Question 86297: I need help with this homework problem. It says: find all real zeros of the function
a) f(x)=-9 +729
b) f(x)= -11+ 26x - 16
The lecture notes are not very good on this and I can't understand the book. The question is not from the book.
Answer by scianci(186)   (Show Source):
You can put this solution on YOUR website! a) f(x)=-9+729
b) f(x)=-11+ 26x - 16
For a), begin by putting it in factored form:
-9+729 = -9( - 81) = -9(x + 9)(x - 9).
Then, set each factor = 0. Each one will generate a solution:
-9 = 0 , x + 9 = 0 , x - 9 = 0
x = 0 , x = -9 , x = 9. These are your real zeros.
For b), apply the rational root theorem and look for a rational root first. By the rational root theorem, the numerator of any rational root will be a factor of 16 [which is your constant term] and the denominator of any rational root will be a factor of 1 [which is your lead coefficient]. So, any rational roots will be from among 1 , -1 , 2 , -2 , 4 , -4 , 8 , -8 , 16 , -16. Try each one in turn until you find one that fits the equation. Start with 1:
 -11 + 26(1) - 16 =
1 - 11 + 26 - 16 = 0. So, x = 1 is a real zero and (x - 1) is a factor of -11+ 26x - 16. Use synthetic diviison to locate its companion quadratic factor:
1 1 -11 26 -16
1 -10 16
1 -10 16 0
So,-11+ 26x - 16 = (x - 1)(( - 10x + 16). Now, break down the - 10x + 16 into (x - 2)(x - 8). So, the factored form is:
(x - 1)(x - 2)(x - 8). Set each of these equal to 0 in turn:
x - 1 = 0 , x - 2 = 0 , x - 8 = 0
x = 1 , x = 2 , x = 8 are the real zeros.
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