How to Solve for x: Factoring Polynomials Homework Statement

[Jen23]

Homework Statement

Solve for x:

3x^3 + 2x^2 + 75x - 50 = 0

The Attempt at a Solution

I have tried substituting multiple values for "x" so that we get a factor, f(x)=0
I cannot seem to find an "x" value that will make this function=0. Is there a way to factor this function or did the book have an error when printing the equation?

 

[SammyS]

Homework Statement

Solve for x:

3x^3 + 2x^2 + 75x - 50 = 0

The Attempt at a Solution

I have tried substituting multiple values for "x" so that we get a factor, f(x)=0
I cannot seem to find an "x" value that will make this function=0. Is there a way to factor this function or did the book have an error when printing the equation?

I suspect that the coefficient of x (the linear term) should have been negative.

 

[fresh_42]

Homework Statement

Solve for x:

3x^3 + 2x^2 + 75x - 50 = 0

The Attempt at a Solution

I have tried substituting multiple values for "x" so that we get a factor, f(x)=0
I cannot seem to find an "x" value that will make this function=0. Is there a way to factor this function or did the book have an error when printing the equation?

It seems it has indeed no easy zeros. Up to degree four all integer polynomials can be factored explicitly. You probably know the formula for quadratic polynomials. Cubic polynomials are a bit harder https://en.wikipedia.org/wiki/Cubic_function#Algebraic_solution and polynomials of degree four are really unpleasant. Are you sure your polynomial is correct?

 

[Jen23]

It seems it has indeed no easy zeros. Up to degree four all integer polynomials can be factored explicitly. You probably know the formula for quadratic polynomials. Cubic polynomials are a bit harder https://en.wikipedia.org/wiki/Cubic_function#Algebraic_solution and polynomials of degree four are really unpleasant. Are you sure your polynomial is correct?

Yes, I am positive I have the equation right, I made sure and checked the book a couple of times and see if I missed anything, but this is exactly how it is written. If this equation was able to be factored, I would have done so until I get a quadratic equation to use the quadratic formula if needed.

However I could not even factor this to begin with. I think there was a book error, but wanted to first hear the opinions of others on this forum before assuming so, just to make sure I was not missing a step to factor it.

 

[fresh_42]

@SammyS was right: with a minus sign at the highest term it is easy to factor. Without it, you either use some numerical algorithms to approximate the zeroes, but two of them are complex, or you use the Wikipedia link I've posted above. What you have to solve is:
$$(x-a_1)(x-a_2)(x-a_3)=3x^3+2x^2+75x-50=0$$
The link says how. Otherwise you could try it on your own, but I'd rather follow Cardano who did this for us in 1545.
https://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

 

[Jen23]

@SammyS was right: with a minus sign at the highest term it is easy to factor. Without it, you either use some numerical algorithms to approximate the zeroes, but two of them are complex, or you use the Wikipedia link I've posted above. What you have to solve is:
$$(x-a_1)(x-a_2)(x-a_3)=3x^3+2x^2+75x-50=0$$
The link says how. Otherwise you could try it on your own, but I'd rather follow Cardano who did this for us in 1545.
https://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

Okay I will give it a shot and try to solve using the link provided, I haven't done this before but I will try it!

 

[FactChecker]

You can tell that there is at least one real root because the term 3x³ will force it very negative for large negative x and very positive for large positive x. If you plot it, you can see that there is one zero. Numerically, the root is at x=0.6448517225759918. I don't know how to find it analytically. (CORRECTION: This result matches the result from the formula in post #9, r₁ = 0.644851722575991. I got it from Maxima. To get such an accurate result, it must have used that formula without saying so.)

As @SammyS suggested, if the linear term is -75x, then it factors: (x-5)*(x+5)*(3*x+2). So that is probably what they had in mind. I don't know how to factor it (I used Maxima). Maybe @SammyS can help there. The constant term, 50, has prime factors of 5,5,2. So that is probably how to find the factors of the modified polynomial.

 

[Ray Vickson]

You can tell that there is at least one real root because the term 3x³ will force it very negative for large negative x and very positive for large positive x. If you plot it, you can see that there is one zero. Numerically, the root is at x=0.6448517225759918. I don't know how to find it analytically.

As @SammyS suggested, if the linear term is -75x, then it factors: (x-5)*(x+5)*(3*x+2). So that is probably what they had in mind. I don't know how to factor it (I used Maxima). Maybe @SammyS can help there. The constant term, 50, has prime factors of 5,5,2. So that is probably how to find the factors of the modified polynomial.

