Question 804181: What are the real zeros of f(x)=x^3-x^2-37x-35
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Properties of Polynomials:
The Fundamental Theorem of Algebra guarantees that an  -th degree polynomial function will have zeros, counting multiplicities, some or all of which may be complex roots.
Complex roots always appear in conjugate pairs, so:
If is even, the polynomial may have 0, 2, 4, ... , n real number zeros.
If is odd, the polynomial is guaranteed to have at least one real number zero, but may have 1, 3, 5, ..., n real number zeros.
Rational Roots Theorem: If a polynomial has a rational number zero, then it will be of the form  where  is an integer factor of the constant cooefficient,  , and  is an integer factor of the lead coefficient, 
Corollary to The Fundamental Theorem of Algebra: If  is a zero of a polynomial, then ) is a factor of the polynomial.
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\ =\ x^3\ -\ x^2\ -\ 37x\ -\ 35)
is a third degree polynomial function, hence there is either 1 real zero or there are 3 real zeros. According to the Rational Roots Theorem, if any of these zeros are rational numbers, they are  ,  , or 
We test using Synthetic Division. First test 1:
1 | 1 -1 -37 -35
| 1 0 -37
----------------------
1 0 -37 -72 Last result non-zero, 1 is not a zero.
Try -1
-1 | 1 -1 -37 -35
| -1 2 35
----------------------
1 -2 -35 0 Last result is zero, -1 is a zero
Hence ) is a factor and ) is the other factor. But is a factorable quadratic.
I'll leave it to you to finish this.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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