Question

If x^100-2ax^99+b is exactly divisible by x^2-1 find the values of a and b ​

Answer 1

Answer:

a=0 & b= -1

Step-by-step explanation:

exactly divisible by x²-1

so two roots(zeros) are-

x²-1=0

x²=1

x= +1

first we take x=+1

(1)¹⁰⁰-2a(1)⁹⁹+b=0 (it should be 0 bcz 1 is a zero of polynomial)

1-2a+b=0...........(1)

Now we take x= -1

(-1)¹⁰⁰-2a(-1)⁹⁹+b=0

1-2a(-1)+b=0

1+2a+b=0..........(2)

add (1)+(2)

2+2b=0

2b= -2

b= -1

put b= -2 in eq.(1)

1-2a-1=0

2a=0

a=0

Ans. a=0 & b= -1