Here, (a2-x2)×7x2=(a+x)×(a-x)×7x2= numerator (Art. 106), and 3a × (a-x)= denomina tor; see Ex. 15, (Art. 79). (a+x)×(a-x)×7x2 Hence, the product is 3ax(a-x) (dividing the numerator and denominator by a-x) 7x2(a+x)_7ax2+7x3 Then, (5a+x)×(3a-x)=15a2-2ax-x2= new numerator, and 5×3=15= denominator: There 157. But, when mixed quantities are to be multiplied together, it is sometimes more convenient to proceed, as in the multiplication of integral quantities, without reducing them to improper frac tions. Ex. 5. Multiply x2-x+ by +2. x2-+ x+2 +x3x2+x +2x2-x+ x2+x2-x+ Ex. 9. It is required to find the continual pro 3α 2x2 و a+b and Ex. 10. It is required to find the continued pro- duct of 2 and a-y a-x Ans. atx. Ex. 11. It is required to find the continued pro duct of a2-2 a2-b2 Ex. 12. Multiply x2-2x+1 by x2-x. Ans. 2004-03 2 8 +2. To divide one fractional quantity by another. RULE. 158. Multiply the dividend by the reciprocal of the divisor, or which is the same, invert the divisor, and proceed, in every respect, as in multiplication of algebraic fractions; and the product thus found will be the quotient required. When a fraction is to be divided by an integral quantity; the process is the reverse of that in multiplication; or, which is the same, multiply the denominator by the integral, (Art. 120), or divide the numerator by it. The latter mode is to be preferred, when the numerator is a multiple of the is the quotient required. 5(-x) 5 |