The quadratic polynomial x^4 + 5x^3 + 8x^2 + 8x -8 can be factored into two prime quadratics of the form (x^2 + ax + b)(x^2 + cx + d). Find (a,b,c,d) if a, b, c, and d are intergers.

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- Nov 12th 2006, 08:23 AMMr_GreenFactoring
The quadratic polynomial x^4 + 5x^3 + 8x^2 + 8x -8 can be factored into two prime quadratics of the form (x^2 + ax + b)(x^2 + cx + d). Find (a,b,c,d) if a, b, c, and d are intergers.

- Nov 12th 2006, 08:33 AMearboth
Hello, Mr. Green,

you got $\displaystyle x^4 + 5x^3 + 8x^2 + 8x -8 $ and

$\displaystyle (x^2 + ax + b)(x^2 + cx + d)=x^4+(a+c)x^3+(b+ac+d)x^2+(bc+ad)x+bd$

By comparison you get the system of equations:

a+c = 5

b+ ac + d = 8

bc+ad = 8

bd = -8

You have 4 variables and 4 equations, so it should be solvable.

EB - Nov 12th 2006, 08:51 AMMr_Greenhow?
how do u solve that system of equations?

- Nov 12th 2006, 09:03 AMearboth
Hello, Mr. Green,

honestly: by trial and error, because to do this system of non-linear equations looks like a life sentence.

1. You got: all a,b,c,d should be integers. Now take the last equation and solve for b = -8/d. That means d can only be in {-8, -4, -2, -1, 1, 2, 4, 8}

2. Choose a value for d; then you have b, then you can calculate c with the 3rd equation and last a.

You'll get pretty soon the only possible solution - but I didn't "solve" this system of equations.

EB

for confirmation only: [a, b, c, d] = [2, 4, 3, -2] - Nov 12th 2006, 10:21 AMtopsquark
This is a pretty mean problem in general. I do have a way to cut down on some of the work, though.

I presume if you are doing this you are familiar with modular mathematics? (This seems to be my favorite trick recently.)

First note that b = -8, -4, -2, -1, 1, 2, 4, or 8.

Now write this polynomial (mod 3). (I tried it first with mod 2, but didn't get quadratic factors.)

$\displaystyle x^4 + 5x^3 + 8x^2 + 8x -8 \equiv x^4 + 2x^3 + 2x^2 + 2x + 1$ (mod 3)

= $\displaystyle (x^4 + 2x^3 + x^2) + (x^2 + 2x + 1)$ (mod 3)

= $\displaystyle x^2(x^2 + 2x + 1) + (x^2 + 2x + 1)$ (mod 3)

= $\displaystyle (x^2 + 0 \cdot x + 1)(x^2 + 2x + 1)$ (mod 3)

Comparing the first factor with $\displaystyle x^2 + ax + b$ we get b = 1 (mod 3).

Thus b = -8, -2, 1, or 4. Because d = 1, 4, -8, or -2 respectively we can arbitrarily decide that b = 1 or 4.

The rest can be done by applying these two possibilities to the problem to see what happens and significantly cuts down the work in finding a and c.

It's not a great solution, but could cut down the time if you have another problem like this.

-Dan