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
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
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
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