Subject: Re: Factoring
Date: Wed, 06 Nov 1996 18:26:15 -0500
From: Alex Bogomolny

Dear Bjorn:

Let me try with one caveat. Should anything be unclear you promise to point this out to me. Also, it's going without saying that you'll try a few examples so that if there are still difficulties you'll be able to be more specific, OK?

Before I start let me remark on your terminology. You mention binomials and trinomials. Binomials have two terms and are linear. You do not generally factor a binomial. Trinomials have three terms; they are generally polynomials of degree 2 and may have two factors. I personally prefer the notion of degree and seldom refer to polynomials as bi- or trinomials.

One major point to understand. A polynomial P(x) has a factor (x-a) if and only if a satisfies P(a)=0, i.e. a is a root of P(x)=0. Sometimes you simplify this to "a root of P". In other words, two problems

• Find all roots of a given polynomial P
• Factor a given polynomial P

are absolutely equivalent. I would guess you have not yet studied complex numbers. So you should know that sometimes a polynomial does not have real roots. In this case it's impossible to factor it into a product of monomials like (x-a) with real a. But skip this for now.

Quadratic polynomials

Quadratic polynomials are the simplest. For, as I have already mentioned, there is no need to factor the linear (first degree) ones, right? So let P(x)=ax²+bx+c. P(x) has at most two (real) roots. Both can be found with a simple formula

x₁=(-b+(b²-4ac))/2a and x₂=(-b-(b²-4ac))/2a

You proceed in four steps

• First identify a, b, and c.
• Then compute b²-4ac.
  • If it's negative you can't factor P into real linear factors.
  • If it's positive you'll be able to find two linear factors.
  • If it's zero, the polynomial has a double root and the two linear factors coincide.
• Use the formula to find the roots.
• Write the answer P(x)=a(x-x₁)(x-x₂).

Step one, examples.

1. P(x)=2x²+7x-4. a=2, b=7, c=-4
2. P(x)=1-2x+x². a=1, b=-2, c=1
3. P(x)=x²+2x+2. a=1, b=2, c=2
4. P(x)=3x-4x². a=-4, b=3, c=0

Step two, compute b²-4ac.

1. 7²-4*2*(-4)=49+32=81. It's positive. Two roots here.
2. (-2)²-4*1*1=4-4=0. It's zero. Two equal roots
3. 2²-4*1*2=4-8=-4. It's negative. No real roots.
4. 3²-4*(-4)*0=9. Two roots here.

Step three, use the formula to find the roots.

1. • x₁=(-7+81)/2*2=(-7+9)/4=1/2.
   • x₂=(-7-81)/2*2=(-7-9)/4=-4.
2. x₁=x₂=(-(-2)+0)/2=2/2=1
3. No roots.
4. • x₁=(-3+9)/2*(-4)=(-3+3)/(-8)=0.
   • x₂=(-3-9)/2*(-4)=(-3-3)/(-8)=3/4.

Step four, write the answer P(x)=a(x-x₁)(x-x₂).

1. P(x)=2x²+7x-4=2(x-1/2)(x-(-4))=2(x-1/2)(x+4)=(2x-1)(x+4)
2. P(x)=1-2x+x²=(x-1)(x-1)=(x-1)²
3. P(x)=x²+2x+2. No factoring.
4. P(x)=3x-4x²=(-4)(x-0)(x-3/4)=x(-4x+3)=x(3-4x)

In the last case P(x)=3x-4x² and it's pretty obvious at the outset that x could be factored out. There is nothing much I can add concerning second degree polynomials. Go ahead and try a few. See how this formula works. With polynomials of degree three the situation is more complex although a formula does exists that leads to three roots. Let first get the second degree polynomials over with.

Remark

Chances are you were taught to somehow guess the factors. Do not. At least not from the beginning.

Regards,
Alex

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