
Concept explainers
a.
Find the real zeros of the function f(x)=25x3−55x2−54x−18 using graphical method.
a.

Answer to Problem 68E
3
Explanation of Solution
Given:
Function: f(x)=25x3−55x2−54x−18
Calculation for graph:
Consider f(x)=25x3−55x2−54x−18
Values of x | Values of f (x) |
0 | -18 |
1 | -102 |
-1 | -44 |
2 | -146 |
-2 | -330 |
By taking different values of x, the graph can be plotted.
Graph:
Interpretation:
By observing graph, it is clear that the curve of the function meets x-axis at x=3.
Hence, the real zeros of the function are x=3.
Conclusion:
Therefore, the real zeros of given polynomial function are x=3.
b.
By using real zeros of the function f(x)=25x3−55x2−54x−18 , find the imaginary roots.
b.

Answer to Problem 68E
3,−2+√2i5,−2−√2i5
Explanation of Solution
Given:
Function: f(x)=25x3−55x2−54x−18
Calculation:
From above answer, the real zeros of function are x=3.
Now divide given function with above zero.
Using synthetic division,
325−55−54−18756018252060 ←Remainder
So, quotient is 25x2+20x+6 and remainder is 0.
So, f(x) factors as
⇒f(x)=(x−3)(25x2+20x+6)
To find other zeros, put 25x2+20x+6=0,
The above polynomial equation is a quadratic equation.
The zeros of a quadratic equation ax2+bx+c=0 can be found using the formula
x=−b±√b2−4ac2a
Applying above formula,
⇒x=−20±√(20)2−4(25)(6)2(25)⇒x=−20±√400−60050⇒x=−20±√−20050⇒x=−20±10√−250⇒x=−20±10√2i50⇒x=−2±√2i5⇒x=−2+√2i5,−2−√2i5
Conclusion:
Therefore, the zeros of given polynomial function are x=3,−2+√2i5 and −2−√2i5.
Chapter 2 Solutions
Precalculus with Limits: A Graphing Approach
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