Question

By remainder theorem, find the remainder when p(x) is divided by g(x)
(i)
${x}^{3} - 2 {x}^{2} - 4x - 1 \: \:$
$g(x) - x + 1$
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Answer 1

Answer:

since g(x)=-x+1=0(Remainder theorem)

=-x=-1 ,so x=1

By putting the value of x

We get, p(x)=p(1)

So, p(x)=x^3-2x^2-4x-1

=1^3-2×1^2-4×1-1=1-4-4-1=-8

Answer 2

Answer :-

• Remainder => -6

To Find :-

• The remainder if x³ - 2x² - 4x - 1 is divided by - x + 1.

Step By Step Explanation :-

We know that p(x) => x³ - 2x² - 4x - 1 and

g(x) => -x + 1

We need to find the remainder.

So let's do it !!

Using remainder theorem ⤵

$\sf \: g(x) \implies -x + 1 = 0 \\ \\ \sf g(x) \implies (- x )= (- 1) \\ \\ \bf\dag By \: cancelling \: minus \:sign \downarrow \\ \\ \sf g(x) \implies \: x = 1$

Now, x = 1

By substituting the value of x = 1

$\implies\sf {x}^{3} - 2 {x}^{2} - 4x - 1 \\ \\ \implies\bf \: x = 1 \\ \\\implies\sf {(1)}^{3} - 2 {(1)}^{2} - 4(1) - 1 \\ \\\implies\sf 1 - 2 - 4 - 1 \\ \\\implies\sf 1 - 7 \\ \\ \implies\sf - 6$

Therefore if x³ - 2x² - 4x - 1 is divided by - x + 1 then remainder => -6

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