Prism Glow - Mathematics - Solving a Quartic Polynomial

Solution of a Quartic Polynomial

This page exists solely to demonstrate the steps taken to solve a quartic polynomial, "quartic" referring to a polynomial of the fourth degree.

The quartic polynomial we wish to solve is:

${ 4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6\quad =\quad 0 }$

There are a series of steps that will lead to a solution of a quartic polynomial. Actually, there is a generalized formula that can be applied, similar to the quadratic formula but much more complex, that will solve a quartic equation. We are not going to study that here. This series of steps uses several different rules and principles that are interesting in their own right.

Steps to Solve a Quartic Equation

1.) Calculate the number of solutions, both real and complex combined. This is equal to the highest power to which the variable is raised in the Polynomial.
2.) Use the Rational Zeroes Theorem to identify possible roots of the equation.
3.) Use Descartes' Rule of Signs to determine possible combinations of counts of positive real, negative real and complex roots.
4.) Make a rough graph of the equation to determine which values in the rational zeroes list are the closest to where the graph crosses the x-axis.
5.) Use the Rule of Upper and Lower Bounds of Roots to narrow down which rational zeroes might apply.
6.) Use Newton's Approximation of Zeroes to get values to required precision for approximated zeroes.
7.) Use approximated zeroes with synthetic division as needed to factor out a quadratic polynomial.
8.) Use the Quadratic Formula to solve for the remaining two zeroes (often the complex zeroes).

We will go through each step to show how they are used to solve our example quartic polynomial.

1.) Calculate the Number of Solutions, Both Real and Complex

This is the simplest part of the solution. The polynomial is arranged with its terms in descending order by exponent.
${ 4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6\quad =\quad 0 }$
Since this is a quartic equation, the first term will be of the fourth power. Therefore, there will be four zeroes to this equation. They could be positive, negative or complex. The mix has yet to be determined. But in total there will be four. (If there are complex solutions, they will always come in pairs and be conjugates of each other.)

2.) Use the Rational Zeroes Theorem to Identify Possible Roots of the Equation

Rational zeroes of an equation are only possible zeroes. They are a convenient place to start to look for solutions. The true zeroes of the equation may be in the set of the rational zeroes or they may not be.

We assign the variable "p" to the constant of the equation and the variable "q" to the coefficient on the first term of the equation. Then for each of "p" and "q", the factors are calculated. The fraction "p/q" is then calculated, making sure to include all possibilities, both positive and negative.

$\underline { Possible\quad Rational\quad Zeroes } \\ \\ q\quad \quad \quad \quad \quad \quad \quad \quad \quad p\\ { 4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6\quad =\quad 0 }\\ \\ factors\quad of\quad p:\quad \pm 1,\quad \pm 2,\quad \pm 3,\quad \pm 6\\ factors\quad of\quad q:\quad \pm 1,\quad \pm 2,\quad \pm 4\\ \\ \frac { p }{ q } :\quad \left\{ \pm \frac { 1 }{ 1 } ,\quad \pm \frac { 1 }{ 2 } ,\quad \pm \frac { 1 }{ 4 } \\ \pm \frac { 2 }{ 1 } \\ \pm \frac { 3 }{ 1 } ,\quad \pm \frac { 3 }{ 2 } ,\quad \pm \frac { 3 }{ 4 } \\ \pm \frac { 6 }{ 1 } \right$
The results obtained are useful to narrow down possible x values that are zeroes of the polynomial. However, as stated earlier, there are no guarantees that any of these p/q values will be actual solutions.

3.) Use Descartes' Rule of Signs to Determine Possible Combinations of Counts of Positive Real, Negative Real and Complex Roots

With the equation arranged with its terms in descending order, check to see how many times the sign of the terms changes from term to term from beginning to end. In this case there is one change of sign between terms. So there will be one real positive root of this equation.

Secondarily, we need to manipulate the equation by replacing "-x" everywhere there is an "x" and recalculate the equation. Then, the same as the first time through, we check to see how many times there is a change of sign between successive terms. Again, we determine that here there is one sign change. That gives us one negative real root.

