Question

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

Answer 1

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`

Answer 2

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`

Answer 3

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.