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Could someone please explain how to solve this: $$x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$$ not the answer only, but a step-by-step solution.

I tried to solve it, with the help of Khan Academy, but still I have no idea how to correctly solve it.

Thank you so much in advance!

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    $\begingroup$ How to solve a quartic equation? Three ways: (i) Numerically; or (ii) Use the Ferrari-Cardano procedure, or one of its kin; or (iii) Pray that by design it is one of the relatively few such equations that collapse, because there are some very simple roots. Prayer works for quartics in homework exercises. $\endgroup$ Jan 30, 2012 at 21:06
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    $\begingroup$ Actually, 4 ways: (iv) Use a computer. Mathematica solves this equation in 0.001764 seconds: {{x -> -2}, {x -> 2}, {x -> 5}, {x -> 5}}. $\endgroup$ Sep 4, 2015 at 16:21

5 Answers 5

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Since the polynomial has integer coefficients, the rational root theorem applies. Thus any rational root must be of the form $x=\pm p/q$, where $p$ divides the constant term 100 and $q$ divides the leading coefficient 1. In this case, the only possibility for $q$ is 1. This tells you that any rational root must be a divisor of $100=2^2*5^2$. It turns out that this polynomial does have rational roots, after which you find one you can perform polynomial division to get a complete factorization.

For instance, we have the potential rational roots $x=\pm2,\pm5,\pm10,\pm20\pm25,\pm50,\pm100$. We could plug in $x=5$ and verify that this is a root. Then, $$ \frac{x^4 - 10x^3 + 21x^2 + 40x - 100}{x-5} = x^3-5x^2-4x+20. $$ Since all the roots are rational, repeating this process will generate all of them. Not every polynomial with integer coefficients has rational roots (for instance $x^2-2=0$), so this won't always be the case.

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  • $\begingroup$ Thank You, it was very helpful! $\endgroup$
    – Uzdawi
    Jan 30, 2012 at 19:07
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$$x^4-10x^3 +25x^2-(4x^2-40x+100)=0$$

$$\Rightarrow x^2(x^2-10x+25)-4(x^2-10x+25)=0 $$

$$\Rightarrow (x^2-4)(x^2-10x+25)=0 $$

$$\Rightarrow (x-2)(x+2)(x-5)^2=0 $$

$$\Rightarrow x_1=-2 , x_2=2 , x_{3,4}=5$$

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    $\begingroup$ How did you even see that :O $\endgroup$
    – Airdish
    Mar 12, 2016 at 15:28
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First note that $x= \pm 2$ are the roots of the equation. Hence using the remainder theorem, we can proceed as follows:

$$ x^3(x-2) - 8x^2(x-2) +5x(x-2) + 50(x-2)=0$$ $$\qquad \Rightarrow (x-2)(x^3 -8x^2 +5x +50) =0$$ $$ (x-2)( x^2(x+2) -10x(x+2) +25(x+2)) =0 $$ $$\qquad \Rightarrow (x^2-4)(x^2-10x+25)=0 $$ Hence we can solve it as

$$ (x^2-4)(x-5)^2 =0 $$ or $$x=\{ -2,+2,5 \}$$

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The idea is to represent the polynomial $p(x) = x^4 - 10x^3 + 21x^2 + 40x - 100$ as a product of simple factors. By Vietta formula the free terms equals to the product of roots. Thus, if we confine ourselves to integer roots, they should divide $100$. Start with $x=2$, and verify that $p(2) = 16 - 10 \times 8 + 21 \times 4 + 40 \times 2 - 100 = 16 -80 + 84 - 80 + 100 = 0$.

Now applying long division we compute $$ p(x) = (x-2)\left( x^3 - 8 x^2 + 5 x + 50 \right) $$ Now try $x=5$. $5^2(5-8) + 5 ( 5+ 10) = 5( -15 + 15) = 0$. Apply long division again: $$ p(x) = (x-2)(x-5) \left(x^2 - 3 x - 10\right) = (x-2)(x+2)(x-5)^2 $$

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  • $\begingroup$ Thank You for explaining it so clearly! $\endgroup$
    – Uzdawi
    Jan 30, 2012 at 19:11
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    $\begingroup$ very last line should have $x^2-3x-10$? instead of $x^3$ $\endgroup$
    – snulty
    Mar 16, 2016 at 19:35
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The "prayer" method (I'm being a little facetious here, but I found that deadpan comment by André on the OP low-key hilarious) has already been addressed and the asker seems to have understood it.