Now, understanding that there must be four roots of this equation, and seeing that we have shown that one is positive and one is negative, that leaves two unaccounted for. Those will be complex roots, roots each with both a real and an imaginary component. These complex roots always come in pairs so that is another check to make sure the answer makes sense.

$\underline { Descartes'\quad Rule\quad of\quad Signs } \\ \\ Base\quad equation\quad is\quad quartic\quad -\quad 4\quad roots\quad total\\ f\left( x \right) =4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6\\ \\ Positive\quad real\quad roots\\ { f\left( x \right) =4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 1\quad sign\quad change\\ \\ Negative\quad real\quad roots\\ f\left( -x \right) =4{ (-x) }^{ 4 }+3(-{ x) }^{ 2 }-2(-x)-6\quad \\ f\left( -x \right) =4{ x }^{ 4 }+3{ x }^{ 2 }+2x-6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 1\quad sign\quad change\\ \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad Roots\\ \begin{matrix} Positive & Negative & Complex \\ 1 & 1 & 2 \end{matrix}$

4.) Make a Rough Graph of the Equation to Determine which Values in the Rational Zeroes List are the Closest to Where the Graph Crosses the X-Axis.

A helpful step can be to either sketch a graph by hand or use a graphing calculator. The images below came from Microsoft Math. Here we can get a pretty good idea where the graph crosses the x-axis. It is close enough where we can actually see that the rational zeroes we identified earlier are not going to be integer solutions of this equation. We can see that the software has also given us some very rough approximate solutions. But we can do better than that, and by applying the techniques illustrated below will do just that.

5.) Use the Rule of Upper and Lower Bounds of Roots to Narrow Down which Rational Zeroes Might Apply

Looking at the graph that was generated we can see that the real zeroes of the equation are very close to positive 1 and between -1 and -3/4.

If we use some of the rational zeroes we calculated earlier with the Rule of Upper and Lower Bounds, we can narrow our possibilities down to something between two of the candidates. When we use one of the rational zeroes as the divisor in a synthetic division setup, we can observe the results and make a conclusion based upon that. For a positive divisor, if the numbers which appear at the bottom, the quotient, are all positive, that means that the divisor is larger than the zero of the graph. Once we find that case, then test a lower rational zero only to find that the quotient does not have all positive signs, we have bound the upper range for the zero of the graph.

Likewise if we use some of the negative rational zeroes we calculated earlier as the divisors, we can put limits on what the negative zero may be. There the quotient will end up alternating the signs between terms, positive and negative. Once that condition has been reached, we know that there can be no more greater negative number than the divisor that will be a zero of the equation.

$\underline { Rule\quad of\quad Upper\quad and\quad Lower\quad Bounds\quad of\quad Roots } \\ \\ Upper\quad Limit\quad of\quad Positive\quad Root\quad -\quad Start\quad with\quad 1\\ f\left( x \right) =4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6\\ \\ \quad \quad 1\quad |\overline { 4\quad \quad 0\quad \quad 3\quad \quad -2\quad \quad -6 } \\ \quad \quad \quad \quad \quad \underline { \quad \quad 4\quad \quad 4\quad \quad \quad 7\quad \quad \quad 5 } \\ \quad \quad \quad \quad \quad 4\quad 4\quad \quad 7\quad \quad \quad 5\quad \quad -1\quad \leftarrow \quad Not\quad all\quad the\quad same\quad sign\\ \\ \frac { 3 }{ 2 } \quad |\overline { 4\quad \quad 0\quad \quad 3\quad \quad -2\quad \quad -6 } \\ \quad \quad \quad \quad \quad \underline { \quad \quad 6\quad \quad 9\quad \quad 18\quad \quad 24 } \\ \quad \quad \quad \quad \quad 4\quad 6\quad \quad 12\quad 16\quad \quad 18\quad \leftarrow \quad All\quad positive\\ There\quad cannot\quad be\quad a\quad positive\quad root\quad greater\quad than\quad \frac { 3 }{ 2 } .\\ \frac { 3 }{ 2 } \quad <\quad x\quad <\quad 1\\ \\ Lower\quad Limit\quad of\quad Negative\quad Root\quad -\quad Start\quad with\quad -1\\ \\ \quad -1\quad |\overline { 4\quad \quad 0\quad \quad 3\quad \quad -2\quad \quad -6 } \\ \quad \quad \quad \quad \quad \underline { \quad -4\quad \quad 4\quad \quad -7\quad \quad \quad 9\quad } \\ \quad \quad \quad \quad \quad 4\quad -4\quad 7\quad \quad -9\quad \quad \quad 3\quad \leftarrow \quad Alternating\quad signs\quad (but\quad is\quad -1\quad the\quad largest\quad negative?)\\ \\ \frac { -1 }{ 2 } \quad |\overline { 4\quad \quad 0\quad \quad 3\quad \quad -2\quad \quad -6 } \\ \quad \quad \quad \quad \quad \underline { \quad \quad -2\quad \quad 1\quad \quad -2\quad \quad \quad 2 } \\ \quad \quad \quad \quad \quad 4\quad -2\quad \quad 4\quad \quad -4\quad \quad -4\quad \leftarrow \quad Not\quad alternating\quad signs,\\ \quad \\ There\quad cannot\quad be\quad a\quad negative\quad root\quad less\quad than\quad -1\quad \\ \quad but\quad there\quad can\quad be\quad one\quad less\quad than\quad \frac { -1 }{ 2 } .\\ -1\quad <\quad x\quad <\quad \frac { -1 }{ 2 }$

