# Polynomials - prove that (2x⁴-6x³+3x³+3x-2) is exactly divisible

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.)

### Comments for Polynomials - prove that (2x⁴-6x³+3x³+3x-2) is exactly divisible

 Sep 12, 2011 Factor Polynomial by: Staff ----------------------------------------------- Part II Is x = +½ a rational root ? f(x) = 2x^4 - 3x^3 + 3x - 2 f(x) = 2*(½)^4 - 3*(½)^3 + 3*(½) - 2 f(x) = 2*(1/16) - 3*(1/8) + 3*(½) - 2 f(x) = 2/16 – 3/8 + 3/2 - 2 f(x) = -3/4 NO, x = +½ is NOT a root Is x = -½ a rational root ? f(x) = 2x^4 - 3x^3 + 3x - 2 f(x) = 2*(-½)^4 - 3*(-½)^3 + 3*(-½) - 2 f(x) = 2*(1/16) - 3*(-1/8) + 3*(-½) - 2 f(x) = 2/16 + 3/8 - 3/2 - 2 f(x) = -3 NO, x = -½ is NOT a root Is x = +2 a rational root ? f(x) = 2x^4 - 3x^3 + 3x - 2 f(x) = 2*2^4 - 3*2^3 + 3*2 - 2 f(x) = 2*16 - 3*8 + 3*2 - 2 f(x) = 32 - 24 + 6 - 2 f(x) = 12 NO, x = +2 is NOT a root Is x = -2 a rational root ? f(x) = 2x^4 - 3x^3 + 3x - 2 f(x) = 2*(-2)^4 - 3*(-2)^3 + 3*(-2) - 2 f(x) = 2*(16) - 3*(-8) + 3*(-2) - 2 f(x) = 32 + 24 - 6 - 2 f(x) = 48 NO, x = -2 is NOT a root Two rational roots have been identified rational roots = {+1, -1} original polynomial 2x^4 - 3x^3 + 3x - 2 Factored (x - 1) * (x + 1) * (another factor) Is another factor = x^2 - 3x + 2 ? (x^2 - 3x + 2 is the factor listed in original problem statement) Ordinarily, you would divide to obtain the expression for “another factor”. However, the problem states that division is not to be used to solve this problem. This does not prevent you from using multiplication. If you substitute x^2 - 3x + 2 for “another factor”, and then multiply, you can see if the result is the original polynomial which appears in the problem statement. f(x) = (x - 1) * (x + 1) * (x^2 - 3x + 2 ) f(x) = (x - 1) * (x + 1) * (x - 2 )* (x - 1) Obviously x^2 - 3x + 2 is not a valid factor since its factored form is (x -2 )* (x - 1). We have already established that neither x = -2 or x = +2 are not valid rational roots of: f(x) = 2x^4 - 6x^3 + 3x^3 + 3x-2 Therefore, (2x^4-6x^3+3x^3+3x-2) is NOT exactly DIVISIBLE by (x^2-3x+2). Thanks for writing. Staff www.solving-math-problems.com

 Sep 12, 2011 Factor Polynomial by: Staff Part IThe question: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 - 2Combine like terms to simplify2x^4 - 3x^3 + 3x - 2Factor the polynomial2x^4 - 3x^3 + 3x - 2Use the "RATIONAL ROOT TEST" to find any possible rational roots.(a_n)x^n + (a_n-1)x^(n-1) + . . . + (a_1)x + a_01. find a_n and a_0a_n (the first coefficient) = 2a_0 (the constant) = 22. Determine factors of a_n and a_0Factors of a_n = 1, 2Factors of a_0 = 1, 23. 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) = 0Is x = +1 a rational root ?f(x) = 2x^4 - 3x^3 + 3x - 2 f(x) = 2*1^4 - 3*1^3 + 3*1 - 2f(x) = 2 - 3 + 3 - 2f(x) = 0YES, x = +1 is a ROOTIs 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) = 0YES, x = -1 is a ROOT-----------------------------------------------

 Sep 17, 2011 hey by: the guy who said the question it is divisible but i can't prove it

 Sep 17, 2011 Polynomial Division by: Staff Clarification 2x^4 - 6x^3 + 3x^3 + 3x – 2 Is DIVISIBLE by (2x² - 3x + 2) It is NOT DIVISIBLE by (x² - 3x + 2). (Apparently there is a typographical error in the original question which was submitted. The problem submitted did explicitly state that (x² - 3x + 2) is valid. It is not. The coefficient of 2 was not included with x² as it should have been.) 2x^4 - 6x^3 + 3x^3 + 3x – 2 = ( x - 1 )( x + 1 )( 2x² - 3x + 2 ) Thanks for writing. Staff www.solving-math-problems.com