How do you find the zeros of f(x) = x^3 − 12x^2 + 47x − 60?

Jun 15, 2016

$x = 3$, $x = 4$ and $x = 5$.

Explanation:

$f \left(x\right) = {x}^{3} - 12 {x}^{2} + 47 x - 60$

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$\underline{\text{Discriminant}}$

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

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

In our example, $a = 1$, $b = - 12$, $c = 47$, $d = - 60$ and we find:

$\Delta = 318096 - 415292 - 414720 - 97200 + 609120 = 4 > 0$

Since $\Delta > 0$ we can deduce that $f \left(x\right)$ has $3$ distinct Real zeros.

The discriminant allows us to distinguish the cases:

• $\Delta > 0$ - The cubic has $3$ distinct Real zeros.
• $\Delta = 0$ - The cubic has $3$ Real zeros, at least two of which coincide.
• $\Delta < 0$ - The cubic has $1$ Real zero and a Complex conjugate pair of non-Real zeros.

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$\underline{\text{Rational zeros}}$

By the rational root theorem, any rational zeros of $f \left(x\right)$ 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.

Note that $f \left(- x\right) = - {x}^{3} - 12 {x}^{2} - 47 x - 60$ has coefficients which all have the same sign. So $f \left(x\right)$ has no negative zeros.

Hence the only possible rational zeros are the positive factors of $60$, namely:

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

We could try each of these in turn to find our zeros, but let's try something different...

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$\underline{\text{Tschirnhaus transformation}}$

We can simplify the cubic, eliminating the square term by making a linear substitution - known as a Tschirnhaus transformation. In our example, we substitute $t = x - 4$ and find that the resulting cubic not only has no square term, it has no constant term either...

Note that:

${\left(x - 4\right)}^{3} = {x}^{3} - 12 {x}^{2} + 48 x - 64$

So:

${x}^{3} - 12 {x}^{2} + 47 x - 60 = {\left(x - 4\right)}^{3} - \left(x - 4\right)$

Let $t = x - 4$

Then

$f \left(x\right) = {\left(x - 4\right)}^{3} - \left(x - 4\right)$

$= {t}^{3} - t$

$= t \left({t}^{2} - 1\right)$

$= t \left({t}^{2} - {1}^{2}\right)$

$= t \left(t - 1\right) \left(t + 1\right)$

$= \left(x - 4\right) \left(x - 5\right) \left(x - 3\right)$

Hence the zeros of $f \left(x\right)$ are $x = 3$, $x = 4$ and $x = 5$.