# Another factoring trick

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

## Find all roots of $x^4+5x^2-29406$ given that $29406=13^2\cdot 2\cdot 87$

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 $\boxed{x={\pm13,\pm i\sqrt{174}}}$ Note by Trevor Arashiro
5 years, 10 months ago

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Solve It

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

Solution

Let'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{aligned} y=u+v &\quad ...(1) \\ 3uv=9181548 &\quad ...(2) \\ u^3+v^3=4384726128 &\quad ...(3) \end{aligned} 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: $u=\sqrt{2192363064+6000i\sqrt{662796041466}}$ $v=\sqrt{2192363064-6000i\sqrt{662796041466}}$ Using complex numbers let's take the principal cube roots for $u$ and $v$: $u=\sqrt{|u|} cis \left(\dfrac{\arg u}{3}\right) \approx 1622.698958+653.730901i$ $v=\sqrt{|v|} cis \left(\dfrac{\arg v}{3}\right) \approx 1622.698958-653.730901i$ 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

- 5 years, 10 months ago

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

- 5 years, 10 months ago

Thanks Trevor!

- 5 years, 10 months ago