. So the calculat is 8 cm b 8 cm. w

5 Eample : The fmula f the number of games to be plaed in a soccer league where each team is to pla each other twice is N, where is the number of teams in the league and N is the number of games to be plaed. If a league wants to limit the games to a total of games, how man teams can be in the league? F this eample we are given a fmula which we must interpret. Since N is the number of games to be plaed and we want a total of games, N. Putting that into the fmula we get an equation we can solve. ( )( ) N Since represents the number of teams, it cannot be negative. Therefe, the league must have teams to pla a total of games. R. Eercises Perfm the indicated operations... ( ) ( ) ( ) ( ) ( 7 ) ( ) ( ) ( 6)... ( 9) ( ) 6. ( 9) ( ) 7. ( ) ( ) 8. ( ) ( 7 6 ) 9. ( ) ( ) 0. ( ) ( ). ( 6 ) ( ). ( ) ( ). ( ) ( ). ( ) ( ). ( ) 6. ( ) ( )( ) ( )( ) ( ) ( ) ( 0)( 0) ( )( )... ( ). ( )( 6 ). ( 9)( ) 6. ( )( ) 7. ( )( ) 8. ( )( ) 9. ( )( )( ) 0. ( )

6 ) ( )( ). (.. ( )( ). ( )( ). ( )( ) 6. ( )( ) Fact completel u u 6u d 7 7d z 6a b 0a b 8ab 8 t 7t 6. a a 60a a b 88ab a b a 7ab b 9 z 6a b a b a b ( ) ( ) ( ) ( t ) 6t( t ) 9( t t ) Solve the following ( 6 ) ( ) 6 7. The length of a rectangle is meters less than times the width. The area is square meters. Find the dimensions. 7. The width of a rectangle is feet less than times the length. The area is square feet. Find the dimensions. 7. The length of a rectangle is inches me than times the width. The area is square inches. Find the dimensions. 76. The area of a triangle is 0 square feet. The height is feet less than the base. Find the dimensions. 77. The area of a triangle is 9 square meters. The height is 7 meters me than the base. Find the dimensions. 78. The area of a triangle is square inches. The base is 9 inches less than the height. Find the dimensions.

7 Find the time it takes the object to hit the ground, where h is in feet, and t is in seconds. 79. h t 6t 80. h 6t 6t 8. h 76t 6t 8. h 6t 8t 6 8. h 6t 0t A ball rolls down a slope and travels a distance d is feet. 8. A ball rolls down a slope and travels a distance d is feet. t d 6t feet in t seconds. Find t when t d 6t feet in t seconds. Find t when

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