Note. The student should notice that the only numerical values
that need be substituted for x are the factors of the last term of the
expression, and that we change the sign of the factor substituted
before connecting it with x. Thus, in Ex. 1, the factor 3 gives a
divisor x— 3, and in Ex. 2, the factor — 1 gives a divisor x + 1.
Ex. 3. Without actual division, show that 5x5—6x3+1 is divisible by x-1.
If the expression is divisible by x—1, it will become 0, or “ vanish,’’ when 1 is substituted for x. Making this substitution, we obtain 0 ; hence the division is possible.
EXAMPLES X. r.
Without actual division, show that x — 2 is a factor of each of the following expressions :
1. x3-5x+2. 8. x3 — 7x2 + 16x — 12: 2. x3+ x2-4x—4. 4. x3 - 8x2+ 17x — 10.
Determine by inspection whether x + 3 is a factor of any of the following expressions :
5. x3 —7x+6. 7. x3+6x +6. 6. x3 +6x2+ 11x+6. 8. x3 + 3x2 +x+3. 9. Show that 32x10— 33x5 +1 is divisible by x — 1.
Resolve into factors .
10. 2x3 + 4x2-2x —4. 11. 3x3 —6x2 —3x+6.
106. We shall employ the Factor Theorem in giving general proofs of the statements made in Art. 62.
We suppose n to be a positive integer.
(I.) xn — yn ts always divisible by x— y.
By Art. 105, xn—yn is exactly divisible by x—y if the substitution of y for x in the expression xn — yn gives zero as a result. Making this substitution we have xn - yn =yn—yn=0. Therefore xn — yn is always divisible by x — y.
(II.) xn — yn is divisible by x+y when n is even.
If this be true, the substitution of —y for x in the expression xn — yn gives zero as a result. Making this substitution we have
xn - yn =(-yn)—yn
