SOLUTION: Let P(x)=x4 −2x3 −10x2 +6x+45 ▪ Use the Rational Zero Theorem to list all the possible rational zeros. ▪ Then find all zeros exactly (rational, irrational, and imaginary).

Question 1156301: Let P(x)=x4 −2x3 −10x2 +6x+45
▪ Use the Rational Zero Theorem to list all the possible rational zeros. ▪ Then find all zeros exactly (rational, irrational, and imaginary).
Hint: Use the Rational Zero Theorem, a graphing calculator, and synthetic division if needed.
Answer by ikleyn(52780) (Show Source): You can put this solution on YOUR website!
.


Since the leading coefficient is 1, the Remainder theorem provides this list of possible zeros 
(all of them are divisors of the constant term 45, in this case)

   +/-1, +/-3, +/-5, +/-9, +/-15, +/-45.



Next, the plot below


    


    Plot y = 


shows the root x= 3 of the multiplicity at least 2.


So, I divide    by  ,  and I get the quotient  .


This quotient is a quadratic polynomial with negative discriminant, so it has no real roots.


Therefore, factoring over real numbers is


     = .


The quadratic polynomial  x^2 + 4x + 5  has no rational roots.

It has no real roots, too, since its discriminant d = (-4)^2 - 4*1*5 = 16 - 20 = -4 is negative.


It has two complex roots   = .


ANSWER.  The roots of the given polynomial are  x= 3  of the multiplicity  2  and 

         two complex roots    and     of the multiplicity 1 each.

Solved.

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