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

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Question 1156241: 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(52778) About Me  (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 = x%5E4+-+2x%5E3+-+10x%5E2+%2B+6x+%2B+45

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

So, I divide x%5E4+-+2x%5E3+-+10x%5E2+%2B+6x+%2B+45 by %28x-3%29%5E2, and I get the quotient x%5E2+%2B+4x+%2B+5.

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

Therefore, factoring over real numbers is

x%5E4+-+2x%5E3+-+10x%5E2+%2B+6x+%2B+45 = %28x-3%29%5E2%2A%28x%5E2+%2B+4x+%2B+5%29.

You may go forward to find complex zeroes of the quadratic quotient.