Question

pls guys help me in this question .

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Answer 1

✧$\huge \mathcal \red {\underline{\underline{A}}}$ $\huge \mathcal \green {\underline{\underline{N}}}$ $\huge \mathcal \pink {\underline{\underline{S}}}$ $\huge \mathcal \blue {\underline{\underline{W}}}$ $\huge \mathcal \orange {\underline{\underline {E}}}$ $\huge \mathcal \purple {\underline{\underline{R}}}$✧

❥ Given :

➛ $p(x) = 3x^{2} + x - 1$

➛ $g(x) = x + 1$

❥ To Find :

The remainder when $\: p(x) = 3x^{2} + x - 1 \:$ is divided by $\: g(x) = x + 1$

❥ Solution :

➛ $g(x) = x + 1$

→ $x + 1 = 0$

→ $x = - 1 \: \: \: \: \: \: [Zero \: of \: g(x) = x + 1]$

Put the value of x in $\: p(x) = 3x^{2} + x - 1$

➛ $p(x) = 3x^{2} + x - 1$

→ $p(x) = 3(-1)^{2} + (-1) - 1$

→ $p(x) = 3(1) - 1 - 1$

→ $p(x) = 3 - 2$

→ $p(x) = 1$

$\implies \textbf {Hence, the remainder when} \: p(x) = 3x^{2} + x - 1 \: \textbf {is divided by} \: g(x) = x + 1 \: \textbf {is 1} .$

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Answer 2

Answer:

$\bf \large\underbrace\orange{Question:}$

• Find the remainder when 3x² + x - 1 is divisible by x + 1.

$\bf\underbrace\orange{Answer:}$

Given :-

• p(x) = 3x² + x - 1
• g(x) = x + 1.

To find :-

• The remainder when 3x² + x - 1 is divisible by x + 1.

Concept used :-

• Remainder Theorem :- : This theorem states: if f(x) is a polynomial in x, then the remainder on dividing f(x) by x − a is f(a).

$\bf\underbrace\orange{Solution:}$

Here,

p(x) = 3x² + x - 1

g(x) = x + 1.

Now, We have to find the remainder when p(x) = 3x² + x - 1

is divided by g(x) = x + 1.

Put x + 1 = 0

⟹ x = - 1.

So, remainder when p(x) = 3x² + x - 1 is divided by g(x) = x + 1 is p( - 1).

So, p(- 1) = 3( - 1)² + ( - 1) - 1

⟹ p( - 1) = 3 - 1 - 1

⟹p ( - 1) = 1.

So, the remainder when 3x² + x - 1 is divisible by x + 1 is 1.

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