Question

solve this fast as possible

Answer 1

Hey Friend ☺

☆ POLYNOMIALS ☆

f( x ) = x^4 - 2x^3 + 3x^2 - ax + b

when it is divided by x - 1 the remainder is 5

using remainder theorem

f ( 1 ) = 5

f ( 1 ) = x^4 - 2x^3 + 3x^2 - ax + b

》5 = ( 1 )^4 - 2 ( 1 )^3 + 3 ( 1 )^2 - a( 1 ) + b

》5 = 1 - 2 + 3 - a + b

》5 = 2 - a + b

》- a + b = 5 - 2

》- a + b = 3. ... ( l )

When f ( x ) is divided by x + 1 remainder is 19

Using remainder theorem

f ( - 1 ) = 19

》f ( - 1 ) = x^4 - 2x^3 + 3x^2 - ax + b

》f ( - 1 ) = ( - 1 )^4 - 2 ( - 1 )^3 + 3 ( - 1 )^2 - a ( - 1 ) + b

》19 = 1 + 2 + 3 + a + b

》19 = 6 + a + b

》a + b = 19 - 6

》a + b = 13. .... ( 'll )

Adding equations ( l ) & ( 'll ) we get ,

2b = 16

》b = 16/2

》b = 8

Putting b = 8 in equation ( 'll ) we get

a + 8 = 13

》a = 13 - 8

》a = 5

So the f ( x ) = x^4 - 2x^3 + 3x^2 - 5x + 8

When f ( x ) is divided by ( x - 2 ) the remainder is given by

f ( 2 )

f ( 2 ) = x^4 - 2x^3 + 3x^2 - 5x + 8

= ( 2 )^4 - 2 ( 2 )^3 + 3 ( 2 )^2 - 5 ( 2 ) + 8

= 16 - 2 × 8 + 3 × 4 - 10 + 8

= 16 - 16 + 12 - 10 + 8

= 2 + 8

= 10

So the remainder is 10.

Hope it helps you ..!!.

✌

Answer 2

Hi there !!

Given,

f(x) = x⁴ - 2x³ + 3x² - ax + b

Zero of x-1 = 1

Using remainder Theorem,
Remainder = 5

(1)⁴ - 2(1)³ + 3(1)² - a(1) + b = 5

1 - 2 + 3 - a + b = 5

2 - a + b = 5

-a + b = 5 - 2

-a + b = 3 ________ (I)

Remainder 2 = 19.
Zero of 1 = -1

Using remainder Theorem,

(-1)⁴ - 2(-1)³ + 3(-1)² - a(-1) + b = 19

1 + 2 + 3 + a + b = 19

6 + a + b = 19

a + b = 19 - 6

a + b = 13 ________(ii)

Adding eq 1 and 2,

-a + b + a + b = 3 + 13

2b = 16

b = 16/2

b = 8

Substituting b = 8 in any of the equation so as to find a ,

a + b = 13
a + 8 = 13
a = 13 - 8
a = 5

Thus,
a = 5
b = 8

when x - 2 = 0, x = 2

Substituting the values and using remainder Theorem,
we've

f(2) = 2⁴ - 2(2)³ + 3(2)² - 5(2) + 8

= 16 - 16 + 12 - 10 + 8

= 10

Thus,
the remainder is 10


Hope it helped :-)