# How do you find all the zeros of \[{x^3} - {x^2} - 4x - 6\]?

Answer

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**Hint**: Here in this question, the given equation in the form of cubic equation we have to find the factors. This equation can be solved by using factor theorem or synthetic division. First find the one factor root by a guesswork or trial and error method and write the quadratic equation by using the method of synthetic equation and further simplify the quadratic equation by factorization for finding other two factors.

**Complete step by step solution:**

A cubic function or equation is a third-degree polynomial. A general polynomial function has the form: \[f(x) = a{x^n} + b{x^{n - 1}} + c{x^{n - 2}}...v{x^3} + w{x^2} + zx + k\]

Here, x is the variable, n is simply any number (and the degree of the polynomial), k is a constant and the other elements or letters are constant coefficients for each power of x. so a cubic function has n=3, and is simply: \[f(x) = a{x^3} + b{x^2} + c{x^1} + d\]

The easiest way to solve a cubic equation involves a bit of guesswork and an algorithmic type of process called synthetic division. Though it is basically the same as the trial-and-error method for cubic equation solutions.

Consider the given cubic equation

\[ \Rightarrow {x^3} - {x^2} - 4x - 6\]

Try to find one of the roots by guessing.

You have to guess one of the values of x, let the first factor be \[x = 3\], then

\[ \Rightarrow {\left( 3 \right)^3} - {\left( 3 \right)^2} - 4\left( 3 \right) - 6\]

\[ \Rightarrow \left( {27} \right) - \left( 9 \right) - 4\left( 3 \right) - 6\]

\[ \Rightarrow 27 - 9 - 12 - 6\]

\[ \Rightarrow 27 - 27\]

\[ \Rightarrow 0\]

This means \[x = 3\] is one of the roots of the cubic equation. The answer will be obtained without much thought and it is also time-consuming (especially if we have to go to higher factors before finding a root). Luckily, when we’ve found one root, you can solve the rest of the equation easily.

The key is incorporating the factor theorem. This states that if x = s is a solution, then \[\left( {x\;-\;s} \right)\] is a factor that can be pulled out of the equation. For this situation, \[s = 3\], and so \[\left( {x - 3} \right)\] is a factor.

Next find the quadratic equation by the method of synthetic division. First, write down the coefficients of the original or given equation in the first row of a table, with a dividing line and then the known root on the right:

\[\begin{array}{*{20}{c}}

{x = 3} \\

{}

\end{array}\left| \!{\underline {

{\begin{array}{*{20}{c}}

1&{ - 1}&{ - 4}&{ - 6} \\

{}&{}&{}&{}

\end{array}} }} \right. \]

Leave one spare row, and then add a horizontal line below it. First, take the first number (1 in this case) down to the row below your horizontal line

\[

\begin{array}{*{20}{c}}

{x = 3} \\

{}

\end{array}\left| \!{\underline {

{\begin{array}{*{20}{c}}

1&{ - 1}&{ - 4}&{ - 6} \\

\downarrow &{}&{}&{}

\end{array}} }} \right. \\

\begin{array}{*{20}{c}}

1&{}&{}&{}

\end{array} \\

\]

Continue the steps of synthetic division

\[

\begin{array}{*{20}{c}}

{x = 3} \\

{}

\end{array}\left| \!{\underline {

{\begin{array}{*{20}{c}}

1&{ - 1}&{ - 4}&{ - 6} \\

\downarrow &{ + 3}&{ + 6}&{ + 6}

\end{array}} }} \right. \\

\begin{array}{*{20}{c}}

1&{2}&{2}&{0}

\end{array} \;

\]

The fact that the last answer is zero tells you that you’ve got a valid root, so if this isn’t zero, then you’ve made a mistake somewhere.

Now, the bottom row tells you the factors of the three terms in the second set of brackets, so the cubic equation is written as:

\[ \Rightarrow \left( {x - 3} \right)\left( {{x^2} + 2x + 2} \right)\]

This is the most important stage of the solution, and you can finish from this point onwards by factorization. This is basically the same problem as factoring a quadratic equation.

Consider the quadratic equation present in second bracket

\[ \Rightarrow {x^2} + 2x + 2\]

We solve the above equation by using the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]

So we have a = 1, b = 2 and c = 2

\[ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4(1)(2)} }}{{2(1)}}\]

On simplifying

\[ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 - 8} }}{2}\]

\[ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt { - 4} }}{2}\]

As we know that \[\sqrt { - 1} = i\], so the above inequality is written as

\[ \Rightarrow x = \dfrac{{ - 2 \pm 2i}}{2}\]

On further simplifying we have

\[ \Rightarrow x = - 1 \pm 1i\]

Therefore, factors of cubic equations are

\[ \Rightarrow \left( {x - 3} \right)\left( {x + 1 - i} \right)\left( {x + 1 + i} \right)\]

Hence, the factors of the given cubic equation \[{x^3} - {x^2} - 4x - 6\] is \[\left( {x - 3} \right)\], \[\left( {x + 1 + i} \right)\] and \[\left( {x + 1 - i} \right)\].

**So, the correct answer is “\[\left( {x - 3} \right)\], \[\left( {x + 1 + i} \right)\] and \[\left( {x + 1 - i} \right)\].”.**

**Note**: The cubic equation contains 3 factors or roots. To find or solve the cubic equation, solved by using the synthetic division method. The one value or root of the given equation determined by the synthetic division. Then the quadratic equation will remain, to the quadratic equation the sum product rule or by standard formula is applied to determine the other two roots.

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