Question

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hay user ----------------------------------------------------------------------------
Without actual division, prove that (2x^4 - 6x³ +3x²+3x -2) is exactly divisible by (x²-3x+2)

don't give me copied answers
best of luck :-D

Answer 1

hay!!

let f(x) = 2x^4 - 6x³ +3x²+3x-2
let g(x)= (x²-3x+2)
=> x²-2x-x+2
=> x(x-2) -(x-2) =(x-2) (x-1)

Now, f(x) will be exactly divisible by g(x) if it's exactly divisible by (x-2) as well as (x-1).

For this we must have f(2)=0 and f(1)=0.

Now, f(2) => (2×2^4-6×2³+3×2²+3×2-2)

=> (32-48+12+6-2)=0

and, f(1) => (2×1^4-6×1³+3×1²+3×1-2)

=> (2-6+3+3-2)=0

f(x) is exactly divisible by (x-2)as well as (x-1)
so, f(x) is exactly divisible by (x-2)(x-1)

Hence,f(x) is exactly divisible by (x²-3x+2).

I hope it's help you

Answer 2

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

Given g(x) = x^2 - 3x + 2.

Now,

g(x) = x^2 - 3x + 2

       = x^2 - 2x - x + 2 = 0

       = (x - 1)(x - 2) = 0

       x = 1 (or) x = 2.

Now,

We have to show that x = 1 and x = 2 are the factors of f(x).

When x = 1:

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

     = 2 - 6 + 3 + 3 - 2

     = - 4 + 6 - 2

     = 0.   ------------------------ (1)


When x = 2:

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

      = 32 - 48 + 12 + 6 - 2

      = -16 + 12 + 6 - 2

      = 2 - 2

      = 0.     ------------------ (2)


From (1) & (2), we can say that f(x) is exactly divisible by g(x).
     

Hope this helps!