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(x-2) is a factor of (x^3+ax^2+bx+6) leaves remainder 3 on divided by (x-3) |
Answer» ★ Answer :given : (x - 2) is a factor of x³ + ax² + bx + 6 and leaves 3 as remainder when divided by (x - 3) Solution : if x - 2 is a factor then, F(2) should be 0 f(x) = x³ + ax² + bx + 6 f(2) = (2)³ + a(2)² + 2b + 6 = 8 + 4a + 2b + 6 14 + 4a + 2b = 0 4a + 2b = -14 2(2a + b) = -14 2a + b = -7 ..... (1) given : f(3) = 3 f(3) = (3)³ + a(3)² + 3b + 6 = 27 + 9a + 3b + 6 = 33 + 9a + 3b = 3 9a + 3b = 3-33 9a + 3b = -30 3(3a + b) = -30 3a + b = -10 ..... (2) (1) - (2) (2a + b) - (3a + b) = -7-(-10) 2a + b - 3a -b = -7 +10 - a = 3 → a = -3 by putting VALUE of a in equation 1... 2a + b = -7 2(-3) + b = -7 -6 + b = -7 b = -7+6 → b = -1 therefore, a = -3 and b= -1.. equation = x³ + (-3)x² + (-1)x + 6 = x³ - 3x² - x + 6 |
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