The other day, while helping my friend on homework, I was stuck on a problem. And by stuck I mean I didn't want to guess rational solutions and then synthetic to find if they worked because that's too main stream and I had already done that to change it from a 4th degree to a third.

So I just happened to come across a little trick which I'll detail in this note on how to factor Polynomials such as \(2x^4-19x^3+57x^2-64x+20\).

Let me begin by saying: This trick isn't the best for two reasons: It doesn't work all that often, and it may be a waste of time. +it's already kinda similar to RR and remainder theorem.

But on the other hand, when it works, it can be very helpful.

We begin by synthetic division on the polynomial \(2x^4-19x^3+57x^2-64x+20\) Divide the entire polynomial by \(x-2\) and we get \(2x^3-15x^2+27x-10=0\)

Move the 10 over to the other side

\(x(2x^2-15x+27)=10\)

\(x(2x+3)(x-9)=10\)

From here, we prime factorize 10 to \(1\cdot 2\cdot5\).

Assuming x is rational, it's quite clear that x=1 or 2 or 5 since we have an x in the front of our equation. Thus we can plug in x= 1,2,5 to see which work. After doing so, we find that 2 and 5 both work.

Now, you might be asking "well how is this quicker". And the thing is that \(\textbf{you don't have to multiply and add everything out}\) all you have to do is check that each binomial is a factor of 10.

This can become especially helpful when the constant term \(a_0\) becomes quite large and you are stuck pluging in large numbers which when raised to the 5th power such as \(6^5\) can become quite strenuous to do by hand.

Here's another example

Move the factors to the right.

\(x^2(x^2+5)=13^2\cdot 2 \cdot 87\)

A little intuition\manipulation yields that \(2\cdot87=13^2+5\)

Looking at our expression now gives is that \(x^2(x^2+5)=13^2(13^2+5)\)

Now it's quite obvious that \(\pm13\) are roots just by looking at the equation.

Thus after dividing our equation by \(x^2-169\) we get

\((x^2-169)(x^2+2\cdot87)=0\)

\((x-13)(x+13)(x-i\sqrt{174})(x+i\sqrt{174})\)

This our answer is \(\boxed{x={\pm13,\pm i\sqrt{174}}}\)

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## Comments

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\(\large{1000x^3-1254x^2-496x+191}\)

SolutionLet's make the substitution \(x=\dfrac{y+1254}{3000}\) to delete the quadratic term: \(1000\left(\dfrac{y+1254}{3000}\right)^3-1254\left(\dfrac{y+1254}{3000}\right)^2-496\left(\dfrac{y+1254}{3000}\right)+191=0\) Expand, clear denominators and simplify: \(y^3-9181548y-4384726128=0\) Now, try to match that equation with the identity \((u+v)^3-3uv(u+v)-(u^3+v^3)=0\), an equation system will be formed: \(\quad \begin{eqnarray} y=u+v &\quad ...(1) \\ 3uv=9181548 &\quad ...(2) \\ u^3+v^3=4384726128 &\quad ...(3) \end{eqnarray}\) Divide \((2)\) by \(3\) and cube both sides: \(u^3v^3=28667113297167468096\) Now, make an equation in \(z\) with roots \(u^3\) and \(v^3\): \((z-u^3)(z-v^3)=0 \Longrightarrow z^2-(u^3+v^3)z+u^3v^3=0\) \(z^2-4384726128z+28667113297167468096=0\) Using the quadratic formula we get two values of \(z\): \(z=2192363064 \pm 6000i\sqrt{662796041466}\) Hence: \(u=\sqrt[3]{2192363064+6000i\sqrt{662796041466}}\) \(v=\sqrt[3]{2192363064-6000i\sqrt{662796041466}}\) Using complex numbers let's take the principal cube roots for \(u\) and \(v\): \(u=\sqrt[3]{|u|} cis \left(\dfrac{\arg u}{3}\right) \approx 1622.698958+653.730901i\) \(v=\sqrt[3]{|v|} cis \left(\dfrac{\arg v}{3}\right) \approx 1622.698958-653.730901i\) From the equation system, there are three possible values for \((u,v)\). They are: \((u,v)\), \((uw,vw^2)\) and \((uw^2,vw)\), where \(w\) is any primitive 3rd root of unity. Can you check the other values that don't work? So, we have three solutions for \(y\): \(y_1=u+v \approx 3245.397916\) \(y_2=uw+vw^2 \approx -2754.994093\) \(y_3=uw^2+vw \approx -490.4038231\) Hence: \(x_1=\dfrac{y_1+1254}{3000} \approx 1.4997\) \(x_2=\dfrac{y_2+1254}{3000} \approx -0.5003\) \(x_3=\dfrac{y_3+1254}{3000} \approx 0.2545\)

\(\Large{Credit~goes~to}\) @Alan Enrique Ontiveros Salazar

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I knew about this! Its a method to solve cubic! Right? I read it from "Hall and Knight". You might like to refer it.

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Thanks Trevor!

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