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
Nov 26, 2016

Answer:

We have two zeros for #x^3+9x^2+15x+7#, #x=-1# with a multiplicity of two and #x=-7#

Explanation:

We can use the factor theorem here. According to factor theorem if for some #x=a#, #f(a)=0#, then #(x-a)# is a factor of #f(x)#.

Now for #x=-1#, we have #f(-1)=(-1)^3+9(-1)^2+15(-1)+7#

= #-1+9-15+7=0#, hence

#(x+1)# is a factor of #x^3+9x^2+15x+7#. Now dividing latter by #(x+1)#

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

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

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

= #(x+1)(x(x+7)+1(x+7)#

= #(x+1)(x+1)(x+7)#

Hence apart from #x=-1#, which appears twice, for #x=-7# too we have #f(x)=0#

Hence, we have two zeros for #x^3+9x^2+15x+7#, #x=-1# with a multiplicity of two and #x=-7#

Nov 26, 2016

Answer:

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

#+-1#, #+-7#

Actually the zeros are: #-1# with multiplicity #2# and #-7# with multiplicity #1#.

Explanation:

Given:

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

Note that #f(x)# is expressed in standard form, with descending powers of #x# and integer coefficients. So we can apply the rational roots theorem:

Any rational zero of #f(x)# is expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #7# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros of #f(x)# are:

#+-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 #f(x)# are positive, hence #f(x)# has no positive Real zeros.

So the only rational zeros we need consider are #-1# and #-7#.

We find:

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

So #x=-1# is a zero and #(x+1)# a factor:

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

Notice that #x=-1# is also a zero of #x^2+8x+7#, since #1-8+7 = 0#, so #(x+1)# is a factor again:

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

Notice that the remaining zero is the expected #x=-7#

Nov 26, 2016

Answer:

Double root #-1 and -7#.

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, #(x - 1)# is not a factor.

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

#x = = 1# is a root of f(x).

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 #x^2 + 8x + 7 = 0# are# -1 and - 7#.

So, the #-1, -1 and -7# are the zeros and all are rational.