# Roots of Polynomial Equations

Baby Mandelbrot found by calculating where the orbit of the critical point (the root of the polynomial equation z2 + c = c) does not tend to infinity

In Chapter 3, we discovered when we vertically translated, reflected or scaled linear (degree 1) equations, the transformed equations were equivalent, in that they had the same solutions.  This allowed us to solve easier equations instead, which gave us the same solutions as the original equations .  These transformations also explained the steps behind solving the linear equations algebraically.

 Given the linear equation  2x + 3 = 9, we would translate the polynomial and the output value down 3 (subtract 3 from both sides) to get  2x = 6, and then scale the transformed polynomial and output value by 1/2 (divide both sides by 2) to get  x = 3.

In other words, we would geometrically transform the polynomial into x, a 45º diagonal line through the origin where the input values and output values are always the same (get the x alone).

In this chapter, the question is, will the same approach work for quadratic (degree 2) and/or higher degree equations?

 Given a quadratic equation, such as  6×2 – 5x + 3 = 7  , will translating and scaling, transform the polynomial to x?

Go on to begin answering this question.

To solve  \$latex 6x^{2} – 5x + 3 = 7 \$ using the same approach as in Chapter 3, we would first vertically translate the polynomial and the output value down 3 to get \$latex 6x^{2} – 5x = 4\$ . The next step would be to vertically scale the translated polynomial and the translated output value.

Below is a snapshot of an applet that will allow you to vertically translate and then vertically scale the above example.

Run the applet  “quadexample.html”  , to see if vertically translating and scaling the polynomial and output value, works for solving quadratic equations, as it does for solving linear equations.

After recording and defending your answer, go on to the next page.

As we discovered, scaling makes the parabola (the graph of a quadratic) wider, but no matter how much we scale, it will never become a line.  Even if scaling could turn a parabola into a line, there are two solutions, not one.  The best this approach can do, is give an estimate of the values of the solutions.

From looking at the graph of  \$latex 6x^{2} – 5x + 3 = 7 \$ on the right, it’s clear there are 2 solutions (there are two input values that have output values of 7). Whatever the solutions are, we would algebraically write them both as, x = value 1  and  x = value 2.

However, x doesn’t just represent the two missing input values, x also represents two linear polynomials.  This means  there are two equivalent linear equations with the same solution values. In other words, the reason the above approach didn’t work, is because we  need a way to transform the polynomial into two lines, not one.

Below are snapshots of an applet, which will randomly generate a quadratic equation, two equivalent linear equations, and show all three graphs.

In the applet  “quadlines.html”  , follow the directions after the 1st bullet, and answer the question after the 2nd bullet.

Use a copy of the pdf  appletnotes  to show your work and to record your answer to the given question.

After neatly showing your work, and recording your answer to the given question, go on to the next page.

In Chapter 2, we found out that all quadratic (degree 2) polynomials can be factored into two linear (degree 1) polynomials, which is how a quadratic equation can be written as two linear polynomials.  Knowing that two linear factors exist is one thing though, being able to find them is another.  So for now, as in Chapter 2, we’ll only consider a quadratic factorable, if the two linear polynomial factors have integer coefficients and constants.  Even if we’re able to factor the quadratic polynomial in an equation into two linear polynomials, then what?

Solve the following factored quadratic equation algebraically (without graphing).

 \$latex 6x^{2} + 7x – 20 = -17 \$ (2x + 5)(3x – 4) = -17 What two input values for x, will result in an output value of -17?

Use a copy of the pdf  booknotes, to record your work and the steps which led you to the two solutions.

After finding both values for x which will make the above equations true, go on to the next page.

Chances are, it took a lot of guessing and checking to find the two solutions to the problem on the previous page.

Below is a snapshot of an applet, which will give us a hint for finding an easier way to solve factored quadratic equations.

Run the applet  “quadtranslate2.html”  , until you have an answer to the given question.

Use a copy of the pdf  appletnotes1g, to record and defend your answer.

After recording and defending your answer to the given question, go on to the next page.

Below is a snapshot of an applet, which will either confirm or allow you to modify your answer from the last applet.

Run the applet  “zpp.html”  , and modify your answer as needed, to the question from the last applet (shown below).