Typically (in applications, at least) we often don't bother using the exact formulas in cases where they would be needed; numerical solutions are easy to get and are often much more useful than the horrible "exact" formulas. That is certainly true in this example. Maple gets the real root as
$$r_1 = -\frac{2}{9} + \frac{1}{9} \left( 8092 + 45 \, \sqrt{181527} \right)^{1/3} -\frac{671}{9 \left(8092 + 45 \, \sqrt{181527} \right)^{1/3}} $$
Or, we can write $r_1 \doteq 0.6448517208$

 

[Jen23]

Typically (in applications, at least) we often don't bother using the exact formulas in cases where they would be needed; numerical solutions are easy to get and are often much more useful than the horrible "exact" formulas. That is certainly true in this example. Maple gets the real root as
$$r_1 = -\frac{2}{9} + \frac{1}{9} \left( 8092 + 45 \, \sqrt{181527} \right)^{1/3} -\frac{671}{9 \left(8092 + 45 \, \sqrt{181527} \right)^{1/3}} $$
Or, we can write $r_1 \doteq 0.6448517208$

Yes it does become more complicated finding the roots! Thanks for showing how we got x=0.6448517208

 

[Ray Vickson]

Yes it does become more complicated finding the roots! Thanks for showing how we got x=0.6448517208

No, I did not show how to get $x=0.6448517208$; I just said that giving a numerical answer may be more useful and more revealing than giving a formula. There are several different ways of getting such a numerical answer, and I did not tell you which method I used.

 

[FactChecker]

An interesting observation (to me at least):
The calculation of the formula in @Ray Vickson 's post #9 is r₁ = 0.644851722575991. That matches the result, 0.6448517225759918, that I got from Maxima. I thought that Maxima used a numerical method to get that, but I can not believe a numerical method would be that accurate. So Maxima must have used the formula in post #9 without saying so.

 

[SammyS]

An interesting observation (to me at least):
The calculation of the formula in @Ray Vickson 's post #9 is r₁ = 0.644851722575991. That matches the result, 0.6448517225759918, that I got from Maxima. I thought that Maxima used a numerical method to get that, but I can not believe a numerical method would be that accurate. So Maxima must have used the formula in post #9 without saying so.

Of course, many numerical methods can be that accurate.

 

[FactChecker]

Of course, many numerical methods can be that accurate.

15 significant digits? That surprises me. But I'll take your word for it.

 

[Ray Vickson]

An interesting observation (to me at least):
The calculation of the formula in @Ray Vickson 's post #9 is r₁ = 0.644851722575991. That matches the result, 0.6448517225759918, that I got from Maxima. I thought that Maxima used a numerical method to get that, but I can not believe a numerical method would be that accurate. So Maxima must have used the formula in post #9 without saying so.

I set Digits:=30 in Maple and asked for a numerical solution; it gave me x = 0.644851722575991745711718821354.

To 30-digit accuracy the analytical solution evaluates to x = 0.644851722575991745711718821368, differing from the numerical solution in the last two decimal places.

I have also tried this out for 300-digit accuracy, and it works just as well, with the two results differing in the last couple of decimal places.

I don't have access to Maxima, so I don't know if it allows similar choices.

 

[FactChecker]

I set Digits:=30 in Maple and asked for a numerical solution; it gave me x = 0.644851722575991745711718821354.

To 30-digit accuracy the analytical solution evaluates to x = 0.644851722575991745711718821368, differing from the numerical solution in the last two decimal places.

I have also tried this out for 300-digit accuracy, and it works just as well, with the two results differing in the last couple of decimal places.

I don't have access to Maxima, so I don't know if it allows similar choices.

Wow! I have no experience with things that take anything above 8 or 9 significant digits. I guess that as long as the derivatives are well behaved, tremendous accuracy is achievable. I should have known that. And this example is a low degree polynomial.

 

[FactChecker]

I have also tried this out for 300-digit accuracy, and it works just as well, with the two results differing in the last couple of decimal places.

I don't have access to Maxima, so I don't know if it allows similar choices.

Out of curiosity, I tried it in Maxima. It allowed 300 digit calculations in the analytical function evaluation and agreed with your numerical answer in all 30 digits. It was limited to 52 digits in the numerical solution calculation and the answer only agreed with its analytical solution (= your numerical solution) to 16 digits.

300 digit analytical answer
= 0.644851722575991745711718821354 368946777483550532246586919913822137285753667222920895012271785282603834732942934997588065535950042085632010513979772346793854308607842067571967862361238390505691684766784820282308565184516416050093768810856396694744187779273009848280822084118560236572890618404276685671

PS. These programs are fun to use.

 

[SammyS]

As @SammyS suggested, if the linear term is -75x, then it factors: (x-5)*(x+5)*(3*x+2). So that is probably what they had in mind. I don't know how to factor it (I used Maxima). Maybe @SammyS can help there. ...

Actually if you change the sign of anyone term in the polynomial, there will be at least one rational zero.