6.) Use Newton's Approximation of Zeroes to Get Values to Required Precision for Approximated Zeroes

Use of Newton's Approximation requires knowledge of how to take the derivative of the equation. This is indeed Calculus but is very basic. It is so much simpler than other methods it is well worth the small investment of time to understand how to mechanically get a derivative from an equation. After just 2 or 3 iterations of applying the starting candidate for zero and then feeding back the result, an exceptionally good approximation of the actual zero may be calculated. As a matter of fact, the accuracy can be as good as you want it to be if you do not mind continuing to run Newton's Approximation.

In our case for the positive zero we used 1 to start with and for the negative zero we used -1 to start. After 4 iterations the answers were by far accurate enough for our purposes and bested MS Math's approximations by several decimal places.

7.) Use Approximated Zeroes with Synthetic Division as Needed to Factor Out a Quadratic Polynomial

Once we have calculated these positive and negative zeroes it is a good idea to test the results. Using our calculated zeroes in polynomial division we can see that the remainders we get are very small decimals. This shows that the solutions we have are close approximations to the exact zeroes. And, by dividing the quotient of the work done on the original equation by the divisor that is the negative zero we get a quadratic equation for our new quotient. This quadratic equation may be used to get the complex zeroes to complete the solution. That will be shown in the next step.

$\quad \quad \quad \quad \quad \quad \quad \quad \quad 4{ x }^{ 3 }\quad +4.1876{ x }^{ 2 }+7.382998x+5.730308\\ x-1.0469\quad |\overline { 4{ x }^{ 4 }\quad +0{ x }^{ 3 }\quad \quad \quad \quad +3{ x }^{ 2 }\quad \quad \quad \quad \quad \quad -2x\quad \quad \quad \quad \quad \quad \quad -6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 4{ x }^{ 4 }\quad -4.1876{ x }^{ 3 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 4.1876{ x }^{ 3 }+{ 3x }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 4.1876{ x }^{ 3 }-4.383998{ x }^{ 2 }\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 7.383998{ x }^{ 2 }-2x\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 7.383998{ x }^{ 2 }-7.730308x\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 5.730308x-6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 5.730308x-5.999059\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -0.000941$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 4{ x }^{ 3 }\quad -3.4105{ x }^{ 2 }+5.907843x-7.037140\\ x+0.85262\quad |\overline { 4{ x }^{ 4 }\quad +0{ x }^{ 3 }\quad \quad \quad \quad +3{ x }^{ 2 }\quad \quad \quad \quad \quad \quad -2x\quad \quad \quad \quad \quad \quad \quad \quad -6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { \quad 4{ x }^{ 4 }\quad +3.4105{ x }^{ 3 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -3.4105{ x }^{ 3 }+{ 3x }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -\underline { 3.4105{ x }^{ 3 }-2.907843{ x }^{ 2 }\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 5.907843{ x }^{ 2 }-2x\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 5.907843{ x }^{ 2 }+5.037145x\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -7.037140x-6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -7.037140x-6.000010\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -0.000010$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 4{ x }^{ 2 }\quad +0.77710{ x }\quad +6.720427\\ x+0.85262\quad |\overline { 4{ x }^{ 3 }\quad +4.1876{ x }^{ 2 }\quad +7.382998x\quad +5.730308 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { \quad 4{ x }^{ 3 }\quad +3.4105{ x }^{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -0.77710{ x }^{ 2 }+{ 7.382998x }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -0.77710{ x }^{ 2 }+0.662571{ x }\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 6.720427{ x }+5.730308\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 6.720427{ x }+5.729970\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.000338$

