Q 69. Roots of the equation given below are: x^4-4x^3+ 6x^2-4x+1=0 Ops: A. 3,1,3,1 B. 1,1,1,1 C. 1,2,1,2 D. 2,2,2,2
Question
Q 69. Roots of the equation given below are: x^4-4x^3+ 6x^2-4x+1=0 Ops: A. 3,1,3,1 B. 1,1,1,1 C. 1,2,1,2 D. 2,2,2,2
Solution
The roots of the equation x^4 - 4x^3 + 6x^2 - 4x + 1 = 0 can be found by factoring the equation or using the rational root theorem. However, this equation is a special case known as a binomial theorem. It is the expansion of (x-1)^4. Therefore, the roots of the equation are 1, 1, 1, 1. So, the correct answer is B. 1,1,1,1.
Similar Questions
If f(x) is a 4th degree polynomial, and has 3 real numbers as roots, the other root is?Question 4Select one:a.0b.Realc.Complexd.1e.None of these
The real root of the equation 3x x− + =4 0 is −1.796 to three decimal places.Determine the real root for each of the following.(a) ( ) 3− −x x1 1( )− + =4 0(b) 38 2x x− + =4 0
Match the following roots with their meaning.Group of answer choicesphaeo
Which of the following is a root of the equation x3 + 4x2 + x – 6 = 01–102
A polynomial has zeros at -4, 2 and 1. Which of the following could be that polynomial?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.