ALGEBRA ALGEBRA Sol. (3) Here, x + a=0x=- a .. f(-a) = 0 = (-a)3 + a (-a)2 - 2 (-a) + a + 6 = 0 3a=-6=a=-2 9. For what value of k, (2x4 + 3x3 + 2kx2 + 3x + 6) is exactly divisible by (x + 2)? (1) 1 (2) 2 fuer- (3) - 2 (4) - 1 Sol. (4) Here, x + 2 = 0 => x=- 2 By Factor Theorem, f(-2) = 0 HYS 2(-2)4 + 3 (-2)3 + 2k(-2)2 + 3(-2) + 6 = 0 - 32 -24 + 8k- 6 + 6 = 8k+ 8 = 0 = 8k=-85k=-1 10. For what values of a and b, (x3 - 10x + ax + b) is exactly divisible by (x - 1) and (x-2) ? (1) a= 23, b=-14 (2) a=-23, b = 14 (3) a= 21, b=-14 (4) a=-21, b= 15 Sol. (1) f(x) = x3 - 10x + ax + b By Factor Theorem, f(1) = 1 - 10 + a+b= a + b - 9 T.: x-1 = 0 = x= 1] :: f(1) = 0 = a + b = 9 f2) = 8 - 40 + 2a + b = 2a + b-32 :: f(2) = 0 2a + b = 32 From equation (ii) – equation (i), a=23 Example : 36ab-60abc = 12 a+b (3a-50) Example : x(x - y)2 + 3xy (x - y) = x(x - y) [(x, y)2 + 3xyl 170 = x (x - y) (x + y2 - 2xy + 3xyl = x (x - y) (x + y + xy) Este Method 2 : Sometimes in a given expression it is not possible to take out a common factor di- rectly. However the terms of the expres- sion are grouped in such a manner that we may have a common factor. This can now be factorised as discussed above. Example : Factorise : 6ab-b2 + 12ac-2bc Sol. : 6ab-b2 + 12ac-2bc=(ab+ 12ac) - (b2 +2bc) = 6a (b + 2c) - b (b + 2C) - (b + 2c) (a - b) Example : Factorise : x + 18x + 813... Sol. x2 + 18x + 81 = x2 + 18x + 92 = x + 2 x 9 x x + 92 = (x + 9)2 Example : Factorise : 64x2 - 16x + 1 Sol. : 64x2 - 16x + 1 = (8x)2 - 16x + 12 = (8x)2 - 2. (8x). 1 + 12 = (8x - 1)2 DAIS Example : Factorise : 81 - 64.x2 Sol. 81 - 64x2 = 92 - (8x)2 = (9 + 8x) (9- 8x) Mothod 3 : Factorisation of Quadratic Trinomials Case I : Polynomial of the form x + bx + c. We find integers pand q such that p+q= band pq= 0 = = 33) ... (i) - + ... (ii) ...
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