Polynomial Functions of Higher Degree with Modeling Worksheet

Polynomial Functions of Higher Degree with Modeling Worksheet
  • Page 1
 1.  
The dimensions of a piece of tin are 18ft and 14ft. From this piece of tin an open box with a volume of 180ft³ is to be made by cutting a square of the same size from each corner and folding up the edges. Find the polynomial equation to find the length x of the side of the square piece that is to be removed.
a.
x³ - 32x² + 252x - 180 = 0
b.
x³ - 16x² + 63x - 45 = 0
c.
2x³ - 32x² + 126x - 45 = 0
d.
x² - 32x + 72 = 0


Solution:


Draw a diagram showing the square corners to be cut from the piece of tin.
[Analyse and understand the Problem.]

Volume of box = 180 ft3

As x represents the length of a side of each square which is to be removed, then 18 - 2x represents the length of the box and 14 - 2x represents the width.

x (18 - 2x) (14 - 2x) = 180
[Use Volume = Length × breadth × height.]

252x - 64x² + 4x³ = 180
[Expand.]

x³ - 16x² + 63x - 45 = 0
[Simplify.]


Correct answer : (2)
 2.  
A unit cube is removed from a cube of side n cm. If the volume of the remaing cube is 728 cm3, then find the side of the cube.
a.
81
b.
9
c.
8
d.
10


Solution:

The volume of the cube with side x cm = n3 cm3
[Write the volume of the cube.]

The volume after the removal of a unit cube = n3-(1)3 = n3 - 1
[Write the volume of the remaining part in terms of n.]

n3 -1 = 728
[The volume of the remaining part is 728 cm3.]

n3 - (9)3 = 0

(n - 9)(n2 + 9n + 81) = 0
[Factor.]

n = 9
[n2 + 9n + 81 = 0 has no real solution.]

The side of the cube is 9 cm


Correct answer : (2)
 3.  
Which of the following is the at most number of local extrema of a polynomial function of degree 5?
a.
5
b.
10
c.
25
d.
4


Solution:

A polynomial function of degree n has at most n - 1 local extrema.

So, the atmost number of local extrema of a polynomial function of degree 5 is 4.


Correct answer : (4)
 4.  
Choose the largest number of zeros that a polynomial of degree 12 can have.
a.
144
b.
24
c.
12
d.
11


Solution:

A polynomial function of degree n has at most n zeros.

So, the at most number of zeros of a polynomial function of degree 12 is 12.


Correct answer : (3)
 5.  
Find the solutions of (x + 3)3 (x + 4) (x - 6) = 0.
a.
- 3, - 4, - 6
b.
- 3, - 4, 6
c.
- 3, 4, 6
d.
- 3, 4, - 6


Solution:

(x + 3)3 (x + 4) (x - 6) = 0

x + 3 = 0, x + 4 = 0, or x - 6 = 0

x = - 3, x = - 4, or x = 6

So, the solutions of the equation are x = - 3, - 4, and 6.


Correct answer : (2)
 6.  
Write the at most number of local extrema of the polynomial function f (x) = 5x3 - 28x2 + 96x + 73.
a.
6
b.
7
c.
3
d.
2


Solution:

The degree of the polynomial function f (x) = 5x3 - 28x2 + 96x + 73 is n = 3.

The at most number of local extrema of the given polynomial function is n - 1 = 3 - 1 = 2.
[A polynomial function of degree n has at most n - 1 local extrema.]


Correct answer : (4)
 7.  
Find the atmost number of local extrema possessed by the polynomial f (x) = - x5 + 6x4 + 36x3 - 4x2 - 90x + 41.
a.
5
b.
1
c.
4
d.
3


Solution:

The degree of the polynomial function f (x) = - x5 + 6x4 + 36x3 - 4x2 - 90x + 41 is n = 5.

The atmost number of local extrema possessed by the given polynomial function is n - 1 = 5 - 1 = 4.
[A polynomial function of degree n has at most n - 1 zeros.]


Correct answer : (3)
 8.  
Find the zeros of the function f (x) = 8x2 - 10x + 3 algebraically.
a.
1 2 and 3 4
b.
- 1 2 and - 3 4
c.
1 2 and - 3 4
d.
- 1 2 and 3 4


Solution:

8x2 - 10x + 3 = 0
[Solve the related equation f(x) = 0.]

8x2 - 6x - 4x + 3 = 0
[Factor.]

(2x - 1) (4x - 3) = 0

2x - 1 = 0 or 4x - 3 = 0

x = 1 / 2 or x = 3 / 4

So, the zeros of f(x) are = 1 / 2 and 3 / 4.


Correct answer : (1)
 9.  
Find the maximum number of zeros possible for the polynomial f (x) = x3 - 30x2 + 93x - 77.
a.
2
b.
1
c.
3
d.
4


Solution:

The degree of the polynomial function f (x) = x3 - 30x2 + 93x - 77 is n = 3.

The maximum number of zeros possessed by the given polynomial function is n = 3.
[A polynomial function of degree n has at most n zeros.]


Correct answer : (3)
 10.  
Find the maximum number of zeroes possible for the polynomial f (x) = x4 - 23x3 + 93x2 - 238x + 90.
a.
1
b.
2
c.
3
d.
4


Solution:

The degree of the polynomial function f (x) = x4 - 23x3 + 93x2 - 238x + 90 is n = 4.

The maximum number of zeroes possessed by the given polynomial function is n = 4.
[A polynomial function of degree n has at most n zeros.]


Correct answer : (4)

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