8.) Use the Quadratic Formula to Solve for the Remaining Two Zeroes (Often the Complex Zeroes)

Taking the quotient calculated in the last step and putting it into the quadratic formula will result in our getting two complex numbers that are conjugates of each other. These are the last two zeroes of our quartic equation.

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 4{ x }^{ 2 }\quad +0.77710{ x }\quad +6.720427\quad =\quad 0\\ \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad Apply\quad the\quad Quadratic\quad Formula\quad to\quad derive\quad x\\ \quad \quad \quad \quad \quad \quad \quad Let\quad a\quad =\quad 4,\quad b\quad =\quad 0.77710,\quad c\quad =\quad 6.720427\\ \\ \quad \quad \quad \quad \quad \quad \quad x\quad =\quad \frac { -b\quad \pm \quad \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { -0.77710\quad \pm \quad \sqrt { { (0.77710) }^{ 2 }-4(4)(6.720427) } }{ 2(4) } \\ \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { -0.77710\quad \pm \quad \sqrt { -106.93865 } }{ 8 } \\ \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \quad -0.09714\quad \pm \quad 1.2926i$

This gives us all the zeroes of our original equation and completes the solution:

X = 1.0469

X = -0.85262

X = -0.09714 + 1.2926i

X = -0.09714 - 1.2926i

Mathematics > Solution of a Quartic Polynomial
Solution of a Quartic Polynomial This page exists solely to demonstrate the steps taken to solve a quartic polynomial, "quartic" referring to a polynomial of the fourth degree. The quartic polynomial we wish to solve is: ${ 4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6\quad =\quad 0 }$ There are a series of steps that will lead to a solution of a quartic polynomial. Actually, there is a generalized formula that can be applied, similar to the quadratic formula but much more complex, that will solve a quartic equation. We are not going to study that here. This series of steps uses several different rules and principles that are interesting in their own right. Steps to Solve a Quartic Equation 1.) Calculate the number of solutions, both real and complex combined. This is equal to the highest power to which the variable is raised in the Polynomial. 2.) Use the Rational Zeroes Theorem to identify possible roots of the equation. 3.) Use Descartes' Rule of Signs to determine possible combinations of counts of positive real, negative real and complex roots. 4.) Make a rough graph of the equation to determine which values in the rational zeroes list are the closest to where the graph crosses the x-axis. 5.) Use the Rule of Upper and Lower Bounds of Roots to narrow down which rational zeroes might apply. 6.) Use Newton's Approximation of Zeroes to get values to required precision for approximated zeroes. 7.) Use approximated zeroes with synthetic division as needed to factor out a quadratic polynomial. 8.) Use the Quadratic Formula to solve for the remaining two zeroes (often the complex zeroes). We will go through each step to show how they are used to solve our example quartic polynomial. 1.) Calculate the Number of Solutions, Both Real and Complex This is the simplest part of the solution. The polynomial is arranged with its terms in descending order by exponent. ${ 4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6\quad =\quad 0 }$ Since this is a quartic equation, the first term will be of the fourth power. Therefore, there will be four zeroes to this equation. They could be positive, negative or complex. The mix has yet to be determined. But in total there will be four. (If there are complex solutions, they will always come in pairs and be conjugates of each other.) 