“Since the plan is to factor the quadratic on the left side of the equation, into two linear polynomials, which equivalent vertically translated quadratic equation would you choose to solve, and why?”

After confirming or modifying your answer to the above question from the last applet, go on to the next page.

By vertically translating the polynomial and the output value in a quadratic equation, until the output value is a zero, turns a “guess and check” problem into a “go right to the solutions” problem.  This is because the only way we can have a zero for a product, is if at least one of the factors is a zero (this is called the zero product property).  Below is the example from earlier in the chapter.

 \$latex 6x^{2} + 7x – 20 = -17 \$  (2x + 5)(3x – 4) = -17 What two input values for x will result in an output value of -17?

Instead of factoring and trying to solve the equation as it was given, we now know we can vertically translate, and then solve an equivalent equation which has an output value of zero.

 \$latex 6x^{2} + 7x – 20 = -17 \$ \$latex 6x^{2} + 7x – 3 = 0 \$ \$latex (3x – 1)(2x + 3) = 0 \$

What’s nice about this is, we can use the zero product property to go right to the solutions, because the only way we can get an output value of zero, is if  3x – 1 = 0  or  2x + 3 = 0  .

 3x – 1 = 0                            2x + 3 = 0  3x = 1                            2x = -3 x = 1/3                           x = -3/2

Below is a snapshot of an applet which gives examples of using the zero product property to solve quadratic equations.

Run the applet  “quadtranslate3.html”  until you can solve a quadratic equation, without moving the slider.

Use a copy of the pdf  appletnotes1g  to copy down an example and record your work.

After successfully solving an example and recording your work neatly, go on to the next page.

Solutions to polynomials, set equal to zero, are called roots or zeros.  When solving polynomial equations though, we can use the term solutions, roots, or zeros, to represent the values that make the equations true (for every polynomial equation with an output value not equal to zero, there’s an equivalent polynomial equation with an output value that is ).

To the right are snapshots of a program, which will randomly generate 7 quadratic equations, in factored form.

Run  program 401  until you can successfully solve at least 5 of the 7 factored quadratic equations.

Use a copy of the pdf  programnotes  to record the equations, your work, and any discoveries.

After successfully solving at least 5 of the 7 problems, and neatly recording your work, go on to the next page.

Below are snapshots of a program, which will randomly generate 10 quadratic equations, with integer roots.

Run  program 402  until you can successfully solve at least 8 of the 10 quadratic equations by factroring.

Use a copy of the pdf  programnotes to record the equations, your work, and any discoveries.

After successfully solving at least 8 of the 10 equations, and neatly recording your work, go on to the next page.

Below are snapshots of a program, which will randomly generate 10 quadratic equations, with rational roots.

Run  program 403  until you can successfully solve at least 8 of the 10 quadratic equations.

Use a copy of the pdf  programnotes  to record the equations, your work, and any discoveries.

After successfully solving at least 8 of the 10 equations, and neatly recording your work, go on to the next page.

As we discovered in chapter 2, sometimes the best we can do, is factor out the greatest common factor.

To the right is a snapshot of a program, which will randomly generate 10 quadratic equations, with rational roots.

Run  program 404  until you can successfully solve at least 8 of the 10 quadratic equations.

Use a copy of the pdf  programnotes  to record the equations, your work, and any discoveries.

After successfully solving at least 8 of the 10 equations, and neatly recording your work, go on to the next page.

Below are examples of a program, which will randomly generate 10 sets of integers. The challenge is to find the simplest polynomial, whose roots, are those sets of integers.

 What is the simplest polynomial with a root of 5? x = 5 x – 5 = 0 x – 5What is the simplest polynomial with roots of 1 and -3? x = 1 or x = -3 x-1=0 or x+3 = 0 (x – 1) (x + 3) = 0 \$latex x^{2} + 2x – 3 = 0 \$ \$latex x^{2} + 2x – 3\$

Run  program 405  until you can successfully determine at least 8 of the 10 polynomials.

Use a copy of the pdf  programnotes  to neatly record your work and any discoveries.

After successfully determining at least 8 of the polynomials, and neatly recording your work, go on to the next page.

Below are examples of a program, which will randomly generate 10 sets of rational numbers. The challenge is to find the simplest polynomial, whose roots, are those sets of rational numbers.

