﻿ Polynomial Functions of Higher Degree with Modeling Worksheet | Problems & Solutions

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. $x$³ - 32$x$² + 252$x$ - 180 = 0 b. $x$³ - 16$x$² + 63$x$ - 45 = 0 c. 2$x$³ - 32$x$² + 126$x$ - 45 = 0 d. $x$² - 32$x$ + 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.]

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

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.

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.

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.

6.
Write the at most number of local extrema of the polynomial function $f$ ($x$) = 5$x$3 - 28$x$2 + 96$x$ + 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.]

7.
Find the atmost number of local extrema possessed by the polynomial $f$ ($x$) = - $x$5 + 6$x$4 + 36$x$3 - 4$x$2 - 90$x$ + 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.]

8.
Find the zeros of the function $f$ ($x$) = 8$x$2 - 10$x$ + 3 algebraically.
 a. $\frac{1}{2}$ and $\frac{3}{4}$ b. - $\frac{1}{2}$ and - $\frac{3}{4}$ c. $\frac{1}{2}$ and - $\frac{3}{4}$ d. - $\frac{1}{2}$ and $\frac{3}{4}$

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.

9.
Find the maximum number of zeros possible for the polynomial $f$ ($x$) = $x$3 - 30$x$2 + 93$x$ - 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.]

10.
Find the maximum number of zeroes possible for the polynomial $f$ ($x$) = $x$4 - 23$x$3 + 93$x$2 - 238$x$ + 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.]