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### Two Methods of Factoring Quadratics

```
Date: 04/14/98 at 21:44:35
From: Derek

I was gone for a week and missed the lesson on how to factor quadratic
trinomials. Could you please show me an example?
```

```
Date: 04/15/98 at 11:35:11
From: Doctor Rob

A quadratic trinomial has the form a*x^2 + b*x + c, for some
expressions a, b, c, and x. In the simplest case, a, b, and c are
constants, and x is an unknown. For example, you might want to factor
3*x^2 - 5*x - 2. Here a = 3, b = -5, and c = -2. You want to factor it
into two linear factors:

a*x^2 + b*x + c = (r*x+s)*(t*x+u)

where r, s, t, and u are expressions of the same kind as a, b, and c.
In the example, it is true that 3*x^2 - 5*x - 2 = (3*x + 1)*(x - 2).
For the general equation to be true, if you expand the righthand
side, you get:

a*x^2 + b*x + c = (r*t)*x^2 + (r*u+s*t)*x + (s*u)

This means that you want the following three equations to be true:

a = r*t
b = r*u + s*t
c = s*u

You know a, b, and c, and you want to find r, s, t, and u, if
possible. In the example, r = 3, s = 1, t = 1, and u = -2, and
a = 3 = r*t, b = -5 = 3*(-2) + 1*1, and c = -2 = 1*(-2). One way to
proceed is as follows.

Multiply together a and c. Now try to factor that product into two
factors whose sum is b. In the example, a*c = 3*(-2) = -6.  You want
to write -6 as a product of two factor whose sum is b = -5.

Why would this help?  Because a*c = (r*t)*(s*u) = (r*u)*(s*t), and
b = r*u + s*t. The two factors of a*c whose sum is b are r*u and s*t.
Once you have done this, you can write b as the sum of those two
factors, and regroup the terms, factor each group, then factor the
whole:

a*x^2 + b*x + c = r*t*x^2 + (r*u + s*t)*x + s*u
= (r*t*x^2 + r*u*x) + (s*t*x + s*u)
= r*x*(t*x + u) + s*(t*x + u)
= (r*x + s)*(t*x + u)

In the example, the two factors of -6 which add up to -5 are -6 and 1.
Then:

3*x^2 - 5*x - 2 = 3*x^2 - 6*x + 1*x - 2
= 3*x*(x - 2) + 1*(x - 2)
= (3*x + 1)*(x - 2)

Here is another example: factor 2*x^2 - 7*x + 6.

Multiply together 2*6 = 12. Now try to factor 12 into two numbers
whose sum is -7. The two numbers are -4 and -3. Then:

2*x^2 - 7*x + 6 = 2*x^2 - 4*x - 3*x + 6
= 2*x*(x - 2) - 3*(x - 2)
= (2*x - 3)*(x - 2)

If you cannot find a way to factor a*c into two factors that add up
to b, then the trinomial cannot be factored.

Example:  x^2 + x + 1.  Note that a*c = 1, and you need to find two
factors of 1 that add up to b = 1. The only pairs of numbers whose
product is 1 are 1 and 1, or -1 and -1, and the sums are 2 and -2.
This is not possible, so this expression cannot be factored.

Sometimes the expression I have labeled x is not just a variable. For
example, it may be a power of a variable. To factor t^6 - t^3 - 30,
we can take x = t^3. Then we have x^2 - x - 30. We need to find a
factorization of -30 into two numbers whose sum is -1. They are
-6 and 5, and the result is (x - 6)*(x + 5) = (t^3 - 6)*(t^3 + 5).

Another method is to "complete the square." The idea of this method is
to express:

a*x^2 + b*x + c = a*[(x + d)^2 - e]

If e is a perfect square, then we can factor the part inside brackets
as the difference of two squares. By expanding the righthand side,

a*x^2 + b*x + c = a*x^2 + 2*a*d* x + a*d^2 - a*e

and to make the "x" terms the same, we need to pick d = b/(2*a). The
procedure goes like this. Figure out d = b/(2*a). Factor a out of all
terms. Inside the remaining factor, add d^2 to make a perfect square
(x+d)^2, and subtract d^2 to make the equation still true.  Figure out
e, and decide if it is a perfect square. If not, the trinomial is not
factorable. If so, factor the difference of squares. Simplify.

Example:  Factor 3*x^2 - 5*x - 2.  d = -5/6

3*x^2 - 5*x - 2 = 3*[x^2 - 5/3*x - 2/3]
= 3*[x^2 - 5/3*x + 25/36 - 25/36 - 2/3]
= 3*[(x - 5/6)^2 - 49/36]

Since e = 49/36 = (7/6)^2 is a perfect square, this is going to work:

= 3*[(x - 5/6)^2 - (7/6)^2]
= 3*[(x - 5/6) - 7/6]*[(x - 5/6) + 7/6]
= 3*(x - 2)*(x + 1/3)
= (x - 2)*(3*x + 1)

Example:  Factor x^2 + x + 1.  d = 1/2.

x^2 + x + 1 = x^2 + x + 1/4 - 1/4 + 1
= (x + 1/2)^2 + 3/4

Since e = -3/4 is not a perfect square, this trinomial is not
factorable.

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Polynomials
Middle School Algebra
Middle School Factoring Expressions

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