Question

If f(x) = x4 – 3 x3 + 5 x2 – a x + b is a polynomial such that when it is divided by x – 1 and x + 1, the remainders are 10 and 18 respectively. Find the values of a and b. Also, determine the remainder when f(x) is divided by (x – 2).​

Answer 1

Step-by-step explanation:

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

∴f(1)=5 and f(−1)=19

⇒(1)

4

−2×(1)

3

+3×(1)

2

−a×1+b=5

and (−1)

4

−2×(−1)

3

+3×(−1)

2

−a×(−1)+b=19

⇒1−2+3−a+b=5

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

⇒2−a+b=5 and 6+a+b=19

⇒−a+b=3 and a+b=13

Adding these two equations, we get

(−a+b)+(a+b)=3+13

⇒2b=16⇒b=8

Putting b=8 and −a+b=3, we get

−a+8=3⇒a=−5⇒a=5

Putting the values of a and b in

f(x)=x

4

−2x

3

+3x

2

−5x+8

The remainder when f(x) is divided by (x-2) is equal to f(2).

So, Remainder =f(2)=(2)

4

−2×(2)

3

+3×(2)

2

−5×2+8=16−16+12−10+8=10

Answer 2

Step-by-step explanation:

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)=x4−2x3+3x2−ax+b

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

f(1)=14−2×13+3×12−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!