= (a - b)2 + 2 (a - b) (c − d) + (c — d)2 -- = (a2 — 2 ab+b2) + 2 a (c− d) — 2b (c− d) + (c2 — 2 cd+d2) =a22ab+b2 + 2 ac- 2 ad-2bc+2bd+c2 - 2 cd+d2 = a2 + b2 + c2+ d2-2ab+2ac2ad2bc+2bd2cd. Here the same law holds as before, the sign of each double product being + or -, according as the factors composing it have like or unlike signs. The same is true for any polynomial. Hence we have the following rule: RULE 4. The square of a polynomial is the sum of the squares of the several terms and twice the product obtained by multiplying each term into all the terms that follow it. 113. Product of Two Binomials. The product of two binomials which have the form x + a, x + b, should be carefully noticed and remembered. 1. (x+5) (x + 3) = x (x + 3) + 5 (x + 3) = x2 + 3x + 5x+15 = x2+8x+15. 2. (x-5) (x-3) = x (x − 3) – 5 (x − 3) = x2-3x-5x+15 = x2-8x+15. 3. (x+5) (x-3) = x (x − 3) + 5 (x − 3) = x2+2x-15. 4. (x − 5) (x + 3) = x (x + 3) − 5 (x + 3) = x2+3x-5x-15 = x22x-15. Each of these results has three terms. The first term of each result is the product of the first terms of the binomials. The last term of each result is the product of the second terms of the binomials. The middle term of each result has for the coefficient of x the algebraic sum of the second terms of the binomials. The intermediate step given above may be omitted, and the products written at once by inspection. Thus, 1. Multiply x+8 by x + 7. 8+7= 15; 8 x 7 = 56. ..(x+8) (x+7)= x2 + 15x+56. 2. Multiply x 8 by x 7. (8)+(-7)=-15; (-8) (→ 7) = + 56. .. (x − 8) (x − 7) = x2 – 15 x + 56. 3. Multiply x7y by x+6y. -7y+6y=y; (− 7y) (6 y) = — 42 y2. · · (x − 7 y) (x + 6 y) = x2 — xy — 42 y2. 4. Multiply x2+6 (a + b) by x2 - 5 (a + b). − 5 (a + b) + 6 (a + b) = (a + b) ; × - 5 (a + b) x 6 (a + b) = − 30 (a + b)2. ··· { x2+6(a+b)} { x2 −5 (a+b)}=x*+(a+b)x2−30(a+b)2. 114. In like manner the product of any two binomials may be written. 1. Multiply 2 a-b by 3 a +46. (2 ab) (3 a + 4b) = 6a2 + 8 ab - 3 ab-4ba = 6a2 + 5 ab - 4 b2. 2. Multiply 2x + 3y by 3x-2y. The middle term is 2 x x (-2y) + 3y × 3x=5xy. Division. 115. The following rule for finding any required root of a monomial will be found useful in solving examples in division: Find the required root of the numerical coefficient, and divide the exponent of each letter by the index of the required root. Thus, the square root of 25 x2y1 is 5 xy2. RULE 1. The difference of the squares of two numbers is divisible by the sum of the numbers, and the quotient is the difference of the numbers. The difference of the squares of two numbers is divisible by the difference of the numbers, and the quotient is the sum of the numbers. |