# What are all the possible rational zeros for f(x)=4x^4+3x^3-2x^2+5x-12 and how do you find all zeros?

May 5, 2017

The "possible" rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12$

The actual zeros are irrational and complex.

#### Explanation:

Given:

$f \left(x\right) = 4 {x}^{4} + 3 {x}^{3} - 2 {x}^{2} + 5 x - 12$

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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 $- 12$ and $q$ a divisor of the coefficient $4$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{4} , \pm \frac{1}{2} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12$

That's rather a lot of possibilities to try, so let us find out more about the nature of these zeros.

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Descartes' Rule of Signs

The signs of the coefficients of $f \left(x\right)$ are in the pattern $+ + - + -$. With $3$ changes of sign, that means that $f \left(x\right)$ has $3$ or $1$ real positive zero.

The signs of the coefficients of $f \left(- x\right)$ are in the pattern $+ - - - -$. With $1$ change of sign, that means that $f \left(x\right)$ has exactly $1$ real negative zero.

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Turning points

Let us first find where the turning points of $f \left(x\right)$ are to help guide our search. Turning points occur where the derivative $f ' \left(x\right)$ of $f \left(x\right)$ is zero.

$f ' \left(x\right) = 16 {x}^{3} + 9 {x}^{2} - 4 x + 5$

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 16$, $b = 9$, $c = - 4$ and $d = 5$, so we find:

$\Delta = 1296 + 4096 - 14580 - 172800 - 51840 = - 233828$

Since $\Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

So $f ' \left(x\right)$ has only one real zero and $f \left(x\right)$ has only one turning point.

Therefore $f \left(x\right)$ has one real negative zero and one real positive zero.

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Back to the rational possibilities...

Trying some of the possible rational zeros, we find:

$f \left(- 2\right) = 10$

$f \left(- \frac{3}{2}\right) = - 13.875$

$f \left(1\right) = - 2$

$f \left(\frac{3}{2}\right) = 21.375$

Hence we can deduce that there is an irrational zero in $\left(- 2 , - \frac{3}{2}\right)$ and one in $\left(1 , \frac{3}{2}\right)$.

There are no rational zeros.

graph{4x^4+3x^3-2x^2+5x-12 [-5, 5, -20, 20]}

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Where do we go from here?

to be continued...

May 6, 2017

Use a numerical method (Durand-Kerner) to find approximations to the irrational zeros...

#### Explanation:

Given:

$f \left(x\right) = 4 {x}^{4} + 3 {x}^{3} - 2 {x}^{2} + 5 x - 12$

In my first answer, we found that the zeros are not rational.

We can find approximations to the zeros using an algorithm known as Durand-Kerner.

Suppose the $4$ zeros are $p , q , r , s$.

Choose initial approximations:

${p}_{0} = {\left(0.4 + 0.9 i\right)}^{0}$
${q}_{0} = {\left(0.4 + 0.9 i\right)}^{1}$
${r}_{0} = {\left(0.4 + 0.9 i\right)}^{2}$
${s}_{0} = {\left(0.4 + 0.9 i\right)}^{3}$

Then iterate using the formulas:

${p}_{i + 1} = {p}_{i} - f \frac{{p}_{i}}{\left({p}_{i} - {q}_{i}\right) \left({p}_{i} - {r}_{i}\right) \left({p}_{i} - {s}_{i}\right)}$

${q}_{i + 1} = {q}_{i} - f \frac{{q}_{i}}{\left({q}_{i} - {p}_{i + 1}\right) \left({q}_{i} - {r}_{i}\right) \left({q}_{i} - {s}_{i}\right)}$

${r}_{i + 1} = {r}_{i} - f \frac{{r}_{i}}{\left({r}_{i} - {p}_{i + 1}\right) \left({r}_{i} - {q}_{i + 1}\right) \left({r}_{i} - {s}_{i}\right)}$

s_(i+1) = s_i-f(s_i)/((s_i-p_(i+1))(s_i-q_(i+1))(s_i-r_(i+1))

This can be done using a computer program, such as this one written in C++ ...

After a few iterations, the approximations settle down to give:

$p \approx 1.0707$

$q \approx - 1.85404$

$r \approx 0.0166701 - 1.22921 i$

$s \approx 0.0166701 + 1.22921 i$