The polynomial $x^{3} - (a + 1) x^{2} - x + 4$ and $x^{3} + (b + 2) x^{2} + x - 6$ leaves a remainder of 0, when divided by (x - 1). Find a, b

a = 3

b = 4

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a = 3

b = 9

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a = 3

b = 2

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a = 1

b = 4

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Solution

The correct option is C
a = 3

b = 2

Let f(x) = $x^{3}$ - (a+1) $x^{2}$ - x + 4

g(x) = $x^{3}$ + (b + 2) $x^{2}$ + x + 6

f(1) = g(1) = 0

f(1) = $1^{3}$ - (a + 1)1 - 1 + 4 = 0

$\Rightarrow$ -(a + 1) + 4 = 0

$\Rightarrow$ a = 3

g(x) = $x^{3}$ + (b + 2) $x^{2}$ + x - 6

g(1) = 1 + (b + 2)1 + 1 - 6 = 0

$\Rightarrow$ -4 + b + 2 = 0

$\Rightarrow$ b = 2

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