# 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
3 years, 7 months ago

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Thanks Trevor!

- 3 years, 6 months ago

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: $$\quad \begin{eqnarray} y=u+v &\quad ...(1) \\ 3uv=9181548 &\quad ...(2) \\ u^3+v^3=4384726128 &\quad ...(3) \end{eqnarray}$$ 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

- 3 years, 7 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.

- 3 years, 7 months ago