Question

Without actual division prove that 2x⁴-6x³+3x²+3x-2 is exactly divisible by x²-3x+2

Answer 1

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Let f(x) = x4 + 2x3 -2x2 + 2x - 3

g(x) = x2 + 2x - 3

= x(x + 3) - 1(x + 3) = (x - 1) (x + 3)

Now f(x) will be exactly divisible by g(x) if it is exactly divisible by (x - 1) as well as (x + 3)

i.e. if f(1) = 0 and f ( -3) = 0

Now f(1) = 14 + 2.13 -2.12 + 2.1 - 3

= 1 + 2 - 2 + 2 - 3 = 0

=> (x - 1) is a factor of f(x)

f ( -3) = (-3)4 + 2.(-3)3 -2.(-3)2 + 2.(-3) - 3

= 81 - 54 - 18 - 6 - 3 = 0

=> (x + 3) is a factor of f(x).

=> (x - 1) (x + 3) divides f (x) exactly

Therefore, x2 + 2x - 3 is a factor of f(x)

Answer 2

Given f(x) = 2x^4 - 6x^3 + 3x^2 + 3x - 2

x^2 - 3x + 2 = x^2 - x - 2x + 2

                    = x(x - 1) -2(x - 1)

                    = (x - 1)(x - 2)


If (x - 1) and (x - 2) are the factors of f(x).Then f(x) is divisible by x^2 - 3x + 2.

if f(1) = 0 and f(2) = 0, then f(x) is exactly divisible by x^2 - 3x + 2.

f(1) = 2(1)^4 - 6(1)^3 + 3(1)^2 + 3(1) - 2

      = 2 - 6 + 3 + 3 - 2

     = 0.   -------- (1)


f(2) = 2(2)^4 - 6(2)^3 + 3(2)^2 + 3(2) - 2

      = 2(16) - 6(8) + 3(4) + 6 - 2

      = 32 - 48 + 12 + 6 - 2

      = 0.    ------- (2)


From (1) & (2), we get

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


Therefore f(x) is exactly divisible by x^2 - 3x + 2.


Hope this helps!