Question

By remainder theorem find the remainder when P(x) is divided by g(x),where P(x)= x3 -x2-4x-1 g(x)=x+1​

Answer 1

Answer:

$\tt{SOLUTION:-}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt{By \: Remainder \: Theorem}$

$\small\tt\purple{x+1 = 0}$

$\boxed{\small\tt\purple{x = - 1}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt{Now \: substitute \: the \: value \: of \: x \: in \: p(x) \: which \: is \: } \small \tt \purple{-1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\small\tt{ p(x) \: \: \: = {x}^{3} - {x}^{2} - 4x - 1}$

$\small\tt{ p( - 1) = { - 1}^{3} - { (- 1)}^{2} - 4( - 1) - 1}$

$\small\tt{ \: \: \: \: \: \: \: \: \: \: \: \: \: = { - 1} { - 1} + 4 - 1}$

$\small\tt{ \: \: \: \: \: \: \: \: \: \: \: \: \: = - 2 + 4 - 1}$

$\small\tt{ \: \: \: \: \: \: \: \: \: \: \: \: \: = 2 - 1}$

$\tt \purple{ Remainder = 1}$

Answer 2

Answer:

Solution :-

By remainder theorem

$x \: + \: 1 \: = \: 0. \\ \\ x \: = \: - 1.$

Now subsitude the value of x in p(x) which is -1.

$p(x) \: = \: x ^{3} \: - \: x \: ^{2} \: - \: 4x \: - 1 \\ p( - 1) \: = - 1 ^{3} \: - \: ( - 1) ^{2} \: - \: 4( - 1) \: - 1 \\ = \: - \: 1 \: - \: 1 \: \ + \: 4 \: - \: 1 \\ = \: - \: 2 \: \ + \: 4 \: - \: 1 \\ = \: 2 \:-\:1$

Remainder = 1 .

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