Question 141202: How many real zeroes are there in f(x) = x^3 + 27
a. 3
b. 1
c. None, you cannot factor this polynomial
d. None of the above
I figured b. 1 because the factors of 27 are 1, 3, 9, 27 and the factors of 1 is 1. So they only have one factor in common.
Found 2 solutions by nabla, vleith:
Answer by nabla(475)   (Show Source):
You can put this solution on YOUR website! First, note that the statement is the same as:
f(x)=x^3+3^3=(x+3)(x^2-3x+9) by the formula for the sum of cubes.
Attempting to find the zeros, we set f(x)=0:
(x+3)(x^2-3x+9)=0
This can only be when
x=-3 or x^2-3x+9=0
We can see that the second will not have any real roots because the discriminant is negative b^2-4ac=9-4(1)(9)=-27.
Thus, your choice of b is correct, but I would recommend using the method I have done here to determine the amount of roots of a polynomial.
Answer by vleith(2983) (Show Source):
You can put this solution on YOUR website! See http://www.algebra-online.com/sum-difference-cubes-1.htm
Then take another stab at it.
You picked the correct answer, but for the wrong reasons.
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