BIG idea: Function

You can represent quantities using variables and algebraic expressions. You can represent some relationships between quantities using equations.

Task 1

The polynomial 2 x 3 + 9 x 2 + 4 x − 15 2 x cubed , plus , 9 x squared , plus 4 x minus 15  represents the volume in cubic feet of a rectangular holding tank at a fish hatchery. The depth of the tank is ( x − 1 ) open x minus 1 close  feet. The length is 13 feet.

1. Use synthetic division to help you factor the volume polynomial. How many linear factors should you look for? What are they?
2. Assume the length is the greatest dimension. Which linear factor represents the 13-ft length? What are the dimensions of the tank? What is its volume? What is the value of x? Do you get the same volume if you substitute the value of x into 2 x 3 + 9 x 2 + 4 x − 15 ? 2 x cubed , plus , 9 x squared , plus 4 x minus 15 question mark

BIG idea: Equivalence

You can use the Binomial Theorem and properties of algebra to rewrite some powers.

Task 2

Show that the following equation is true for all values of a and b.

[ ( a − b ) + 1 ] 5 = a 5 − 5 a 4 ( b − 1 ) + 10 a 3 ( b − 1 ) 2 − 10 a 2 ( b − 1 ) 3 + 5 a ( b − 1 ) 4 − ( b − 1 ) 5 left bracket . open , eh minus b , close . plus 1 . right bracket to the fifth . equals , eh to the fifth , minus 5 , eh to the fourth . open , b minus 1 , close . plus 10 , eh cubed . open , b minus 1 , close squared . minus 10 , eh squared . open , b minus 1 , close cubed . plus 5 eh . open , b minus 1 , close to the fourth . minus . open , b minus 1 , close to the fifth

BIG idea: Solving Equations and Inequalities

A polynomial P ( x ) p open x close  of degree n , n ≥ 1 , n comma n greater than or equal to 1 comma  and its related polynomial function y = P ( x ) y equals p open x close  have n complex zeros. The zeros are identical to the n complex roots of the related polynomial equation P ( x ) = 0 . p open x close equals 0 .

Task 3

Four of these five polynomial functions have identical zeros. The fifth has exactly two zeros in common with each of the other functions. Write this fifth function as a product of its linear factors.

P 1 ( x ) = x 3 − 6 x 2 + 11 x − 6 P 2 ( x ) = x 4 − 5 x 3 + 7 x 2 − 5 x + 6 P 3 ( x ) = x 4 − 7 x 3 + 17 x 2 − 17 x + 6 P 4 ( x ) = x 4 − 8 x 3 + 23 x 2 − 28 x + 12 P 5 ( x ) = x 4 − 9 x 3 + 29 x 2 − 39 x + 18 table with 5 rows and 1 column , row1 column 1 , p sub 1 . open x close , equals , x cubed , minus 6 , x squared , plus 11 x minus 6 , row2 column 1 , p sub 2 . open x close , equals , x to the fourth , minus 5 , x cubed , plus 7 , x squared , minus 5 x plus 6 , row3 column 1 , p sub 3 . open x close , equals , x to the fourth , minus 7 , x cubed , plus 17 , x squared , minus 17 x plus 6 , row4 column 1 , p sub 4 . open x close , equals , x to the fourth , minus 8 , x cubed , plus 23 , x squared , minus 28 x plus 12 , row5 column 1 , p sub 5 . open x close , equals , x to the fourth , minus 9 , x cubed , plus 29 , x squared , minus 39 x plus 18 , end table