Question

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

Answer 1

heya......


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)


tysm......@kundan

Answer 2

Hay!!

Dear friend -

without actual division prove that (2x^4-6x^3+3x^2+3x-2) is exactly divisible by (x³-3x+2)

Here is ur answer

$letf(x) = 2x {}^{4} - 6x {}^{3} + 3 {x}^{2} + 3x + 2$

$letg(x) = ( {x}^{2} - 3x + 2) \\ = {x}^{2} - 2x - x + 2 \\ = x(x - 2) - (x - 2) \\ = (x - 2)(x - 1)$

Now, f(x) will be exactly divisible by g(x) if it is 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