# What are all the possible rational zeros for f(x)=5x^4-46x^3+84x^2-50x+7 and how do you find all zeros?

May 19, 2017

The "possible" rational zeros are:

$\pm \frac{1}{5} , \pm 1 , \pm \frac{7}{5} , \pm 7$

The actual zeros are:

$1 , 1 , \frac{1}{5} , 7$

#### Explanation:

Given:

$f \left(x\right) = 5 {x}^{4} - 46 {x}^{3} + 84 {x}^{2} - 50 x + 7$

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Rational Roots Theorem

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are 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 $5$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{5} , \pm 1 , \pm \frac{7}{5} , \pm 7$

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Descartes' Rule of Signs

Note that the pattern of the signs of the coefficients is $+ - + - +$. By Descartes' Rule of Signs, since this has $4$ changes of sign, we can deduce that $f \left(x\right)$ has $4$, $2$ or $0$ positive real zeros.

The pattern of the signs of the coefficients of $f \left(- x\right)$ is $+ + + + +$. With no changes of sign, we can deduce that $f \left(x\right)$ has no negative real roots.

So the only possible rational zeros are:

$\frac{1}{5} , 1 , \frac{7}{5} , 7$

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Sum of coefficients shortcut

Note that the sum of the coefficients of $f \left(x\right)$ is zero. That is:

$5 - 46 + 84 - 50 + 7 = 0$

Hence we can deduce that $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

$5 {x}^{4} - 46 {x}^{3} + 84 {x}^{2} - 50 x + 7 = \left(x - 1\right) \left(5 {x}^{3} - 41 {x}^{2} + 43 x - 7\right)$

Looking at the coefficients of the remaining cubic, we find:

$5 - 41 + 43 - 7 = 0$

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

$5 {x}^{3} - 41 {x}^{2} + 43 x - 7 = \left(x - 1\right) \left(5 {x}^{2} - 36 x + 7\right)$

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Let's try the remaining rational possibilities with the remaining quadratic:

We find:

$5 {\left(\textcolor{b l u e}{\frac{1}{5}}\right)}^{2} - 36 \left(\textcolor{b l u e}{\frac{1}{5}}\right) + 7 = \frac{5}{25} - \frac{36}{5} + 7 = \frac{1 - 36 + 35}{5} = 0$

So $x = \frac{1}{5}$ is a zero and $\left(5 x - 1\right)$ a factor:

$5 {x}^{2} - 36 x + 7 = \left(5 x - 1\right) \left(x - 7\right)$

So the last zero is $x = 7$