Let us consider the function be,
\(f\left( x \right) = {x^3} - 3{x^2} - 10x + 24\)
Use rational zero theorem, to find all the rational zeros of function,
According to the rational zero theorem, if Q(x) is a polynomial function with integral coefficients, then every rational zero of is of the form\(\frac{p}{q}\), a rational number.
\(p: \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\) and\(q: \pm 1\).
Thus, possible value of \(\frac{p}{q}\) in simplest form will be
\[\frac{p}{q} = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\]
\(\begin{array}{c}{x^2} - x - 12 = 0\\{x^2} - 4x + 3x - 12 = 0\\x\left( {x - 4} \right) + 3\left( {x - 4} \right) = 0\\\left( {x - 4} \right)\left( {x + 3} \right) = 0\end{array}\)
\(f\left( x \right) = {x^3} - 3{x^2} - 10x + 24\)
Use rational zero theorem, to find all the rational zeros of function,
According to the rational zero theorem, if Q(x) is a polynomial function with integral coefficients, then every rational zero of
Where p is a factor of the constant term of the polynomial function and q is a factor of the leading coefficient.
If\(\frac{p}{q}\)is a rational zero of\(f\left( x \right) = {x^3} - 3{x^2} - 10x + 24\)then p is a factor of 24 and qis a factor of 1.
Then, \(p: \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\) and\(q: \pm 1\).
Thus, possible value of \(\frac{p}{q}\) in simplest form will be
\[\frac{p}{q} = \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\]
To find other zeros divide the above polynomial,
\(f\left( x \right) = {x^3} - 3{x^2} - 10x + 24\) with\(x - 2\)
Thus, solve as,
\(\frac{{{\rm{ }}{x^3} - 3{x^2} - 10x + 24{\rm{ }}}}{{\left( {x - 2} \right)}} = {x^2} - x - 12\)
To find other two zeros solve, \({x^2} - x - 12\) by splitting the middle term,\(\begin{array}{c}{x^2} - x - 12 = 0\\{x^2} - 4x + 3x - 12 = 0\\x\left( {x - 4} \right) + 3\left( {x - 4} \right) = 0\\\left( {x - 4} \right)\left( {x + 3} \right) = 0\end{array}\)
Either x= 4, or x = -3
Hence, the zeros of the functions are \({ - 3,2{\rm{ and }}4}\).