15. Show that the square of x + 1 exactly divides (x^3 + x^2 + 4)^3 -(x^3 - 2x + 3)^3.
16. Show that (3 x^2-7x+2)^3-(x^2-8 x+8)^2 is divisible by 2 x-3, and by x + 2.
105. The Factor Theorem. If any rational and integral expression containing x becomes equal to 0 when a is written for x, it is exactly divisible by x - a.
Let P stand for the expression. Divide P by x-a until the remainder no longer contains x. Let R denote this remainder, and Q the quotient obtained. Then
P=Q(x-a)+R.
Since this equation is true for all values of x, we will assume that x equals a. By hypothesis, the substitution of a for x makes P equal to 0; thus,
0=Q(0)+ R. R = ( 0)
As the remainder is 0, the expression is exactly divisible by x-a.
The following examples illustrate the application of this principle:
Ex. 1. Resolve into factors x^3 + 3x^2 - 13x - 15.
By trial we find that this expression becomes 0 when 2 = 3; hence, x-3 is a factor. Dividing by x - 3, we obtain the quotient
x^2+ 6x+5, The factors of this expression are easily seen to be x+ 1 and x + 5;
hence, x^3+ 3x^2-13x -15 =(x - 3)(x+ 1)(x+ 5).
Ex. 2. Resolve into factors x^3 + 6x^2 + 11x + 6.
It is evident that substituting a positive number for x will not make the expression equal to 0. By substituting - 1, however, for x, the expression becomes - 1 +6 - 11 +6, or 0; hence,
x^3 + 6x^2 + 11x + 6
is divisible by x+1. Dividing by x+1, we obtain the quotient x^2+ 5x +6, and factoring this expression we have
x^3 + 6x^2 + 11a +6 =(x + 1)(x + 2)(x +8).
