# How do you find the rational roots of x^4+8x^3+7x^2-40x-60=0?

Jul 23, 2017

The rational roots are: $- 2$ and $- 6$

It also has irrational roots: $\pm \sqrt{5}$

#### Explanation:

Given:

${x}^{4} + 8 {x}^{3} + 7 {x}^{2} - 40 x - 60 = 0$

Let:

$f \left(x\right) = {x}^{4} + 8 {x}^{3} + 7 {x}^{2} - 40 x - 60$

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Rational roots theorem

By the rational roots theorem, any rational zeros of this quartic must be expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 60$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 5 , \pm 6 , \pm 10 , \pm 12 , \pm 15 , \pm 20 , \pm 30 , \pm 60$

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Descartes' Rule of Signs

The pattern of signs of the coefficients of $f \left(x\right)$ is $+ + + - -$. With one change of sign, we can tell that $f \left(x\right)$ has one positive real zero.

The pattern of signs of the coefficients of $f \left(- x\right)$ is $+ - + + -$. With three changes of sign, we can tell that $f \left(x\right)$ has $3$ or $1$ negative real zeros.

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Given that there is at least one and possibly three negative real zeros, let us try the possible negative rational zeros first:

$f \left(- 1\right) = 1 - 8 + 7 + 40 - 60 = - 20$

$f \left(- 2\right) = 16 - 64 + 28 + 80 - 60 = 0$

So $x = - 2$ is a zero and $\left(x + 2\right)$ a factor:

${x}^{4} + 8 {x}^{3} + 7 {x}^{2} - 40 x - 60 = \left(x + 2\right) \left({x}^{3} + 6 {x}^{2} - 5 x - 30\right)$

Note that the ratio between the first and second terms of the remaining cubic is the same as that between the third and fourth terms. So this cubic will factor by grouping:

${x}^{3} + 6 {x}^{2} - 5 x - 30 = \left({x}^{3} + 6 {x}^{2}\right) - \left(5 x + 30\right)$

$\textcolor{w h i t e}{{x}^{3} + 6 {x}^{2} - 5 x - 30} = {x}^{2} \left(x + 6\right) - 5 \left(x + 6\right)$

$\textcolor{w h i t e}{{x}^{3} + 6 {x}^{2} - 5 x - 30} = \left({x}^{2} - 5\right) \left(x + 6\right)$

$\textcolor{w h i t e}{{x}^{3} + 6 {x}^{2} - 5 x - 30} = \left({x}^{2} - {\left(\sqrt{5}\right)}^{2}\right) \left(x + 6\right)$

$\textcolor{w h i t e}{{x}^{3} + 6 {x}^{2} - 5 x - 30} = \left(x - \sqrt{5}\right) \left(x + \sqrt{5}\right) \left(x + 6\right)$

Hence the remaining rational zero is $x = - 6$ and there are two irrational zeros: $x = \pm \sqrt{5}$.