Question

it is given that 2x^3-7x^2+10=(x-1). Q(x)+R. Find the quotient and remainder​

Answer 1

Step-by-step explanation:

here quotient is -5(x+1)

remainder is 5

Answer 2

Step-by-step explanation:

Given :-

2x³-7x²+10=(x-1). Q(x)+R

To find :-

Find the quotient and remainder ?

Solution :-

Given that

2x³-7x²+10=(x-1). Q(x)+R

On comparing with P(x) = g(x).Q(x)+R

P(x) = 2x³-7x²+10

g(x) = (x-1)

On dividing P(x) by g(x) then

x-1)2x³-7x²+10(2x²-5x-5

2x³-2x²

(-) (+)

_________

-5x²+10

-5x²+5x

(+) (-)

____________

-5x+10

-5x +5

(+) (-)

_____________

5

_____________

We can write it as

2x³-7x²+10 =(x-1) (2x²-5x+5)+5

Q(x) = 2x²-5x+5

R = 5

Answer:-

The Quotient for the given problem is 2x²-5x+5

The remainder for the given problem is 5

Used formulae:-

Fundamental Theorem on Polynomials:-

Let P(x) is divided by g(x) then q(x) is the quotient and r(x) is the remainder then

P(x) = g(x)×q(x) +r(x)