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GiGiven that the polynomial (x2 + ax + b) leaves that same remainder when by (x – 1) or (x + 1) What are the values of a and b respectively?
  • a)
    4 and 0
  • b)
    0 and 3
  • c)
    3 and 0
  • d)
    0 and any interger
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
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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Most Upvoted Answer
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.
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GiGiven that the polynomial (x2 + ax + b) leaves that same remainder when by (x – 1) or (x + 1) What are the values of a and b respectively?a)4 and 0b)0 and 3c)3 and 0d)0 and any intergerCorrect answer is option 'D'. Can you explain this answer?
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