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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

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Sep 12, 2011
Factor Polynomial
by: Staff


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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 I

The 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 - 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

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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


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