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ALGEBRA ALGEBRA = p(x) when divided by (x - a) gives the remainder equal to pla). pla) = 0 Remarks (i) (x + a) is a factor of a polynomial iff (if and only if) p (-a) = 0 P (2) 3 = m. (ii) (ax-b) is a factor of a polynomial iff p = 0 a (iii) (ax + b) is a factor of a polynomial p(x) iff. b pl - a = 0 4. If polynomials 2x2 + ax2 + 3x-5 and x3 + x2 - 2x + a are divided by (x - 2), the same remainders are obtained. Find the value of a. (1) -3 (3) 4 (4) -9 Sol. (1) f(x) = 2x2 + ax + 3x - 5 g(x) = x + x - 2x + a By Remainder Theorem, f(2) = (2 23 + ax 22 + 3 2-5) = 17 + 4a Again, g(2) = (23 + 22 – 2x2 + a) = 8 + a :. 17 + 4a= 8 + a 3a = -9 > a=-3 5. If the polynomial f(x) = x4 – 2x3 + 3x2 - ax + b is divided by (x - 1) and (x + 1), the remainders are respectively 5 and 19. The values of a and bare : (1) a = 8, b= 7 (2) a = 5, b = 8 (3) a=8, b = 5 (4) a= 6, b= 8 Sol. (2) f(x) = x4 - 2x3 + 3x - ax + b f(1) = 1 - 2+3 - a + b = 2 - a + b (x - 1 = 0 = x= 1] f-1) = 1 + 2 + 3 + a + b = 6 + a + b (x + 1 = 0 = x=- 1] 2- a+b=5b-a=3 (i) and, 6 + a + b = 19=a+ b = 13 By adding equations (i) and (ii), 2b = 16 b= 8 co = (iv) (x - a) (x- b) are factors of a polynomial p(x) iff pla) = 0 and plb) = 0 SOLVED EXAMPLES = 1. When the polynomial f(x) = x + 2x2 – 3x2 + x-1 is divided by (x-2) what will be the remainder? (1) 21 (2) 22 (3) 23 (4) 29 Sol. (1) Here, X-2 = 0 = x= 2 By Remainder Theorem, when polynomial f(x) is divided by (x - 2), the remainder is f(2). .. f(2) = 24 + 2 x 23 - 3 x 22 + 2 - 1 [Remember : x has been replaced by 2] = 16 + 16 - 12 + 2-1 = 21 .. Remainder = 21 . (ii) : : ...
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