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 2x419x3+57x264x+202x^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 2x419x3+57x264x+202x^4-19x^3+57x^2-64x+20 Divide the entire polynomial by x2x-2 and we get 2x315x2+27x10=02x^3-15x^2+27x-10=0

Move the 10 over to the other side

x(2x215x+27)=10x(2x^2-15x+27)=10

x(2x+3)(x9)=10x(2x+3)(x-9)=10

From here, we prime factorize 10 to 1251\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 you don’t have to multiply and add everything out\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 a0a_0 becomes quite large and you are stuck pluging in large numbers which when raised to the 5th power such as 656^5 can become quite strenuous to do by hand.


Here's another example


Find all roots of x4+5x229406x^4+5x^2-29406 given that 29406=13228729406=13^2\cdot 2\cdot 87

Move the factors to the right.

x2(x2+5)=132287x^2(x^2+5)=13^2\cdot 2 \cdot 87

A little intuition\manipulation yields that 287=132+52\cdot87=13^2+5

Looking at our expression now gives is that x2(x2+5)=132(132+5)x^2(x^2+5)=13^2(13^2+5)

Now it's quite obvious that ±13\pm13 are roots just by looking at the equation.

Thus after dividing our equation by x2169x^2-169 we get

(x2169)(x2+287)=0(x^2-169)(x^2+2\cdot87)=0

(x13)(x+13)(xi174)(x+i174)(x-13)(x+13)(x-i\sqrt{174})(x+i\sqrt{174})

This our answer is x=±13,±i174\boxed{x={\pm13,\pm i\sqrt{174}}}

Note by Trevor Arashiro
4 years, 10 months ago

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Solve It

1000x31254x2496x+191\large{1000x^3-1254x^2-496x+191}

Solution

Let's make the substitution x=y+12543000x=\dfrac{y+1254}{3000} to delete the quadratic term: 1000(y+12543000)31254(y+12543000)2496(y+12543000)+191=01000\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: y39181548y4384726128=0y^3-9181548y-4384726128=0 Now, try to match that equation with the identity (u+v)33uv(u+v)(u3+v3)=0(u+v)^3-3uv(u+v)-(u^3+v^3)=0, an equation system will be formed: y=u+v...(1)3uv=9181548...(2)u3+v3=4384726128...(3)\quad \begin{aligned} y=u+v &\quad ...(1) \\ 3uv=9181548 &\quad ...(2) \\ u^3+v^3=4384726128 &\quad ...(3) \end{aligned} Divide (2)(2) by 33 and cube both sides: u3v3=28667113297167468096u^3v^3=28667113297167468096 Now, make an equation in zz with roots u3u^3 and v3v^3: (zu3)(zv3)=0z2(u3+v3)z+u3v3=0(z-u^3)(z-v^3)=0 \Longrightarrow z^2-(u^3+v^3)z+u^3v^3=0 z24384726128z+28667113297167468096=0z^2-4384726128z+28667113297167468096=0 Using the quadratic formula we get two values of zz: z=2192363064±6000i662796041466z=2192363064 \pm 6000i\sqrt{662796041466} Hence: u=2192363064+6000i6627960414663u=\sqrt[3]{2192363064+6000i\sqrt{662796041466}} v=21923630646000i6627960414663v=\sqrt[3]{2192363064-6000i\sqrt{662796041466}} Using complex numbers let's take the principal cube roots for uu and vv: u=u3cis(argu3)1622.698958+653.730901iu=\sqrt[3]{|u|} cis \left(\dfrac{\arg u}{3}\right) \approx 1622.698958+653.730901i v=v3cis(argv3)1622.698958653.730901iv=\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)(u,v). They are: (u,v)(u,v), (uw,vw2)(uw,vw^2) and (uw2,vw)(uw^2,vw), where ww is any primitive 3rd root of unity. Can you check the other values that don't work? So, we have three solutions for yy: y1=u+v3245.397916y_1=u+v \approx 3245.397916 y2=uw+vw22754.994093y_2=uw+vw^2 \approx -2754.994093 y3=uw2+vw490.4038231y_3=uw^2+vw \approx -490.4038231 Hence: x1=y1+125430001.4997x_1=\dfrac{y_1+1254}{3000} \approx 1.4997 x2=y2+125430000.5003x_2=\dfrac{y_2+1254}{3000} \approx -0.5003 x3=y3+125430000.2545x_3=\dfrac{y_3+1254}{3000} \approx 0.2545

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

Mehul Chaturvedi - 4 years, 10 months ago

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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.

Pranjal Jain - 4 years, 10 months ago

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Thanks Trevor!

Karan Shekhawat - 4 years, 9 months ago

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