Question

how to find remainder theorm questions

Answer 1

hey mate

here l is your answer...

Answer:

find the remainder when x²+x+1 is divided by x+1.

solution:

let f(x)=x²+x+1

then,

x+1=0

x=-1

by remainder theorem,

when f(x) is divided by x+1, remainder is f(-1).

f(-1)

(-1)²+(-1)+1

1-1+1

1

so, the remainder obtained will be 1.

so in this way need to solve the question related to remainder theorem.

some more questions:

find remainder when:

(i) x²+5x+1 is divided by x+4

(ii) x⁴+x²+1 is divided by x-1

(iii) x⁴+2x³+3x²+x+1 is divided by x+6.

I hope this is sufficient for you.

LIKE AS MUCH AS YOU CAN!!!

Answer 2

$\huge\boxed{\fcolorbox{lime}{yellow}{⭐ANSWER⤵࿐}}$

$\sf\large\underline\red{⭐RemainderTheorem :-}$

Remainder Theorem is an approach of Euclidean division of polynomials. According to this theorem, if we divide a polynomial P(x) by a factor ( x – a); that isn’t essentially an element of the polynomial; you will find a smaller polynomial along with a remainder. This remainder that has been obtained is actually a value of P(x) at x = a, specifically P(a). So basically, x -a is the divisor of P(x) if and only if P(a) = 0. It is applied to factorize polynomials of each degree in an elegant manner.

For example: if f(a) = a3-12a2-42 is divided by (a-3) then the quotient will be a2-9a-27 and the remainder is -123.

if we put, a-3 = 0

then a = 3

Hence, f(a) = f(3) = -123

Thus, it satisfies the remainder theorem.

$\sf\large\underline\blue{✴Proof :-}$

Theorem functions on an actual case that a polynomial is comprehensively dividable, at least one time by its factor in order to get a smaller polynomial and ‘a’ remainder of zero. This acts as one of the simplest ways to determine whether the value ‘a’ is a root of the polynomial P(x).

That is when we divide p(x) by x-a we obtain

p(x) = (x-a)·q(x) + r(x),

as we know that Dividend = (Divisor × Quotient) + Remainder

But if r(x) is simply the constant r (remember when we divide by (x-a) the remainder is a constant)…. so we obtain the following solution, i.e

p(x) = (x-a)·q(x) + r

Observe what happens when we have x equal to a:

p(a) = (a-a)·q(a) + r

p(a) = (0)·q(a) + r

p(a) = r

Hence, proved.

$\sf\large\underline\red{⭐Hope It's Help You !!✔ }$