Question

in the expression x3+ax2+bx-6 is completely divisible by x2-x-2 find a and b

Answer 1

Let, f(x) = x³ + ax² + bx - 6

And,

x² - x - 2 = 0

=> x² -2x + x - 2 = 0

=> x(x - 2) + 1(x - 2) = 0

=> (x - 2)(x + 1) = 0

According to remainder theorem, if f(x) is completely divisible by x² - x - 2, then f(x) is also divisible by (x - 2) and (x + 1).

And hence,

f(2) = 0 and f(-1) = 0

So,

8 + 4a + 2b - 6 = 0 ----- (1)

-1 + a - b - 6 = 0 ---------- (2)

On solving (1) and (2), we get

a = (2)

b = (-5)

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Answer 2

Answer:

∴r(x)= 0

synthetic division:

remainders

a+b+3=0         and 2a-4=0

∴a=2

b=-5