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

Nov 26, 2016

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

Explanation:

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

Now for $x = - 1$, we have $f \left(- 1\right) = {\left(- 1\right)}^{3} + 9 {\left(- 1\right)}^{2} + 15 \left(- 1\right) + 7$

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

$\left(x + 1\right)$ is a factor of ${x}^{3} + 9 {x}^{2} + 15 x + 7$. Now dividing latter by $\left(x + 1\right)$

${x}^{3} + 9 {x}^{2} + 15 x + 7 = {x}^{2} \left(x + 1\right) + 8 x \left(x + 1\right) + 7 \left(x + 1\right)$

= $\left(x + 1\right) \left({x}^{2} + 8 x + 7\right)$

= $\left(x + 1\right) \left({x}^{2} + 7 x + x + 7\right)$

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

= $\left(x + 1\right) \left(x + 1\right) \left(x + 7\right)$

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

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

Nov 26, 2016

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

$\pm 1$, $\pm 7$

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

Explanation:

Given:

$f \left(x\right) = {x}^{3} + 9 {x}^{2} + 15 x + 7$

Note that $f \left(x\right)$ 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 $f \left(x\right)$ is expressible in the form $\frac{p}{q}$ 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 $f \left(x\right)$ are:

$\pm 1$, $\pm 7$

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

In addition, note that all of the coefficients of $f \left(x\right)$ are positive, hence $f \left(x\right)$ has no positive Real zeros.

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

We find:

$f \left(- 1\right) = - 1 + 9 - 15 + 7 = 0$

So $x = - 1$ is a zero and $\left(x + 1\right)$ a factor:

${x}^{3} + 9 {x}^{2} + 15 x + 7 = \left(x + 1\right) \left({x}^{2} + 8 x + 7\right)$

Notice that $x = - 1$ is also a zero of ${x}^{2} + 8 x + 7$, since $1 - 8 + 7 = 0$, so $\left(x + 1\right)$ is a factor again:

${x}^{2} + 8 x + 7 = \left(x + 1\right) \left(x + 7\right)$

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

Nov 26, 2016

Double root $- 1 \mathmr{and} - 7$.

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, $\left(x - 1\right)$ 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 ${x}^{2} + 8 x + 7 = 0$ are$- 1 \mathmr{and} - 7$.

So, the $- 1 , - 1 \mathmr{and} - 7$ are the zeros and all are rational.