SOLUTION: I am trying to find all zeroes of P(x) = 6x^4 - 7x^3 - 8x^2 + 5x. Using synthetic division, I see that the upper bound is 2 and the lower bound is -2. Using Descartes rule, I

Question 56782This question is from textbook College Algebra
: I am trying to find all zeroes of P(x) = 6x^4 - 7x^3 - 8x^2 + 5x.
Using synthetic division, I see that the upper bound is 2 and the lower bound is -2.
Using Descartes rule, I found that there are 2 or 0 positive real zeroes and 1 negative real zero. Is this correct? Because shouldn't there be 4 zeroes altogether due to the leading coefficient having a degree of 4? And complex zeroes only come in two's (pos & neg), so I have no way to combine these options to create 4 zeros. Where have I gone wrong? This question is from textbook College Algebra

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
find all zeroes of P(x) = 6x^4 - 7x^3 - 8x^2 + 5x.
------------------------
x(6x^3-7x^2-8x+5)=0
x=0 is a zero.
Graphing I find x=-1 is a zero.
Using synthetic I find the quotient is 6x^2-13x+5
x=[13+-sqrt(169-120)]/12
x=[13+-7]/12
x=5/3 or x=1/2
Cheers,
Stan H.
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