Question

If f(x)=x^4-2x^3+3x^2-ax+b is divided by x-1 and x+1 the remainder are 5 and 19 respectively, then find a and b

Answer 1

There is theorem known as “Polynomial Remainder Theorem” or “ Bezout’s Theorem”. It is Stated as -

A Polynomial f(x) if divided by a linear polynomial (x-a) leaves remainder which equals f(a).

So , getting back to our question -

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

So , when it is divided by (x - 1) it’ll leave a remainder = f(1) = 5 (Given).

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

=> 1 - 2 + 3 - a + b = 5

=> a - b = (-3) …. Eqn(1)

Now , Similarly -

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

=> 1 + 2 + 3 + a + b = 19

=> a + b = 13 …. Eqn(2)

Now , adding equations (1) and (2) , We’ll get -

(a+b) + (a-b) = (-3) + 13

=> 2a = 10 => a = 5

So , (a +b) = 13 implies b = 8

Hence , Values of a and b are 5 and 8 respectively.

Done!

HOPE THIS HELPS...☺☺☺

Answer 2

Answer:

47

Step-by-step explanation:

Given that the equation

f(x) = x4 – 2x3 + 3x2 – ax +b

When f(x) is divided by (x+1) and (x-1) , the remainders are 19 and 5 respectively .

∴ f(-1) = 19 and f(1) = 5

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

⇒ 1 +2 + 3 + a + b = 19

∴ a + b = 13 ——- (1)

According to given condition f(1) = 5

f(x) = x4 – 2x3 + 3x2 – ax

⇒ 14 – 2 3 + 3 2 – a (1) b = 5

⇒ 1 – 2 + 3 – a + b = 5

∴ b – a = 3 —— (2)

solving equations (1) and (2)

a = 5 and b = 8

Now substituting the values of a and b in f(x) , we get

∴ f(x) = x4 – 2x3 + 3x2 – 5x + 8

Also f(x) is divided by (x-3) so remainder will be f(3)

∴ f(x)= x4 – 2x3 + 3x2 – 5x + 8

⇒ f(3) = 34 – 2 × 33 + 3 × 32 – 5 × 3 + 8

= 81 – 54 + 27 – 15 + 8

= 47

Therefore, f(x) = x4 – 2x3 + 3x2 – ax +b when a=3 and b= 8 is 47