52 779
Assignments Done
98,3%
Successfully Done
In October 2017
Your physics homework can be a real challenge, and the due date can be really close — feel free to use our assistance and get the desired result.
Be sure that math assignments completed by our experts will be error-free and done according to your instructions specified in the submitted order form.
Our experts will gladly share their knowledge and help you with programming homework. Keep up with the world’s newest programming trends.

Answer on Calculus Question for shasha

Question #42524
State how many imaginary and real zeros the function has.
f(x) = x5 + 7x4 + 2x3 + 14x2 + x + 7

3 imaginary; 2 real

4 imaginary; 1 real

0 imaginary; 5 real

2 imaginary; 3 real


show work and explain it to me please
Expert's answer
The number of zeros does not exceed 5, because it is equation of degree 5. The closed-form formula for solutions does not exist, though some types of equations can be solved. Possible roots are divisors of 7 by Rational Zeros Theorem. By trial and error method, -7 is a zero of this equation. Divide x5+7x4+2x3+14x2+x+7 by (x+7). So,
x5+7x4+2x3+14x2+x+7=(x+7)(x2+1)2. Zeros of equation x2+1=0 are i and (-i), factor  x2+1 is squared, therefore, zeros of the initial equation are i,i,-i, -i and -7. Thus, the answer is 1 real and 4 complex zeros.       

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question