 What’s the simplest polynomial which has a root of  -8/9  ? x = -8/9 9x = -8 9x + 8 = 0 9x + 8What is the simplest polynomial which has roots of  7/8  and  -3/4  ? x = 7/8  or  x = -3/4 8x = 7  or  4x = -3 8x – 7 = 0  or  4x + 3 = 0 (8x – 7)(4x + 3) = 0 \$latex 32x^{2} – 4x – 21 = 0 \$ \$latex 32x^{2} – 4x – 21 \$

Run  program 406  until you can successfully determine at least 8 of the 10 polynomials.

Use a copy of the pdf  programnotes to neatly record your work and any discoveries.

After successfully determining at least 8 of the polynomials, and neatly recording your work, go on to the next page.

So far we can solve any factorable quadratic equation, by using the zero product property.

In other words, we only know how to factor quadratics, which have linear factors with integer coefficients and constants.

 In 8 guesses or less, we can factor the quadratic 2×2 – 11x – 21( 2x – 1 )( 1x + 21 )  =  2×2 + 41x – 21 ( 2x + 1 )( 1x – 21 )  =  2×2 – 41x – 21 ( 2x – 21 )( 1x + 1 )  =  2×2 – 19x – 21 ( 2x + 21 )( 1x – 1 )  =  2×2 + 19x – 21 ( 2x – 7 )( 1x + 3 )  =  2×2 – 1x – 21 ( 2x + 7 )( 1x – 3 )  =  2×2 + 1x – 21 ( 2x – 3 )( 1x + 7 )  =  2×2 + 11x – 21 ( 2x + 3 )( 1x – 7 )  =  2×2 – 11x – 21

Good luck though, trying to factor the quadratic,  4×2 – 7x – 3  , by guessing!

 ( 8x – 7 + √ 97  )( 8x – 7 – √ 97  )

We can solve any factorable quadratic equation, by vertically translating until the output value is zero, and then use the zero product property.

If the quadratic isn’t factorable though, having an output value of zero, isn’t good enough.

To the right are snapshots of an applet, which will allow  us to see what happens to the quadratic equation, when we continue to use other transformations as well.

Run the applet  “quadtransform.html”  , until you have successfully completed all 3 steps.

Use a copy of the pdf  appletnotes, to record any information which may be revealed, along with an example.

After you have neatly recorded any revealed information, along with an example, go on to the next page.

Now we know we can transform any quadratic equation, into a monic quadratic equation (where the lead coefficient is a positive one), with an output value of zero.

Below are snapshots of an applet, which will give examples of a special type of monic quadratic equation.

Run the applet  “rorschach.html”  , and use the slider until you can answer the 2 questions.

Use a copy of the pdf  appletnotes1g  to record both of your answers, an example, and any discoveries.

After recording an example, and defending your answers to the 2 questions, go on to the next page.

A monic quadratic equation where the coefficient of the x term is zero, and the constant is negative, is referred to as a difference of two squares, which can always be factored into x +/- the square root of the constant.

It even works if the constant isn’t a perfect square.

 x2 – 25 = 0   x2 – 7 = 0 (x + 5)(x – 5) = 0           (x + √7)(x – √7) = 0 x = -5   or   x = 5          x = -√7 or x = √7

We can also solve the equations by translating up and solving the equivalent linear equation (which also works for higher degree polynomial equations).

Below are snapshots of an applet which will give examples of monic polynomial equations of degree 1 through 5.

Run the applet  “polyroots.html”  to explore more examples, and then go on to the next page.

So now we know we can always transform a quadratic equation into a monic quadratic equation, and a monic quadratic equation with no x term, can always be factored.

If we could transform any monic quadratic equation, into a monic quadratic equation with no x term, we’d be all set.

Below are snapshots of an applet which will allow you to discover if this is possible, or if we’ll have to look for another way.

Run the applet  “xminusn.html”  , and use the slider until you can answer the 2 questions.

Use a copy of the pdf  appletnotes1g, to record both answers, and to algebraically simplify an example.

After recording both answers, and defending them by simplifying an example algebraically, go on to the next page.

The good news is, it’s possible to transform a monic quadratic equation into a monic quadratic equation with no x term.  The bad news is, without using an applet with a slider, how are we supposed to know how far to translate horizontally?

