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2x3 - 2x2 - x plus 1?

Updated: 12/13/2022
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13y ago

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2x3 - 2x2 - x + 1 = 2x2(x -1) - (x - 1) = (x - 1)(2x2 -1) = (x - 1){[(sq. root of 2)(x)]2 -12} = (x - 1)[(sq. root of 2)(x) - 1][(sq. root of 2)(x) + 1]

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(x4 - 2x3 + 2x2 + x + 4) / (x2 + x + 1)You can work this out using long division:x2 - 3x + 4___________________________x2 + x + 1 ) x4 - 2x3 + 2x2 + x + 4x4 + x3 + x2-3x3 + x2 + x-3x3 - 3x2 - 3x4x2 + 4x + 44x2 + 4x + 40R∴ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)


How do you factor 2x cubed plus 2 x squared?

2x3 + 2x2 = 2x2 * (x + 1)


What is the factorization of the trinomial -2x3 plus 2x2 plus 12x?

-2x3 + 2x2 + 12x =(-2x) (x2 - x - 6) =(-2x) (x+2) (x-3)


2x3 plus 2x2 - 12x?

Do you want its factorisation? 2x3 + 2x2 - 12x = 2x(x2 + x - 6) = 2x(x + 3)(x - 2).


What is the factorization of the polynomial below -2x3-2x2 plus 12x?

(-2x3 - 2x2 + 12x) = -2x (x2 + x - 6) = -2x (x + 3) (x - 2)


Is x-3 a factor of 2x3-11x2 plus 12x plus 9?

Yes. 2x3 - 11x2 + 12x + 9 = (x - 3)(2x2 - 5x - 3) = (x - 3)(2x2 - 6x + x - 3) = (x - 3)(2x + 1)(x - 3) = (2x + 1)(x - 3)2


X plus 1 divided by 2x cubed plus 3x plus 5?

(x + 1) / (2x3 + 3x + 5) --- Can't depict long division here, but you can work this out by seeing if (x + 1) is a factor of (2x3 + 3x + 5). If it is, then the answer will be 1/(the other factor). In this case, that is true. (2x3 + 3x + 5)/(x + 1) = 2x2 - 2x + 5, so the term term above is equal to: 1 / (2x2 - 2x + 5)


What is the factorization of this trinomial -2x3 plus 2x2 plus 12x?

-2x3 + 2x2 + 12x = -2x(x2 - x - 6) = -2x(x2 + 2x - 3x - 6) = -2x[ x(x + 2) - 3(x + 2) ] = -2x(x - 3)(x + 2)


2x3 plus 2x2 - 12?

(2 x 3) + (2 x 2) - 12 = -2


What is the factorization of -2x3 - 2x2 plus 12x?

-2x3 - 2x2 + 12x = -2x(x2 + x - 6) = -2x(x2 - 2x + 3x - 6) = -2x[ x(x - 2) + 3(x - 2) ] = -2x(x + 3)(x - 2)


What is 2x3 plus 11x equals 6x?

your equation is this... 2x3 + 11x = 6x 2x3 + 5x = 0 x(2x2 + 5) = 0 x = 0 and (5/2)i and -(5/2)i


Factor 2x3 plus 2x2 - 12x?

2x( x2 + x - 6) = 2x (x + 3)(x - 2)