 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 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.

### 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.

• x1=(-7+ 81)/2*2=(-7+9)/4=1/2.
• x2=(-7- 81)/2*2=(-7-9)/4=-4.
1. x1=x2=(-(-2)+ 0)/2=2/2=1
2. No roots.
• 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 |Reply| |Up| |Exchange index| |Contents| |Store|