Question

फाइंड द रिमाइंडर एन एक्स क्यूब प्लस 3 एक्स स्क्वायर प्लस 3 एक्स प्लस वन इज डिवाइडेड बाय एक्स प्लस वन ​

Answer 1

Correct Question :- Find the remainder when x³ + 3x² + 3x + 1 is divided by (x + 1) ?

Solution :-

we know that, according to remainder theorem when polynomial p(x) is divided by (x - a) , remainder will be p(a) .

so,

→ p(x) = x³ + 3x² + 3x + 1

→ p(-1) = (-1)³ + 3(-1)² + 3(-1) + 1

→ p(-1) = (-1) + 3 * 1 + (-3) + 1

→ p(-1) = (-1) + 3 - 3 + 1

→ p(-1) = 3 - 3 + 1 - 1

→ p(-1) = 0

therefore, when + 3x² + 3x + 1 is divided by (x + 1) remainder will be 0 .

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Answer 2

SOLUTION

TO DETERMINE

The remainder when x³ + 3x² + 3x + 1 is divided by x + 1

EVALUATION

Method : I

Let f(x) be the given the polynomial

$\sf{f(x) = {x}^{ 3 } + 3 {x}^{2} + 3x + 1}$

$\sf{ \implies \: f(x) = {(x + 1)}^{ 3 } }$

So f(x) contains the factor x + 1

Therefore f(x) is completely divisible by x + 1

Hence the required Remainder = 0

Method : II

Here the given polynomial is

$\sf{f(x) = {x}^{ 3 } + 3 {x}^{2} + 3x + 1}$

Let g(x) = x + 1

For Zero of the polynomial g(x) we have

g(x) = 0

⇒x + 1 = 0

⇒ x = - 1

So - 1 is zero of the polynomial g(x)

Hence by the Remainder Theorem the required Remainder when f(x) is divided by g(x) is

$\sf{ = f( - 1) }$

$\sf{= {( - 1)}^{ 3 } + 3 \times {( - 1)}^{2} + 3 \times ( - 1) + 1}$

$\sf{= - 1 + 3 - 3 + 1}$

$\sf{= 0}$

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