$3x^3 + 2x^2 + 75x - 50$​

A method known as "Factoring by Grouping" can be used to get the factors.

For instance, starting with $\ 3x^3 - 2x^2 + 75x - 50\,,\$ factor out $\ x^2\$ from the first two terms and 25 from the last two.

$\ 3x^3 - 2x^2 + 75x - 50\,\$

$=x^2(3x-2)+25(3x-2)$
$=(3x-2)(x^2+25)$​

 

[Ray Vickson]

Yes it does become more complicated finding the roots! Thanks for showing how we got x=0.6448517208

Sometimes it is useful to know something about possible roots before you try to find them. Let us look at the two suggested versions of your problem:
(1) $p_1(x) = 3 x^3 + 2x^2 + 75x - 50 = 0$ (the original); and
(2) $p_2(x) = 3 x^3 -2x^2 + 75 x - 50 = 0$ (the suggested alternative).
Here are two helpful general results:

(I) Descarte's Rule of Signs; and (II) The Rational Root Theorem.

(I) The rule of signs states that for a polynomial $p(x) = a_n p^n + a_{n-1} x^{n-1} + \cdots + a_0$ the number of roots in the region $\{ x \geq 0\}$ is either $k$ or $k-2$ or $k-4$ $\cdots$. Here, $k$ is the number of sign changes in the coefficients $a_n, a_{n-1}, \ldots, a_0$.

For the polynomial $p_1$ the coefficients are $+3,+2,+75, -50$ with one sign change, so there is exactly one root $x > 0$. In fact, since $p_1(0)=-50<0$ and $p_1(1) = 30>0$ there must be a root between 0 and 1; and there are no other positive roots. For negative $x$ we can set $x = -t$ where $t > 0$, and get $p_1(x) = p_1(-t) = -3 t^3 + 2 t^2 - 75 t - 50$, with coefficients $-3,+2,-75,-50$ having two sign changes. Thus the number of roots $t > 0$ is either 2 or 0. And it turns out, is actually 0, because the roots are complex, not real.

For the polynomial $p_2$ the coefficients are $+3,-2, +75, -50$, with 3 sign changes; thus, the number of roots $x > 0$ is either 3 or 1. Again, since $p_2(0) = -50<0$ and $p_1(1) = 26 >0$ there is a root between 0 and 1. For $x = -t < 0$ we have $p_2(x) = p_2(-t) = -3 t^3 - 2 t^2 - 75 t - 50$, whose coefficients have no sign changes; thus, there are no negative real roots of $p_2$.

So, Descarte's Rule of signs has given us some useful information to help us along the path to a solution.

(II) Next, consider the Rational Root Theorem. This says that if the coefficients $a_n, a_{n-1}, \ldots, a_0$ are all integers (+ or -), then any rational root of the form $x = p/q$ (with integer $p,q$) must have the property that $p$ is a divisor of the constant coefficient $a_0$ and $q$ is a divisor of the leading coefficient $a_n$. For both $p_1$ and $p_2$ we know there is a root between 0 and 1. If $p_1$ has a rational root $p/q$, $p$ must be a divisor of 50 and $q$ must be a divisor of $3$. That is, either $q=1$ or $q=3$. The value $q=1$ will not work because that would make $x=p/q$ an integer, and we already know that $x=0$ and $x=1$ are not roots. So, any possible integer root would be > 1, and we already know there are none. Therefore, if $p_1$ has a rational root we must have $q = 3$, so $x = p/3$, where $p$ is a divisor of 50. The only two divisors of 50 that give a result < 1 are 1 and 2, giving the possibilities $1/3$ and $2/3$. If you plug these into $p_1$ you will see that they both fail; thus, $p_1$ has no rational roots at all in the region $\{ x > 0 \}$. That means there is no hope for a "nice" factorization; you will need square roots or cube roots, or other unpleasant objects in your factorization.

The same considerations apply to $p_2$: if there is a rational positive root, it must be either $1/3$ or $2/3$. (The rational root theorem guarantees that there are no other possibilities, except for irrational roots.) By direct substitution we find that $x = 2/3$ is a root, so that means that one of the factors is $3x-2$, as Sammy has shown. Note, however, that Sammy's suggested method of "grouping" does not always work, but the rational root theorem always works when there are, in fact, rational roots; and when the theorem fails, you know, 100%, that there are no rational roots.

For more information on these results, see, eg.,
http://www.purplemath.com/modules/drofsign.htm or
https://en.wikipedia.org/wiki/Descartes'_rule_of_signs
for the Rule of Signs, and
http://www.sparknotes.com/math/algebra2/polynomials/section4.rhtml or
https://en.wikipedia.org/wiki/Rational_root_theorem
for the rational root theorem.
In each case the first cited links give gentle introductions, while the second links are more demanding, even including proofs of the results.