SOLUTION: What are the real zeros of f(x)=x^3-x^2-37x-35

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.

============================================================

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

The Out Campaign: Scarlet Letter of Atheism

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