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If the polynomials ax3 + 3x2 - 3x and 2x3 - 5x + a when divided by (x - 4) leave the remainder R1 and R2 respectively. Find the value of a in each of the following cases, if(i) R1 = R2 (ii) R1 + R2=0 (iii) 2R1 - R2 = 0. |
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Answer» Let, p (x) = ax3 + 3x2 - 3 and q (x) = 2x 3-5x+a be the given polynomials. Now, R1 = Remainder when p (x) is divided by (x – 4) = p (4) = a (4)3 + 3 (4)2 – 3 [Therefore, p (x) = ax3 + 3x2 - 3] = 64a + 48 – 3 R1 = 64a + 45 And, R2 = Remainder when q (x) is divided by (x – 4) = q (4) = 2 (4)3 – 5 (4) + a [Therefore, q (x) = 2x3 - 5x + a] = 128 – 20 + a R2 = 108 + a (i) Given condition is, R1 = R2 64a + 45 = 108 + a 63a – 63 = 0 63a = 63 a = 1 (ii) Given condition is R1 + R2 = 0 64a + 45 + 108 + a = 0 65a + 153 = 0 65a = -153 a = \(\frac{-153}{65}\) (iii) Given condition is 2R1 – R2 = 0 2 (64a + 45) – (108 + a) = 0 128a + 90 – 108 – a 127a – 18 = 0 127a = 18 a = \(\frac{18}{127}\) |
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