Question

If p(x)=x³ -ax² +bx+3 leaves a reminder-19 when divided by (x+2) and a reminder 17 when divided by (x-2), price that a+b=6

Answer 1

To prove :- a + b = 6
Given that, when p(x) = x^3 - ax^2 + bx + 3 leaves the remainder (-19), when divided by the (x + 2).
Therefore, by applying the remainder theorem, p(-2) should be equal to -19.
So, p(-2) = x^3 - ax^2 + bx + 3 = (-19)
=> (-2)^3 - a(-2)^2 + b(-2) + 3 = (-19)
=> - 8 - 4a - 2b + 3 = (-19)
=> - 4a - 2b - 5 = (-19)
=> - 4a - 2b = (-19) + 5
=> - 4a - 2b = (-14)
=> -2 (2a + b) = (-14)
=> 2a + b = (-14)/(-2)
=> 2a + b = 7 –––––––––– (1)
Now, the other condition given is when p(x) = x^3 - ax^2 + bx + 3 when divided by (x-2) gives the remainder 17.
So, by applying the remainder theorem p(2) should be equal to 17
So, p(2) = x^3 - ax^2 + bx + 3 = 17
=> (2)^3 - a(2)^2 + b(2) + 3 = 17
=> 8 - 4a + 2b + 3 = 17
=> 11 - 4a + 2b = 17
=> - 4a + 2b = 17 - 11
=> -2 (2a - b) = 6
=> 2a - b = 6/(-2)
=> 2a - b = (-3) –––––––– (2)
Adding equations (1) and (2), we get,
2a + b + 2a - b = 7 + (-3)
=> 4a = 4
=> a = 4/4
=> a = 1
So, substituting the value of 'a' in equation (1),
2a + b = 7
=> 2(1) + b = 7
=> 2 + b = 7
=> b = 7 - 2
=> b = 5
Therefore, a + b = 1 + 5 = 6
Hence, proved.