We know we want the vertex (the lowest or highest point) of the parabola (the graph of a quadratic) to be on the y-axis, so if we knew the x-value of the vertex of the original quadratic, we would know how far to translate.

In the example below, the x-value of the vertex of the original quadratic (in black) is 3, so we know we’ll have to translate horizontally to the left 3.

The problem is, without looking at the above graph, how would we know the x-value of the vertex was 3?

Below are snapshots of an applet, which will allow you to discover a connection (if there is one), between the roots of a monic quadratic, and the x-value of the vertex.

Run the applet  “vertex.html”  until you discover a connection between the roots and the vertex of a monic quadratic.

Use a copy of the pdf  appletnotes1g to record your discovery, along with at least one example.

After recording your discovery, and defending it with at least one example, go on to the next page.

We now know, for any monic quadratic, there is a connection between the two roots and the x-value of the vertex.

Knowing the roots, we can just average them to find the x-value of the vertex, and know how far to horizontally translate.

The problem is, that’s the thing, we’re looking for a way to find the roots of unfactorable quadratic equations.

Hopefully, there’s another way of determining the x-value of the vertex, without having to know what the roots are.

To the right, are snapshots of an applet,  which will allow you to search for a different way of determining the x-value of the vertex, without knowing the values of the two roots of a monic quadratic.

Run the applet  “monicsumprod.html”  to see if you can find another way to determine the x-value of the vertex, without knowing the values of the two roots.

Use a copy of the pdf  appletnotes1g, to record your conclusion, and defend it with an example.

After searching for another way to determine the x-value of the vertex, without knowing the values of the two roots, and defending your conclusion with an example, go on to the next page.

We know the x-value of the vertex of a monic quadratic, is the average of the two roots, due to the symmetry of parabolas.

We also know, the average of two values, is just their sum divided by two.  As we just discovered, this means all we have to do, is take half of the opposite of the x coefficient, to know how far to horizontally translate.  Below is an example.

Below are snapshots of an applet which will show examples of solving monic quadratic equations, by translating horizontally.  After getting used to this approach, we will be able to solve any quadratic equation, with any type of roots.

Run the applet  “quadroots.html”  , until you can successfully find the roots of a monic quadratic equation, by algebraically translating horizontally, without using the slider.

Use a copy of the pdf  appletnotes, to neatly record your work for at least one example.

Challenge:  Try the horizontal translation method on a problem from  program 403  , which won’t be monic to start.

After finding the roots of a monic quadratic equation, by algebraically translating horizontally, and neatly recording your work before using the slider, go on to the next page.

Since translating horizontally can turn any monic quadratic equation into a difference of two squares equation, which can always be factored (even if the roots end up being irrational), we can now solve any quadratic equation.  Of course, if the roots are rational numbers (integers and/or fractions), we can still factor and use the zero product property if we wish.

In other words, we can use the horizontal translation method on a general quadratic equation (one that doesn’t use specific values for the coefficients and constant), to create a formula for solving any quadratic equation.

Below are the first few steps for using the horizontal translation method on a general quadratic equation.

Challenge: continue the steps, until you have a formula for both roots of any quadratic equation.

Use a copy of the pdf  booknotes  to neatly show all of your steps.

To see how you did, open the pdf  quadraticformula, before going on to the next page.

For objects in free fall, near the surface of the earth, there is a quadratic relationship between time and distance.

Ignoring friction, the quadratic  h – 16t2  gives the height in feet of an object, after falling t seconds, from a height of h feet.

Below are snapshots of an applet, which will allow you to model a blue sphere being dropped from a given height.

Run the applet  “gravity.html”  to explore the quadratic relationship between time and distance of an object in free fall.

Choose a height (other than 25 ft), and determine how many seconds (rounded to the hundredths place) it would take for the blue sphere to hit the ground, and then use the “free fall time” slider to check your prediction.

Use a copy of the pdf  appletnotes  to make a sketch, to show all of your work, and to record any discoveries.

Determine how many seconds it would take an object to fall from 142 feet, and then open the pdf  “Bowdoin.pdf”  .

Determine how many seconds it would take an object to fall from 420 feet, and then open the pdf  “tower.pdf”  .

To learn how to make a ruler which measures time, open the pdf  gravityruler

After determining the correct free fall time of the blue sphere, and neatly showing your work,go on to the next page.

