Page:Elementary algebra (1896).djvu/118

100 Rh 9. 3 x^4 - 3 x3 - 2 x2 - X - 1, 9 x4 - 3 x^3 - x - 1. 10. 3 x^3 - 3 ax2 + 2 a2x - 2 a^3, 3 x^3 + 12 ax2 + 2 a2x, + 8 a^3. 11. 2 x3 - 9 ax2 + 9 a2x - 7 a^3, 4 x^3 - 20 ax^2 + 20 a^2x - 16 a^3. 12. 10 x^3 + 25 ax2 - 5 a3, 4 x^3 + 9 ax^2 - 2 a2x - a^3. 13. 24x4y4 + 72x3y2 - 6x2y3 - 90xy4, 6x4y2+13x3y3 - 4x2y4-15xy5. 14. 4x5a2+10x4a3-60x3a4+54x2a5, 24x5a3 + 30 x3a5-126x2a6. 15. 4x5 + 14x^4 + 20x3 + 70x2, 8x^7 + 28x6 - 8x^5 - 12x4 + 56x3. 16. 72x3-12ax2+72a2x-420a3, 18x3 + 42ax2 - 282 a2x+270a3. 17. x5 - x3 - x + 1, x^7 + x^6 + x^4 - 1. 18. 1 + x + x3 - x^5, 1 - x^4 - x^6 + x^7.

122. The statements of Art. 116 may be proved as follows :

I. If F divides A it will also divide mA. For suppose A = aF, then mA = maF. Thus F is a factor of mA.

II. If F divides A and B, then it will divide mA ± nB.

For suppose A = aF, B = bF, then mA ± nB = maF ± nbF = F(ma ± nb).

Thus F divides mA ± nB.

123. We may now enunciate and prove the rule for finding the highest common factor of any two compound algebraic expressions.

We suppose that any simple factors are first removed. (See Example, Art. 117.)

Let A and B be the two expressions after the simple factors have been removed. Let them be arranged in descending or ascending powers of some common letter; also let the highest power of that letter in B be not less than the highest power in A.

Divide B by A, let p be the quotient, and C the remainder. Suppose C to have a simple factor m. Remove this factor, and so obtain a new divisor D. Further, suppose that in order to make A divisible by D it is necessary to