without actual division.prove that 2x^4-6x^3+3x^2+3x-2 is exactly divisible by x^2-3x+2
Answers
Step-by-step explanation:
Let f(x) = 2x⁴ - 6x³ + 3x² + 3x - 2.
Let g(x) = x² - 3x + 2
= x² - 2x - x + 2
= x(x - 2) - 1(x - 2)
= (x - 1)(x - 2)
∴ If (x - 1) and (x - 2) are the factors of p(x), then
f(x) is divisible by x² - 3x + 2.
⇒ If f(1) and f(2) = 0, then f(x) is exactly divisible by x² - 3x + 2.
∴ f(1) = 2(1)⁴ - 6(1)³ + 3(1)² + 3(1) - 2
= 2 - 6 + 3 + 3 - 2
= 0
∴ f(2) = 2(2)⁴ - 6(2)³ + 3(2)² + 3(2) - 2
= 32 - 48 + 12 + 6 - 2
= 0
⇒ (x - 1) and (x - 2) are the factors of f(x).
∴ g(x) = (x - 1) and (x - 2) is a factor of f(x).
Therefore, f(x) is exactly divisible by g(x).
Hope it helps!
The plan is find the roots of x^3 -3x +2. Say r is a root of this.
Then check that r is a root of quartic by substitution.
Now x^3 -3x +2 can be solved by either factoring or guessing (trial and error).
You check a few small integers and you find that 1, -2, +2 are roots.
So you need only plug those values into the quartic and check that the answer is 0.