SOLUTION-* TO THE EXERCISES.] MATHEMATICS. ALGEBRA. 493 



10. (x + l)(x l)(x 2 + l)=(x 2 l)(x 2 + 1) = x 4 1, and Or 4 1)- = x 3 



11. (x ]) 2 =x 2 2x + l .'. x 2 2x + 1 



4 2x 3 +x 2 

 2x 3 -j-4x 2 2x 



x a 2x+l 



x 2 2x + 1 



x 4X 6 + Gx 4 4s 3 + y? 

 2x* + 8x 4 12X 3 + 8x ! 



.-. {(x 1)'} 3 = x GxO+lox 4 20X 3 + 15V Gx+1. 



NOTE. In working the foregoing examples, it will be seen that frequent application has been made of the 

 principles marked 1, 2, 3, at page 4CO. They should always be used in like manner whenever opportunity occurs. 

 The Binomial Theorem, in the next article, is, however, of still wider application, and you should make yourself 

 well acquainted with its form : the eleventh example here worked out in full, is solved in a moment by aid of tlio 

 Binomial Theorem, at you will see by a reference to the table of developments at the middle of page 461. The 

 example in question is to find the sixth power of x 1 ; for { (x I) 2 } 3 is the same as (x 1)*. If you suppose a 

 in the table to be 1, you will see that the development of (x 1) is the same as that found above by actual 

 multiplication ; the directions at page 4G2 show how this development might have been written down at ouce. 



BINOMIAL THEOREM. EXAMPLES FOB EXERCISE. PAGE 462. 



1. To develop (o + y) 4 . The terms without the coefficients are 



a 4 , aV, ay, ay 3 , y 4 . 

 The coefficients, as far as the third, are 



1 4. 3x4 r 



1, 4, _ _ 6 



/. (a + y) 4 - a* + 4a>y + Gay + 4ay + y 4 . 



Instead of finding the terms without the coefficients, and then the coefficients in a separate step, it is just as 

 easy to write down each term in succession in its complete form, attending to the directions at page 461. 



2. (a z) 8 - o s 5a 4 x 



3. (a + 2X) 3 = a 3 + 3o(2x) + 3a(2x) > + (Zxf 



- a 3 + Ga'x + 12ax + Sx 3 . 

 4 (1 x)=l 6x + 15x ! 20x 3 + 15x 4 



In this example the powers of 1, the first term of the binomial 1 x, are, of course, suppressed ; but tho 

 exponents of those powers are mentally employed, as in other cases, to form the coefficients. 



5. 



5. (1 + 3x) 4 = 1 + 4(3.r) + G(3.r)' + 4(3*)' + (3*) 4 



= 1 + 12* + 54x* + lO&e 3 + Six*. 



6. (x 



= X s + 



7. (x 1 + 3 1 )'= W + 6 (xWj/ 5 ) + 10 (xWvV + l<K**yWP + 5 (**) < 3 '/) 4 + (V)* 

 -x" + 15xy 



8. (2a xV> = (2a) 6(2a) i x + 15(2a) t x s 20(2a)'x 3 + loW* 4 6(2a)x 6 + 

 64a 192a<r + 240a 4 x 



9. (x 2y*y = x' 7zW) + 21x(2y)* 35x'(2y) 3 + 35x 3 (2y 2 ) 4 21x ! (2/ J ) s + 7x(2y) 



672xy + 448xy" 1 



DIVISION. CASE I. EXAMPLES FOR EXERCISE. PAGE 463. 



a Sc^ac 2 !/^ 5d* 



- **y. 8 ""Sfo^V __ 18y* 



O 9 _> I A ..* 4 1O ..*.. 1 



10. 



4my ' 2Gy 4 Jz 2 



DIVISION. CASE II. PAGE 463. 

 3x ax+2a'. 



12xy Z 4 + toys' -3*y ^j , 2 s _ y_ 

 * '?' 



SJyz 