2.) Use the Rational Zeroes Theorem to Identify Possible Roots of the Equation Rational zeroes of an equation are only possible zeroes. They are a convenient place to start to look for solutions. The true zeroes of the equation may be in the set of the rational zeroes or they may not be. We assign the variable "p" to the constant of the equation and the variable "q" to the coefficient on the first term of the equation. Then for each of "p" and "q", the factors are calculated. The fraction "p/q" is then calculated, making sure to include all possibilities, both positive and negative. $\underline { Possible\quad Rational\quad Zeroes } \\ \\ q\quad \quad \quad \quad \quad \quad \quad \quad \quad p\\ { 4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6\quad =\quad 0 }\\ \\ factors\quad of\quad p:\quad \pm 1,\quad \pm 2,\quad \pm 3,\quad \pm 6\\ factors\quad of\quad q:\quad \pm 1,\quad \pm 2,\quad \pm 4\\ \\ \frac { p }{ q } :\quad \left\{ \pm \frac { 1 }{ 1 } ,\quad \pm \frac { 1 }{ 2 } ,\quad \pm \frac { 1 }{ 4 } \\ \pm \frac { 2 }{ 1 } \\ \pm \frac { 3 }{ 1 } ,\quad \pm \frac { 3 }{ 2 } ,\quad \pm \frac { 3 }{ 4 } \\ \pm \frac { 6 }{ 1 } \right$ The results obtained are useful to narrow down possible x values that are zeroes of the polynomial. However, as stated earlier, there are no guarantees that any of these p/q values will be actual solutions. 3.) Use Descartes' Rule of Signs to Determine Possible Combinations of Counts of Positive Real, Negative Real and Complex Roots With the equation arranged with its terms in descending order, check to see how many times the sign of the terms changes from term to term from beginning to end. In this case there is one change of sign between terms. So there will be one real positive root of this equation. Secondarily, we need to manipulate the equation by replacing "-x" everywhere there is an "x" and recalculate the equation. Then, the same as the first time through, we check to see how many times there is a change of sign between successive terms. Again, we determine that here there is one sign change. That gives us one negative real root. Now, understanding that there must be four roots of this equation, and seeing that we have shown that one is positive and one is negative, that leaves two unaccounted for. Those will be complex roots, roots each with both a real and an imaginary component. These complex roots always come in pairs so that is another check to make sure the answer makes sense. $\underline { Descartes'\quad Rule\quad of\quad Signs } \\ \\ Base\quad equation\quad is\quad quartic\quad -\quad 4\quad roots\quad total\\ f\left( x \right) =4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6\\ \\ Positive\quad real\quad roots\\ { f\left( x \right) =4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 1\quad sign\quad change\\ \\ Negative\quad real\quad roots\\ f\left( -x \right) =4{ (-x) }^{ 4 }+3(-{ x) }^{ 2 }-2(-x)-6\quad \\ f\left( -x \right) =4{ x }^{ 4 }+3{ x }^{ 2 }+2x-6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 1\quad sign\quad change\\ \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad Roots\\ \begin{matrix} Positive & Negative & Complex \\ 1 & 1 & 2 \end{matrix}$ 4.) Make a Rough Graph of the Equation to Determine which Values in the Rational Zeroes List are the Closest to Where the Graph Crosses the X-Axis. A helpful step can be to either sketch a graph by hand or use a graphing calculator. The images below came from Microsoft Math. Here we can get a pretty good idea where the graph crosses the x-axis. It is close enough where we can actually see that the rational zeroes we identified earlier are not going to be integer solutions of this equation. We can see that the software has also given us some very rough approximate solutions. But we can do better than that, and by applying the techniques illustrated below will do just that. 5.) Use the Rule of Upper and Lower Bounds of Roots to Narrow Down which Rational Zeroes Might Apply Looking at the graph that was generated we can see that the real zeroes of the equation are very close to positive 1 and between -1 and -3/4. If we use some of the rational zeroes we calculated earlier with the Rule of Upper and Lower Bounds, we can narrow our possibilities down to something between two of the candidates. When we use one of the rational zeroes as the divisor in a synthetic division setup, we can observe the results and make a conclusion based upon that. For a positive divisor, if the numbers which appear at the bottom, the quotient, are all positive, that means that the