For elementary homework questions, you would expect at least one root to be rational (two in the case of quartics, because a general cubic is also pretty tough to solve if the roots aren't nice). So the use of rational root theorem is an excellent starting point.

But since the OP asked for a "step-by-step solution", I am presenting here a general method that can be applied to any quartic. My motivations for doing this are to a) show it can be done and b) because I already did it on paper to prove the point and I wanted to preserve the work for posterity and most importantly c) to show students that a perfectly understandable, step-by-step solution to the general quartic (and the general cubic too) are possible. It's just that these are extremely tedious and really not worth doing in most real-world cases. Either you have an artificially constructed equation with at least one or two "easy" roots, or you have a real-world problem with ugly roots, but you can just chuck it into a numerical solver and noone is going to think less of you for it.

With that lengthy preamble out of the way, on with the solution.

The quartic is $x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$

Firstly, any quartic or cubic should be reduced to monic form (lead coefficient one). It's as simple as dividing throughout by the lead coefficient. This quartic already starts out this way, so nothing to be done for this step.

Secondly, make the equation "depressed". You're going to get pretty depressed when you see how tedious the entire process gets, but for now, "depressed" means removing the degree $3$ term in a quartic, and the degree $2$ term in a cubic.

For the quartic $x^4 + bx^3 + cx^2 + dx + e = 0$, the substitution is simply $x = u - \frac b4$. You can actually see this process very simply by using the binomial expansion and watching the cube term cancel out in the end result. Here we will just apply it (here $b = -10$ so $x = u - (-\frac{10}{4}) = u + \frac 52$) :

$(u+\frac52)^4 - 10(u+\frac52)^3 + 21(u+\frac52)^2 + 40(u+\frac52) - 100 = 0$

$u^4 - \frac{33 u^2}2 + 20 u + \frac{225}{16} = 0$

That is the depressed quartic. No degree $3$ term left.

Next step is to express the quartic as the product of two quadratic factors:

$u^4 - \frac{33 u^2}2 + 20 u + \frac{225}{16} = (u^2 + pu + q)(u^2 - pu + s)$,

the idea being that all solutions of those individual quadratics will be roots of the original quartics (and there will be no other roots to find).

Note that the coefficient of the degree $1$ term ($u$) is the same in each factor, with opposite signs. This is done on purpose so the expansion has no cubic term (to correspond to a depressed quartic).

Expanding,

$u^4 - \frac{33}2u^2 + 20 u + \frac{225}{16} = u^4 + (s + q - p^2)u^2 + p(s-q)u + qs$

Comparing coefficients and rearranging we get the following system of equations:

$s + q = p^2 - \frac{33}2$

$s - q = \frac{20}p$

$qs = \frac{225}{16}$

Noting that (identically) $(s+q)^2 - (s-q)^2 = 4qs$, we get:

$(p^2 - \frac{33}2)^2 - (\frac{20}p)^2 = 4(\frac{225}{16})$

which becomes $p^6 - 33 p^4 + 216p^2 - 400 = 0$

Substituting $P = p^2$, we get the cubic equation:

$P^3 - 33P^2 + 216P - 400 = 0$, which we need to solve.

Ordinarily, you would already have "known" to solve a cubic before you learn to solve a quartic. But because I'm doing it this way, I'll go through the detailed solution for the cubic as well.

The first step, just like in the quartic, is to reduce to a monic equation. This is (again) already the case here.

The second step, just like before, is to make a simple linear substitution to get a depressed cubic (without the degree $2$ term, in this case). For a cubic $x^3 + bx^2 + cx + d = 0$, the substitution is $x = (u - \frac b3)$. Note how the denominator of the second term follows the degree of the equation we wish to depress. This is an easy way to remember the substitution.

