Lesson Vocabulary

• sum of cubes
• difference of cubes

Problem 1 Solving Polynomial Equations Using Factors

What are the real or imaginary solutions of each polynomial equation?

1. 2 x 3 − 5 x 2 = 3 x 2 , x cubed , minus 5 , x squared , equals 3 x

   2 x 3 − 5 x 2 − 3 x = 0 Rewrite in the form  P ( x ) = 0. x ( 2 x 2 − 5 x − 3 ) = 0 Factor out the GCF , x . x ( 2 x + 1 ) ( x − 3 ) = 0 Factor  2 x 2 − 5 x − 3. x = 0 or 2 x + 1 = 0 or x − 3 = 0 Zero Product Property x = 0 x = − 1 2 x = 3 Solve each equation for  x . table with 5 rows and 3 columns , row1 column 1 , 2 , x cubed , minus 5 , x squared , minus 3 x , column 2 equals 0 , column 3 cap rewriteintheform p open x close equals 0. , row2 column 1 , x . open . 2 , x squared , minus 5 x minus 3 . close , column 2 equals 0 , column 3 cap factoroutthecap gcap ccap f . comma x . , row3 column 1 , x . open , 2 x plus 1 , close . open , x minus 3 , close , column 2 equals 0 , column 3 cap factor , 2 , x squared , minus 5 x minus 3. , row4 column 1 , x equals 0 or 2 x plus 1 , column 2 equals 0 or x minus 3 equals 0 , column 3 cap zerocap productcap property , row5 column 1 , x equals 0 x , column 2 equals negative , 1 half , x equals 3 , column 3 cap solveeachequationfor . x . , end table

   The solutions are 0 , − 1 2 , 0 comma negative , 1 half , comma  and 3.