divisor is larger than the zero of the graph. Once we find that case, then test a lower rational zero only to find that the quotient does not have all positive signs, we have bound the upper range for the zero of the graph. Likewise if we use some of the negative rational zeroes we calculated earlier as the divisors, we can put limits on what the negative zero may be. There the quotient will end up alternating the signs between terms, positive and negative. Once that condition has been reached, we know that there can be no more greater negative number than the divisor that will be a zero of the equation. $\underline { Rule\quad of\quad Upper\quad and\quad Lower\quad Bounds\quad of\quad Roots } \\ \\ Upper\quad Limit\quad of\quad Positive\quad Root\quad -\quad Start\quad with\quad 1\\ f\left( x \right) =4{ x }^{ 4 }+3{ x }^{ 2 }-2x-6\\ \\ \quad \quad 1\quad \|\overline { 4\quad \quad 0\quad \quad 3\quad \quad -2\quad \quad -6 } \\ \quad \quad \quad \quad \quad \underline { \quad \quad 4\quad \quad 4\quad \quad \quad 7\quad \quad \quad 5 } \\ \quad \quad \quad \quad \quad 4\quad 4\quad \quad 7\quad \quad \quad 5\quad \quad -1\quad \leftarrow \quad Not\quad all\quad the\quad same\quad sign\\ \\ \frac { 3 }{ 2 } \quad \|\overline { 4\quad \quad 0\quad \quad 3\quad \quad -2\quad \quad -6 } \\ \quad \quad \quad \quad \quad \underline { \quad \quad 6\quad \quad 9\quad \quad 18\quad \quad 24 } \\ \quad \quad \quad \quad \quad 4\quad 6\quad \quad 12\quad 16\quad \quad 18\quad \leftarrow \quad All\quad positive\\ There\quad cannot\quad be\quad a\quad positive\quad root\quad greater\quad than\quad \frac { 3 }{ 2 } .\\ \frac { 3 }{ 2 } \quad <\quad x\quad <\quad 1\\ \\ Lower\quad Limit\quad of\quad Negative\quad Root\quad -\quad Start\quad with\quad -1\\ \\ \quad -1\quad \|\overline { 4\quad \quad 0\quad \quad 3\quad \quad -2\quad \quad -6 } \\ \quad \quad \quad \quad \quad \underline { \quad -4\quad \quad 4\quad \quad -7\quad \quad \quad 9\quad } \\ \quad \quad \quad \quad \quad 4\quad -4\quad 7\quad \quad -9\quad \quad \quad 3\quad \leftarrow \quad Alternating\quad signs\quad (but\quad is\quad -1\quad the\quad largest\quad negative?)\\ \\ \frac { -1 }{ 2 } \quad \|\overline { 4\quad \quad 0\quad \quad 3\quad \quad -2\quad \quad -6 } \\ \quad \quad \quad \quad \quad \underline { \quad \quad -2\quad \quad 1\quad \quad -2\quad \quad \quad 2 } \\ \quad \quad \quad \quad \quad 4\quad -2\quad \quad 4\quad \quad -4\quad \quad -4\quad \leftarrow \quad Not\quad alternating\quad signs,\\ \quad \\ There\quad cannot\quad be\quad a\quad negative\quad root\quad less\quad than\quad -1\quad \\ \quad but\quad there\quad can\quad be\quad one\quad less\quad than\quad \frac { -1 }{ 2 } .\\ -1\quad <\quad x\quad <\quad \frac { -1 }{ 2 }$ 6.) Use Newton's Approximation of Zeroes to Get Values to Required Precision for Approximated Zeroes Use of Newton's Approximation requires knowledge of how to take the derivative of the equation. This is indeed Calculus but is very basic. It is so much simpler than other methods it is well worth the small investment of time to understand how to mechanically get a derivative from an equation. After just 2 or 3 iterations of applying the starting candidate for zero and then feeding back the result, an exceptionally good approximation of the actual zero may be calculated. As a matter of fact, the accuracy can be as good as you want it to be if you do not mind continuing to run Newton's Approximation. In our case for the positive zero we used 1 to start with and for the negative zero we used -1 to start. After 4 iterations the answers were by far accurate enough for our purposes and bested MS Math's approximations by several decimal places. 