Question

If x³+ax²+bx+6 has x-2 as a factor and leaves a remainder 3 when devided by x-3,find the values of a and b

Answer 1

Given Equation is f(x) = x^3 + ax^2 + bx + 6.

Given that x - 2 is a factor of f(x).

By the remainder theorem, we know that

x - 2 = 0

x = 2.


Substitute x = 2 in f(x), we get

f(2) = (2)^3 + a(2)^2 + b(2) + 6

       = 8 + 4a + 2b + 6 = 0

        = 14 + 4a + 2b = 0

       = 4a + 2b = -14      ------------------------------ (1)


Now,


Given that x - 3 is also a factor of f(x) and the remainder is 3. 

f(3) = (3)^3 + a(3)^2 + b(3) + 6

3 = 27 + 9a + 3b + 6

3 = 9a + 3b + 33

9a + 3b + 30 = 0

9a + 3b = -30    ---------- (2)


On solving (1) * 9, (2) * 4, we get

36a + 18b = -126

36a + 12b = - 120

------------------------

          6b = -6

            b = -1.



Substitute b = -1 in above equations, we get

36a + 18b = -126

36a + 18(-1) = -126

36a - 18 = -126

36a = -108

a = -3.



Therefore the values of a = -3 and b = -1.



Hope this helps!

Answer 2

a = -3 b = -1

☺Hope this helps☺