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Differential Calculus 201-NYA-05
Vincent Carrier
Factoring
The process of factoring consists in transforming a sum in a product.
The most general method of factoring is to factor common factors.
Example:
16x7 + 8x5 − 24x3 = 8x3 (2x4 + x2 − 3)
Let us now see techniques for expressions of different degrees.
1) Second-Degree Expressions
We have
(x + a)(x + b) = x2 + (a + b)x + ab.
Thus, to factor x2 − 3x − 10 (say), two numbers a and b must be found such that
a + b = −3
ab = −10.
and
It is easy to see that a = −5 and b = 2 so
x2 − 3x − 10 = (x − 5)(x + 2).
If the expression to factor is a difference of squares, then the identity
a2 − b2 = (a − b)(a + b)
can be used.
Examples:
a) x2 + 5x + 4 = (x + 1)(x + 4)
b) x2 + x − 6 = (x − 2)(x + 3)
c) x2 − 2x − 8 = (x + 2)(x − 4)
d) x2 − 6x + 9 = (x − 3)2
e) x2 − 4 = (x − 2)(x + 2)
f)
√
√
5 − x2 = ( 5 − x)( 5 + x)
2) Fourth-Degree Expressions
In general, fourth-degree expressions are very difficult to factor. However, it is possible to factor them if they follow one of the 2 patterns in the examples below.
Examples:
a)
x4 − 81 = (x2 − 9)(x2 + 9) = (x − 3)(x + 3)(x2 + 9)
b)
x4 − 5x2 + 4 = (x2 − 1)(x2 − 4) = (x − 1)(x + 1)(x − 2)(x + 2)
c)
x4 − 8x2 − 9 = (x2 + 1)(x2 − 9) = (x2 + 1)(x − 3)(x + 3)
3) Third-Degree Expressions
Third-degree expressions are also very difficult to factor.
The first technique to factor them is grouping.
Example:
x3 − 3x2 − 4x + 12 = x2 (x − 3) − 4(x − 3)
= (x − 3)(x2 − 4)
= (x − 3)(x − 2)(x + 2)
The second technique consists in using the following theorem.
Factor Theorem: Let p(x) be a polynomial of arbitrary degree. If p(c) = 0, then
x − c is a factor of p(x).
Assume that p(x) is of degree 3. If a number c is such that p(c) = 0, then
p(x) = (x − c)(second-degree expression)
with the missing second-degree expression found by using long division.
Examples:
a)
b)
c)
x3 − 8 = (x − 2)(x2 + 2x + 4)
x3 + 4x − 3 = (x + 1)(x2 − x − 3)
x3 − 2x2 − 5x + 6 = (x − 1)(x2 − x − 6) = (x − 1)(x − 3)(x + 2)