Near the surface of the earth, the trajectory (path) of an object when propelled at an angle, can be described as parabolic.

Below are snapshots of an applet, which will challenge you to describe the trajectory of the water, coming out of the bubbler in front of the school’s 8th grade science room.

Run the applet  “fountain.html”  , until you make the blue parabola match the stream of water.

After finding the quadratic polynomial which describes the path of the water, go on to the next page.

The vertical and horizontal parts of the paths of projectiles are independent of each other, and can be studied separately.

Below are snapshots of an applet, which will let you further explore trajectories near the earth’s surface (ignoring friction).

Run the applet  “trajectories.html”  to see what you can discover.

Use a copy of the pdf  “appletnotes(2g)”  to record your discoveries.

After recording your discoveries, go on to the next page.

Earlier in the chapter, we discovered it’s possible to transform any quadratic equation into a monic one (where the lead coefficient is a positive 1).  We also discovered it’s possible to solve (find the 2 roots for) any monic quadratic equation by translating horizontally.  Not only that, you discovered the same approach can produce a formula for finding the two roots.

Formulas for the roots of cubic (degree 3) and quartic (degree 4) equations exist as well.  It was proven by Niels Abel in the 1800’s though, that it’s impossible to create algebraic formulas based on coefficients, for finding all of the roots of polynomial equations of degree 5 or higher.  Thanks to the Fundamental Theorem of Algebra, wich was first proved by Carl Gauss in 1799, we know for any polynomial equation, a root will always exist.  Actually, a degree n polynomial equation will always have n roots (they could be complex numbers and/or repeats though).  Many methods for finding estimates of roots to higher degree polynomial equations exist.  For now though we’ll rely on the fact, if the input value for a polynomial results in an output value of zero, then that input value is a root.  Two examples are shown below.

 \$latex x^{3} – x^{2} – 4x + 4 \$                              \$latex x^{3} – x^{2} – 4x + 4 \$ \$latex (2)^{3} – (2)^{2} – 4(2) + 4 \$                           \$latex (5)^{3} – (5)^{2} – 4(5) + 4 \$ 8 – 4 – 8 + 4                                                         125 – 25 – 20 + 4 0  ( so 2 is a root )                                                 84  ( so 5 is not a root )

It turns out, there’s a quicker way to determine if an input value is a root to a polynomial. It’s called synthetic division.

Below are snapshots of a program, which will randomly generate 10 synthetic division problems for you.

Run  program 412  , until you discover how to use synthetic division, and get at least 8 of the 10 problems correct.

Use a copy of the pdf  programnotes2g to record your work and discoveries.

Challenge:  show why synthetic division works.  Use x as the input value, and letters for the coefficients/constant.

Challenge:  If an input of 5 results in a -3, and an input of 6 results in a 2, what do you know about the polynomial?

After using synthetic division correctly for at least 8 of the problems, and recording your work, go on to the next page.

To the right is a picture of the Mandelbrot set (the bug shaped object shown in black). Points in the c-plane are used for c in the quadratic  z2 + c , and iterated using a seed of 0.  If the orbit  does not go off to infinity,  the c-value is part of  the Mandelbrot set.

The seed of 0 is  called the critical point. It’s the root of the unique quadratic equation z2 + c = c .

If you thought the picture at the beginning of the chapter was interesting, open the spreadsheet  “bigMset.numbers”  , to explore a 26 ft by 13 ft picture of the Mandelbrot set.

Below is a snapshot of the chapter 4 programs menu program.

Run the program  “4menu.bas”, or its alias to access every program used in chapter 4.

“If I feel unhappy, I do mathematics to become happy.  If I am happy, I do mathematics to keep happy” — Alfred Renyi —

“The tantalizing and compelling pursuit of mathematical problems offers mental obsorption, peace of mind amid endless challenges, repose in activity, battle without conflict…” — Morris Kline —

“The Mandelbrot set is the most complex mathematical object known to mankind.” — Benoit Mandelbrot —

“There can be very little of present-day science and technology that is not dependent on complex numbers in one way or another.” — Keith Devlin —

“Indeed, nowadays no electical engineer could get along without complex numbers, and neither could anyone working in aerodynamics or fluid dynamics.” — Keith Devlin —