Answer:

Step-by-step explanation:

without actual division, prove that (2x^4-6x^3+3x^3+3x-2)is exactly DIVISIBLE by(x^2-3x+2).

(note : here'^'this SYMBOL is being referred as raised to.)

The answer:  

2x^4 - 6x^3 + 3x^3 + 3x - 2

Combine like terms to simplify

2x^4 - 3x^3 + 3x - 2

Factor the polynomial

2x^4 - 3x^3 + 3x - 2

Use the "RATIONAL ROOT TEST" to find any POSSIBLE rational roots.

(a_n)x^n + (a_n-1)x^(n-1) + . . . + (a_1)x + a_0

1. find a_n and a_0

a_n (the first coefficient) = 2

a_0 (the CONSTANT) = 2

2. Determine factors of a_n and a_0

Factors of a_n = 1, 2

Factors of a_0 = 1, 2

3. Determine possible rational roots

(prelim) Possible rational roots = ±{1/1, ½, 2/1, 2/2}

Eliminate duplicates

(final) Possible rational roots = ±{1, ½, 2}

4. Determine if any of the possible rational roots are actually roots where f(x) = 0

Is x = +1 a rational root ?

f(x) = 2x^4 - 3x^3 + 3x - 2  

f(x) = 2*1^4 - 3*1^3 + 3*1 - 2

f(x) = 2 - 3 + 3 - 2

f(x) = 0

YES, x = +1 is a ROOT

Is x = -1 a rational root ?

f(x) = 2x^4 - 3x^3 + 3x - 2  

f(x) = 2*(-1)^4 - 3*(-1)^3 + 3*(-1) - 2  

f(x) = 2*1 - 3*(-1) + 3*(-1) - 2  

f(x) = 2 + 3 - 3 - 2  

f(x) = 0

YES, x = -1 is a ROOT