Question

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

Answer 1

Answer:

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!

Answer 2

Solution :

Let

• f(x) = 2x⁴ - 6x³ + 3x² + 3x - 2
• g(x) = x² - 3x + 2

Now, g(x) = x² - 3x + 2 = x² - 2x - x + 2

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

= (x - 1) (x - 2)

Hence, x - 1 and x - 2 are factors of g(x).

Now, put x = 1 and x = 2 in f(x) :

For x = 1 :

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

= 2 × 1 - 6 × 1 + 3 × 1 + 3 × 1 - 2

= 2 - 6 + 3 + 3 - 2

= - 4 + 6 - 2

= 2 - 2

= 0

For x = 2 :

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

= 2 × 16 - 6 × 8 + 3 × 4 + 3 × 2 - 2

= 32 - 48 + 12 + 6 - 2

= - 16 + 18 - 2

= 2 - 2

= 0

As, f(x) is divisible by both the factors of g(x), i.e., x - 1 and x - 2.

So, f(x) is divisible by (x - 1)(x - 2) which is equal to x² - 3x + 2 = g(x).

Hence, 2x⁴ - 6x³ + 3x² + 3x - 2 is exactly divisible by x² - 3x + 2.