Question

1. Find the remainder when +3 + 3x + 1 is divided by

(1) x+1

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Answer 1

AnsweR :-

First we will find the zero of x-1 that will be -

$x-1 = 0$

$x = 0-1$ ( as 1 as been transposed to another side the sign will also change)

$x = -1$

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$p(x) = 3 + 3x + 1$

$p( - 1) = 3 + 3( - 1) + 1$

$p( - 1) = 3 - 3 + 1$

$p ( - 1) = 1$

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$\therefore$ value of p(-1) = 1

Procedure :-

• First we have to find out the value of x +1 as per your required question. As you can see when 1 got transposed to RHS the sign got changed to negetive. So if we will subracte anything from zero then the value will become in negetive.
• After that we placed -1 in place of x and solved it further. As you can see that there a positive 3 and a negetive 3 so they will get cancelled and only 1 will be remained that is your answer.

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Answer 2

SOLUTION :-

GIVEN :-

$Dividend\Rightarrow p(x) = 3 + 3x + 1\\\\Divisor\Rightarrow g(x) = x + 1$

To find the Remainder of p (x) using the Remainder Theorem -

Let the Divisor, g (x) = 0 => x + 1 = 0

∴ x = (-1)                          [Zero of the Polynomial p (x)]

Substituting x = (-1) in the Polynomial p (x), we get the Remainder as -

$\Rightarrow p (-1) = 3 + 3 (-1) + 1\\\\\Rightarrow p (-1) = 3 + (-3) + 1\\\\\Rightarrow p (-1) = 3 - 3 + 1\\\\= \bold {p( -1) = +1}$

∴ p (-1) = +1 is the Required Remainder of p (x)