# What are all the possible rational zeros for #f(x)=x^3+9x^2+15x+7# and how do you find all zeros?

##### 3 Answers

#### Answer:

We have two zeros for

#### Explanation:

We can use the factor theorem here. According to factor theorem if for some

Now for

=

=

=

=

=

Hence apart from

Hence, we have two zeros for

#### Answer:

By the rational root theorem, the "possible" rational zeros are:

#+-1# ,#+-7#

Actually the zeros are:

#### Explanation:

Given:

#f(x) = x^3+9x^2+15x+7#

Note that

Any rational zero of

That means that the only possible *rational* zeros of

#+-1# ,#+-7#

It could have other zeros, but they would be irrational or non-Real Complex.

In addition, note that all of the coefficients of

So the only *rational* zeros we need consider are

We find:

#f(-1) = -1+9-15+7 = 0#

So

#x^3+9x^2+15x+7 = (x+1)(x^2+8x+7)#

Notice that

#x^2+8x+7 = (x+1)(x+7)#

Notice that the remaining zero is the expected

#### Answer:

Double root

#### Explanation:

In this problem, I could find all the roots and then sort out rational

ones.

The sum of the coefficients in f(x) is not 0. So,

The sum of the coefficients in f(-x) is 0. So, x + 1 is a factor, and so,

Now, #f(x)=( x + 1 ) (x^2 + ax + b) = x^3 + 9x^2+15x+7).

Comparison of the like coefficients gives a = 8 and b = 7.

The roots of

So, the