GiGiven that the polynomial (x2 + ax + b) leaves that same remainder w...
Since x2 + ax + b when divided by x-1 or x+1 leaves the same remainder
So on putting x=1 and x=-1 we get the same value
1+a+b = 1-a+b
2a=0
a=0
here b can take any value as it will always get cancelled out
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GiGiven that the polynomial (x2 + ax + b) leaves that same remainder w...
To find the value of a and b, we can use polynomial long division.
When (x^2 + ax + b) is divided by (x - c), the remainder is the same as when it is divided by (x + c).
Using polynomial long division, we have:
x + c | x^2 + ax + b
| x^2 + cx
_____________
(a - c)x + b
Since the remainder is the same, we have (a - c)x + b = (a + c)x + b.
This equation holds for all values of x, so the coefficients of x on both sides must be equal:
a - c = a + c
Simplifying, we get -c = c.
This implies that c = 0.
So, when (x^2 + ax + b) is divided by (x - 0) or (x + 0), the remainder is the same.
Using polynomial long division with c = 0, we have:
x | x^2 + ax + b
| x^2
_____________
ax + b
Since the remainder is the same, we have ax + b = ax + b.
This equation holds for all values of x, so the coefficients of x on both sides must be equal:
a = a
This implies that a can be any real number.
Therefore, the values of a and b can be any real numbers.