7.) Use Approximated Zeroes with Synthetic Division as Needed to Factor Out a Quadratic Polynomial Once we have calculated these positive and negative zeroes it is a good idea to test the results. Using our calculated zeroes in polynomial division we can see that the remainders we get are very small decimals. This shows that the solutions we have are close approximations to the exact zeroes. And, by dividing the quotient of the work done on the original equation by the divisor that is the negative zero we get a quadratic equation for our new quotient. This quadratic equation may be used to get the complex zeroes to complete the solution. That will be shown in the next step. $\quad \quad \quad \quad \quad \quad \quad \quad \quad 4{ x }^{ 3 }\quad +4.1876{ x }^{ 2 }+7.382998x+5.730308\\ x-1.0469\quad \|\overline { 4{ x }^{ 4 }\quad +0{ x }^{ 3 }\quad \quad \quad \quad +3{ x }^{ 2 }\quad \quad \quad \quad \quad \quad -2x\quad \quad \quad \quad \quad \quad \quad -6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 4{ x }^{ 4 }\quad -4.1876{ x }^{ 3 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 4.1876{ x }^{ 3 }+{ 3x }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 4.1876{ x }^{ 3 }-4.383998{ x }^{ 2 }\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 7.383998{ x }^{ 2 }-2x\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 7.383998{ x }^{ 2 }-7.730308x\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 5.730308x-6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 5.730308x-5.999059\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -0.000941$ $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 4{ x }^{ 3 }\quad -3.4105{ x }^{ 2 }+5.907843x-7.037140\\ x+0.85262\quad \|\overline { 4{ x }^{ 4 }\quad +0{ x }^{ 3 }\quad \quad \quad \quad +3{ x }^{ 2 }\quad \quad \quad \quad \quad \quad -2x\quad \quad \quad \quad \quad \quad \quad \quad -6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { \quad 4{ x }^{ 4 }\quad +3.4105{ x }^{ 3 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -3.4105{ x }^{ 3 }+{ 3x }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -\underline { 3.4105{ x }^{ 3 }-2.907843{ x }^{ 2 }\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 5.907843{ x }^{ 2 }-2x\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 5.907843{ x }^{ 2 }+5.037145x\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -7.037140x-6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -7.037140x-6.000010\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -0.000010$ $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 4{ x }^{ 2 }\quad +0.77710{ x }\quad +6.720427\\ x+0.85262\quad \|\overline { 4{ x }^{ 3 }\quad +4.1876{ x }^{ 2 }\quad +7.382998x\quad +5.730308 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { \quad 4{ x }^{ 3 }\quad +3.4105{ x }^{ 2 } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -0.77710{ x }^{ 2 }+{ 7.382998x }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { -0.77710{ x }^{ 2 }+0.662571{ x }\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 6.720427{ x }+5.730308\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \underline { 6.720427{ x }+5.729970\quad } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.000338$ 8.) Use the Quadratic Formula to Solve for the Remaining Two Zeroes (Often the Complex Zeroes) Taking the quotient calculated in the last step and putting it into the quadratic formula will result in our getting two complex numbers that are conjugates of each other. These are the last two zeroes of our quartic equation. $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 4{ x }^{ 2 }\quad +0.77710{ x }\quad +6.720427\quad =\quad 0\\ \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad Apply\quad the\quad Quadratic\quad Formula\quad to\quad derive\quad x\\ \quad \quad \quad \quad \quad \quad \quad Let\quad a\quad =\quad 4,\quad b\quad =\quad 0.77710,\quad c\quad =\quad 6.720427\\ \\ \quad \quad \quad \quad \quad \quad \quad x\quad =\quad \frac { -b\quad \pm \quad \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { -0.77710\quad \pm \quad \sqrt { { (0.77710) }^{ 2 }-4(4)(6.720427) } }{ 2(4) } \\ \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { -0.77710\quad \pm \quad \sqrt { -106.93865 } }{ 8 } \\ \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \quad -0.09714\quad \pm \quad 1.2926i$ This gives us all the zeroes of our original equation and completes the solution: X = 1.0469 X = -0.85262 X = -0.09714 + 1.2926i X = -0.09714 - 1.2926i