Polynomial Root Theorems

To determine which of these is a root of the given polynomial function, each of the factors will be substituted into the equation $p(x)=0$ to check which results in a true statement. First, the positive factors will be substituted.

Equation	Solution
$(1{)}^4-(1{)}^3-6(1{)}^2+14(1)-12=0$	$\text{-}4\ne 0\times$
$(2{)}^4-(2{)}^3-6(2{)}^2+14(2)-12=0$	$0=0✓$
$(3{)}^4-(3{)}^3-6(3{)}^2+14(3)-12=0$	$30\ne 0\times$
$(4{)}^4-(4{)}^3-6(4{)}^2+14(4)-12=0$	$140\ne 0\times$
$(6{)}^4-(6{)}^3-6(6{)}^2+14(6)-12=0$	$936\ne 0\times$
$(12{)}^4-(12{)}^3-6(12{)}^2+14(12)-12=0$	$18300\ne 0\times$

The value of $x=2$ is a root of the function. Since the polynomial function is of degree $4,$ it is possible that one of the negative factors is also a root, so these values will be substituted next.

Equation	Solution
$(\text{-}1{)}^4-(\text{-}1{)}^3-6(\text{-}1{)}^2+14(\text{-}1)-12=0$	$\text{-}30\ne 0\times$
$(\text{-}2{)}^4-(\text{-}2{)}^3-6(\text{-}2{)}^2+14(\text{-}2)-12=0$	$\text{-}40\ne 0\times$
$(\text{-}3{)}^4-(\text{-}3{)}^3-6(\text{-}3{)}^2+14(\text{-}3)-12=0$	$0=0✓$
$(\text{-}4{)}^4-(\text{-}4{)}^3-6(\text{-}4{)}^2+14(\text{-}4)-12=0$	$156\ne 0\times$
$(\text{-}6{)}^4-(\text{-}6{)}^3-6(\text{-}6{)}^2+14(\text{-}6)-12=0$	$1200\ne 0\times$
$(\text{-}12{)}^4-(\text{-}12{)}^3-6(\text{-}12{)}^2+14(\text{-}12)-12=0$	$21420\ne 0\times$

As can be seen in the table, $\text{-}3$ is also a root of $p(x).$ Since every integer root has to be a factor of $\text{-}12$ and every factor was tested, the only integer roots of the given function are $\text{-}3$ and $2.$

To find the polynomial $q(x),$ which is one degree less than $p(x),$ the polynomial $p(x)$ has to be divided by the binomial $(x-2).$ This division can be performed by synthetic division. First, the coefficients are ordered and the leftmost coefficient goes down, below the horizontal line. Each time, the newest number under the horizontal line is multiplied by $2,$ then the product is aligned with the following number inside the division, and the newest terms inline are added. This procedure is shown below. The resulting polynomial of the division can be written using the quotient. It is important to note that the quotient is one degree less than the original polynomial. Now the given function can be rewritten factoring the binomial $(x-2).$ A similar thing can be done to rewrite $q(x).$ A root of the function $p(x)$ is also a root of $q(x).$ Therefore, the binomial $(x+3)$ can be factored from $q(x)$ using synthetic division again. With this result, the given function can be rewritten again. Finally, to find the remaining two roots, the roots of the polynomial at the rightmost parentheses are needed. This time the Quadratic Formula will be used. First, the coefficients of the quadratic equation are identified. Now these coefficients can be substituted into the Quadratic Formula. There are two possible values for $x$ to solve the polynomial. Since there is a square root of a negative number involved, the results are complex numbers.

$x=\frac{2+\sqrt{\text{-}4}}{2}$	$x=\frac{2-\sqrt{\text{-}4}}{2}$
$x=\frac{2+2i}{2}$	$x=\frac{2-2i}{2}$
$x=1+i$	$x=1-i$

Therefore, the remaining roots of the given polynomial function are the complex numbers $1+i$ and $1-i.$

Considering the Rational Root Theorem, it is possible to find the integer and the rational roots. According to the theorem, the integer roots of the polynomial must be factors of the constant term of the polynomial, which is $2.$ Each of these factors is substituted into the equation $g(x)=0$ to determine which, if any, is a root of the function.

