x^3+3x^2+3x+1 ÷ x-1/2

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Solution

Apply remainder theorem
=>x – 1/2 =0
=> x = 1/2
Replace x by 1/2 we get
=>x³+3x² + 3x + 1
=>(1/2)³ + 3(1/2)² + 3(1/2) + 1
=> 1/8 + 3/4 + 3/2 + 1
Add the fraction taking LCM of denominator we get
=>(1 + 6 + 12 + 8)/8
=>27/8
Remainder is 27/8

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lim x \to 1 \frac{x^{3} + {3 x}^{2} - 6 x + 2}{x^{3} + {3 x}^{2} - 3 x - 1}

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$