In this case (to avoid reusing variables), we will let $P = w - (-\frac{33}{3}) = w + 11$.

We then get:

$(w+11)^3 - 33(w+11)^2 + 216(w+11) - 400 = 0$

$w^3 - 147 w - 686 = 0$

and we have our depressed cubic. (Feeling depressed yet?)

At this point, the cubic solution diverges slightly from what is described above for the quartic. Instead of attempting to express it as a factor of a linear and quadratic, we're going to make another substitution. This one is more tricky, but it's easy to remember if you understand the motivation of it.

The substitution for a general reduced cubic $x^3 + bx + c = 0$ is to put $x = z - \frac{b}{3z}$. Note how tantalisingly close this appears to our earlier linear substitution. The difference is that there is a $z$ term appearing in the denominator here (which makes this a non-linear substitution). The idea here is actually to increase the degree of the depressed cubic to six, but then to realise that it is actually just a quadratic in disguise! Let me show you.

We want to solve $w^3 - 147 w - 686 = 0$.

Let $w = z - (-\frac{147}{3z}) = z + \frac{49}z$

We then get:

$(z + \frac{49}z)^3 - 147(z + \frac{49}z) - 686 = 0$

$z^3 + \frac{117649}{z^3} - 686 = 0$

$z^6 - 686z^3 + 117649 = 0$

That's a sextic (just a power $6$ equation even thought the name sounds rather salacious). But it is actually just a quadratic in disguise. To see this make the next substitution $m = z^3$:

$m^2 - 686m + 117649 = 0$

And now we have our good old fashioned quadratic which any kid should know how to solve. Ordinarily, you are taught to either factorise (in the case of nice roots and coefficients), complete the square or use the quadratic formula. But since I've made such a huge deal about following a set process, let's see if that can be done for the quadratic too, shall we?

First, we reduce it to monic form. That's already the case here.

Second, we make a linear substitution to get a "depressed quadratic". Note that I am putting that in quote marks because I have never actually seen or heard that term being used elsewhere (as I said, other methods to solve the quadratic are taught). So I'll define it here: a "depressed quadratic" is one without a degree $1$ term. The substition for the quadratic $x^2 + bx + c = 0$ is simply $x = u - \frac b2$. Note how it neatly follows the pattern for the quartic and the cubic.

So let's apply that here. We want to solve:

$m^2 - 686m + 117649 = 0$

Put $m = n - (-\frac{686}{2}) = n + 343$, yielding:

$(n+343)^2 - 686(n+343) + 117649 = 0$

$n^2 = 0$

$n = \pm \sqrt 0 = 0$

Here, it is clear we have a repeated root.

Now, reverse all the substitutions. So we have $m = n+343 = 343$.

We then have $z^3 = m$.

At this point, we might be tempted to just take the cube root and write $z = 7$. But that would be mathematically wrong, as we need to consider complex roots as well (that's another difficulty with the general cubic and quartic). There are two complex conjugate numbers called the complex cube roots of unity, written as $\omega = -\frac 12 + i\frac{\sqrt 3}{2}$ and $\omega^2 = -\frac 12 - i\frac{\sqrt 3}{2}$. You get this by solving $x^3 = 1 \implies (x-1)(x^2 + x + 1) = 0$ completely. But I am not going to harp on this point, you can look up the details of the complex roots of unity elsewhere.

So we have $z^3 = 343$ which gives the three possible complex values of $z$, namely $z = 7, 7\omega, 7\omega^2$.

And then we have $w = z + \frac{49}z$ which gives the two distinct possible values for $w$ as $14$ or $-7$. Again, $\omega$ has some nice properties that make this calculation easy by hand, but I won't lengthen the derivation here.

So $w = 14, -7$.

and we get $P = w + 11 = 25, 4$.

and $P = p^2$ so $p = \pm \sqrt P = \pm 5, \pm 2$.

At this stage, I will note that you only need to consider one of the two possible values of the square root of each value. The reason is the symmetry of the quadratic factors of the quartic. Changing to the other sign would simply reverse the factors and make no impact. I have added a note below to illustrate this, but if you are not convinced, you can work through the algebra using the other sign of the square roots as well, you'll see what I mean.