Substitution	Simplify
$3(\text{-}2{)}^3-(\text{-}2{)}^2-6(\text{-}2)+2=0$	$\text{-}14\ne 0\times$
$3(\text{-}1{)}^3-(\text{-}1{)}^2-6(\text{-}1)+2=0$	$4\ne 0\times$
$3(1{)}^3-(1{)}^2-6(1)+2=0$	$\text{-}2\ne 0\times$
$3(2{)}^3-(2{)}^2-6(2)+2=0$	$10\ne 0\times$

No value resulted in a true statement. This means there is no integer root of the function. However, the Rational Root Theorem also allows to look for the rational roots of a polynomial. These roots can be written as follows. In this expression, $p,$ the numerator, is a factor of the constant term $q,$ the denominator, is a factor of the leading coefficient, which in this case is $3.$ Previously the factors of $2$ were presented. Below are the factors of $3.$ Since these factors must be denominators and the integer roots were already considered, the only factors to be considered are $\text{-}3$ and $3,$ which result in the following rational numbers. Again, each of these numbers is substituted into the equation as before to verity which, if any, is a root of the function.

Substitution	Simplify
$3{\left(\text{-}\frac{2}{3}\right)}^3-{\left(\text{-}\frac{2}{3}\right)}^2-6\left(\text{-}\frac{2}{3}\right)+2=0$	$\frac{14}{3}\ne 0\times$
$3{\left(\text{-}\frac{1}{3}\right)}^3-{\left(\text{-}\frac{1}{3}\right)}^2-6\left(\text{-}\frac{1}{3}\right)+2=0$	$\frac{34}{3}\ne 0\times$
$3{\left(\frac{1}{3}\right)}^3-{\left(\frac{1}{3}\right)}^2-6\left(\frac{1}{3}\right)+2=0$	$0=0✓$
$3{\left(\frac{2}{3}\right)}^3-{\left(\frac{2}{3}\right)}^2-6\left(\frac{2}{3}\right)+2=0$	$\text{-}\frac{14}{9}\ne 0\times$

The value $x=\frac{1}{3}$ is a root of the given function. This means that there are $2$ roots remaining. Using synthetic division, the root $\frac{1}{3}$ can be factored from the given function to obtain a quadratic equation. The first step of the synthetic division is to write the coefficients and the root as shown below. Then, the $3$ is moved down, below the horizontal line. In synthetic division, the number is multiplied by the newest number below the horizontal line. It is then added to the following number above the line, and the sum is moved below the horizontal line. The division can then be done. Using this result, it is possible to rewrite the given function. It is important to remember that the resulting polynomial is always one degree less than the original one. The roots of the function are also roots of the polynomial inside the rightmost parentheses. Because of this, this polynomial has to be factored. Now that the polynomial is factored, the given function can be rewritten again. By the Factor Theorem, the roots of the function are $\frac{1}{3},$ $\text{-}\sqrt{2},$ and $\sqrt{2}.$

It is mentioned that the polynomial is of degree $3.$ Considering the Fundamental Theorem of Algebra, the given polynomial has three roots. Two of those roots are already given. All the roots of the polynomial are needed to determine the correct polynomial. Since every coefficient of the polynomial is an integer and the solution $x_2=1-\sqrt{3}$ is an irrational number, by the Irrational Conjugate Root Theorem, the irrational conjugate of $1-\sqrt{3}$ has to be a root of the function. Using the Factor Theorem, it is possible to write the polynomial by multiplying the binomials produced by the roots. To expand this expression clearly, first the two rightmost binomials will be multiplied. Now, this result will be multiplied by the binomial $(x+5).$ With this result, it is possible to write the correct answer. Therefore, Ali will be the winner. Additionally, it can be checked that the polynomials written by the other friends did not meet the conditions Izabella requested. Zain, in particular, feels sad to have not gotten their answer correct.