Algebra Homework: Polynomials, Zeroes, and Wind Problems

This document presents a comprehensive solution to a mathematics homework assignment. The assignment covers several key concepts, including the analysis of a polynomial function, determining its degree, the maximum number of turning points, and the number of zeroes. It also includes the application of the leading coefficient test and the rational zero theorem to identify possible zeroes. Furthermore, the solution applies Descartes's rule of sign to analyze the nature of the roots. The assignment also includes a wind problem that requires the application of direct variation formulas to determine the constant and solve for different wind speeds. The solution provides step-by-step explanations and calculations for each problem, making it a useful resource for students studying algebra and related mathematical concepts.

Running head: MATHEMATICS
MATHEMATICS
Name of the Student
Name of the University
Author Note

1
MATHEMATICS
1. Given Polynomial f ( x ) =2 x4 +3 x3−6 x2−5 x+6
a. Degree of the polynomial
Solution: In a polynomial, the degree of the polynomial is the highest degree from the exponents
of the variables.
According to the question the variable is here x and the highest degree is 4.
Hence, the degree of the polynomial is 4. Ans.
b. Maximum no of turning points.
Solution: For any polynomials the highest numbers of turning points is equal to the (highest
degree of any term – 1).
In this case, the highest degree is 4. Hence the maximum number of turning points are 3.
c. Numbers of zeros the polynomial have
Solution: A polynomial of degree ‘n’ can have ‘n’ numbers of distinct real roots at mots. In this
case the numbers of the zero will be 4.
d. Leading coefficient test
Solution: For the Leading coefficient test, the degree of the function and leading coefficient is
required. Let the leading coefficient is ‘a’, and the degree is ‘n’.
According to the leading coefficient test, certain rules are needed to be followed.
i. If n is odd and a is positive:
The left hand of the graph falls and right hand of the graph rises.

2
MATHEMATICS
ii. If n is odd and a is negative:
The left hand of the graph rises and right hand of the graph falls.
iii. If n is even and a is positive:

3
MATHEMATICS
The left hand of the graph rises and right hand of the graph rises.
iv. If n is even and a is negative:
The left hand of the graph falls and right hand of the graph falls.

4
MATHEMATICS
In this scenario, the function f (x) has degree 4 which is even and leading coefficient +2 which is
positive.
Hence, the left hand of the graph will rise and right hand of the graph will rises
{ x →−∞, then y → ∞
x → ∞, then y → ∞ .
e. Rational Zero Theorem
Solution: According to the rational zero theorem, the ration zero of the function will be in the
form of
± p
q = factors of p
factors of q
In this question we have the value p and q where the p = 6 and q= 2
Hence, ¿>± p
q =± 6 , ±3 ,± 2 ,± 1
±2 , ±1
¿>± p
q =± 6
2 ,± 3
2 , ± 2
2 ,± 1
2 , ± 6
1 ,± 3
1 , ± 2
1 , ± 1
1
¿>± p
q =±3 , ± 3
2 , ±1 , ± 1
2 , ± 6 , ±3 , ± 2, ± 1
After reducing the rational numbers, only the distinct numbers will be the possible zeros.
Hence, all the possible zeros of the equation will be,

5
MATHEMATICS
{± 3
2 ,± 1
2 , ±3 , ± 1, ± 2, ± 6 } Ans.
f. Descartes’s rule of Sign
Solution: According to the Descartes’s rule of sign,
i. The number of positive real zeros is equal to the number of sign changes in the
function f(x) or minus an even integer.
ii. The Number of negative real zeros is equal to the number of sign changes in the
function f (-x) or minus an even integer.
Now there are two sign changes in f ( x )=2 x4 +3 x3−6 x2−5 x+ 6
Now find f (−x ) =2(−x)4+ 3 ¿
f ( −x ) =2 x4 −3 x3 −6 x2 +5 x +6
There are two sign changes in f(-x).
Hence follow the table, all the positive, negative and imaginary zeros should add to give 4 as the
total number of zero of the function is 4.
Positive Negative Imaginary Total
2 2 0 4
=(2-even integer)=(2-
2)=0
=(2-even integer)=(2-
2)=0
4 4
Hence, for the function f(x) there will be 2 positive real zeros and 2 negative real zeros. Ans.

6
MATHEMATICS
2. Wind Problem.
Solution: Given in the question, P= 750 kilowatts = 750000 Watts
n=3
S=37
Using the direct variation formula as an nth power, determine the value of the constant
P=k xn
¿>750000=k 373
¿> k= 750000
373
¿> k= 750000
50653 =14.8
Using the same formula for the different wind speed,
P=14.8 × 403
¿> P=14.8 ×64000=947200 watts=947.2 KW Ans .