Question

How do you find all the zeros of $f\left( x \right) = {x^3} + 11{x^2} + 39x + 29$?

Answer 1

The given function is a cubic polynomial. To find its zeros, we’ll put it to zero and determine its roots. First root can be obtained by hit and trial and then it will be converted into a product of linear and quadratic factors. We’ll solve the quadratic equation to determine the other two roots.

Complete step by step answer:
According to the question, we have to determine the zeros of the function $f\left( x \right) = {x^3} + 11{x^2} + 39x + 29$.
This is a cubic polynomial. To find its zeros, we’ll put it to zero to determine its roots. So we have:
$f\left( x \right) = {x^3} + 11{x^2} + 39x + 29 = 0 \Rightarrow {x^3} + 11{x^2} + 39x + 29 = 0 \\\$
First root can be obtained by hit and trial method. So if we put $x = - 1$ in the equation, we’ll get:
$f\left( { - 1} \right) = {\left( { - 1} \right)^3} + 11{\left( { - 1} \right)^2} + 39\left( { - 1} \right) + 29 \\\ \Rightarrow f\left( { - 1} \right) = - 1 + 11 - 39 + 29 \\\ \Rightarrow f\left( { - 1} \right) = - 40 + 40 = 0$
Thus $x = - 1$ is a zero of the function and $\left( {x + 1} \right)$ is a factor of the polynomial.
Now, the other factor will be a quadratic expression and we know that a cubic polynomial can be written as the product of a linear and a quadratic factor. So let $\left( {a{x^2} + bx + c} \right)$ is the other factor. So we have:
$\Rightarrow {x^3} + 11{x^2} + 39x + 29 = \left( {x + 1} \right)\left( {a{x^2} + bx + c} \right){\text{ }}.....{\text{(1)}}$
If we expand this, we’ll get:
$\Rightarrow {x^3} + 11{x^2} + 39x + 29 = a{x^3} + \left( {a + b} \right){x^2} + \left( {b + c} \right)x + c$
We will compare the coefficients on both sides, we have:
$\Rightarrow a = 1,{\text{ }}c = 29,{\text{ }}a + b = 11{\text{ and }}b + c = 39$
Putting values of $a$ to find the value of $b$, we’ll get:
$1 + b = 11 \\\ \Rightarrow b = 10$
Putting the values of $a,{\text{ }}b{\text{ and }}c$ in equation (1), we’ll get:
$\Rightarrow {x^3} + 11{x^2} + 39x + 29 = \left( {x + 1} \right)\left( {{x^2} + 10x + 29} \right)$
Now, to find the other two zeros of the cubic function, we need to find the roots of the quadratic equation $\left( {{x^2} + 10x + 29} \right)$.
For a quadratic equation $\left( {a{x^2} + bx + c} \right)$, its roots can be determined by the formula:
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Using this formula for the equation $\left( {{x^2} + 10x + 29} \right)$, we’ll get:
$x = \dfrac{{ - 29 \pm \sqrt {{{10}^2} - 4\left( 1 \right)\left( {29} \right)} }}{{2\left( 1 \right)}} \\\ \Rightarrow x = \dfrac{{ - 29 \pm \sqrt {100 - 116} }}{2} = \dfrac{{ - 29 \pm \sqrt { - 16} }}{2} \\\ \Rightarrow x = \dfrac{{ - 29 \pm \sqrt {16} \sqrt { - 1} }}{2}$
The roots will be imaginary as we know $\sqrt { - 1} = i$. So we have:
$\Rightarrow x = \dfrac{{ - 29 \pm 4i}}{2}$
Thus the two roots are $\dfrac{{ - 29 + 4i}}{2}$ and $\dfrac{{ - 29 - 4i}}{2}$

So we have our function as:
$f\left( x \right) = {x^3} + 11{x^2} + 39x + 29 = \left( {x + 1} \right)\left( {{x^2} + 10x + 29} \right)$
And the three zeros of this function are $x = - 1$, $x = \dfrac{{ - 29 + 4i}}{2}$ and $x = \dfrac{{ - 29 - 4i}}{2}$

Note: For a quadratic equation to have real roots, its discriminant must be greater than or equal to zero.
Consider the given quadratic equation:
$y = a{x^2} + bx + c$
Discriminant of this quadratic equation is $D = {b^2} - 4ac$
Thus for the quadratic equation to have real roots, following condition must satisfy:
$\Rightarrow D \geqslant 0{\text{ or}} \\\ \Rightarrow {b^2} - 4ac \geqslant 0$