Going back to the system of simultaneous equations above:

$s + q = p^2 - \frac{33}2$

$s - q = \frac{20}p$

$qs = \frac{225}{16}$

We can add the first two equations to get:

$s + q + s - q = p^2 - \frac{33}2 + \frac{20}p$

$s = \frac 12 (p^2 - \frac{33}2 + \frac {20}p)$

Similarly, we can subtract the first two equations to get:

$s + q - (s - q) = p^2 - \frac{33}2 - \frac{20}p$

$q = \frac 12 (p^2 - \frac{33}2 - \frac {20}p)$

Substituting one relevant positive root for $p$ ($=5$) we get:

$s = \frac{25}4$ and $q = \frac{9}4$.

Now, substituting the other relevant positive root for $p$ ($=2$) we get:

$s = -\frac 54$ and $q = -\frac{45}4$

We have now factorised the quartic in $u$ to:

$(u^2 + 5u + \frac 94)(u^2 -5u + \frac{25}4) = 0$

or

$(u^2 + 2u - \frac{45}4)(u^2 - 2u - \frac {5}4) = 0$

which can be solved easily. We will get to this in a bit.

Note that if we had taken the respective negative roots for $p$, we would have arrived at the exact same factorisations, just in the reverse order. For instance, with $p = -5$, we have $s = \frac 94, q = \frac{25}{4}$, which would've given us the factorisation $(u^2 - 5u + \frac{25}{4})(u^2 + 5u + \frac{9}{4}) = 0$, the same as the first form, just reversed. I hope you can see why it is pointless to consider both square roots for $p$, so just take one set in all cases.

Anyway, solving the quadratics (there are four of them!), we get:

$u^2 + 5u + \frac 94 = 0 \implies u = -\frac 92$ or $u = -\frac 12$

or

$u^2 -5u + \frac{25}4 = 0 \implies u = \frac 52$ (repeated root)

or

$u^2 + 2u - \frac{45}4 = 0 \implies u = -\frac 92$ or $u = \frac 52$

or

$u^2 - 2u - \frac {5}4 = 0 \implies u = -\frac 12$ or $u = \frac 52$

from which we get the unique possible values of $u$ as $-\frac 92, -\frac 12, \frac 52$

And finally (phew!) we reverse the initial substitution $x = u + \frac 52$ to get:

$x = -2, 2, 5$ (with the latter being a repeated root).

The reason I went through this exercise in masochism (other than what I've already stated at the outset) is that you won't have to (unless you want to torture yourself). You can see that the method is extremely irritating, even for a quartic with such nice integral roots. This is why, at the elementary level, you are not usually taught this (even though all the individual steps are perfectly easy to understand for a high-schooler). And homework problems will generally have rational (or often even integral) roots, making the use of rational root theorem coupled with polynomial long division (or even more simply, synthetic division) the obvious choice for these highly artificial problems. But the method I described works for any quartic and cubic, with complex and ugly roots. You can always solve for these roots in horrible looking radicals, often nested ones. But at least they are exact solutions.

There are other known methods to exactly solve the general cubic and the quartic, but for what it's worth, the methods I described here are what I find simplest and most memorable (I don't have to look anything up to replicate the general method). The other methods I've seen (which are probably the original described methods) involve making even more non-intuitive substitutions or adding polynomial expressions that come seemingly out of nowhere (unless you really, really think about the method). That's why I stick with what I've described here.

Finally, you can stop with the quartic. The general quintic (and above) has no general solution in terms of radicals (this is provable by group (Galois) theory). This is why learning the general solution up to the quartic is kind of neat.

Anyway, I hope this long-winded answer is not deleted (or otherwise treated too harshly), as I've tried my best to explain my motivations here. It can be linked if someone else asks for a practical demonstration of a method to solve a general cubic or quartic equation in future questions.

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    $\begingroup$ +1 awesome, thanks for posting this! $\endgroup$
    – user541686
    Mar 27, 2023 at 7:17
  • $\begingroup$ @user541686 Most welcome! $\endgroup$
    – Deepak
    Mar 27, 2023 at 7:58

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