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I assume your problem is: -2x3-2x2+12x

-2x(x2-x-6)

-2x(x-3)(x+2)

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

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

whats the answer for -2x^3+2^2+12x

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

Provided that x- 2 is not zero: 2x3 + x2 - 13x + 6 = 2x3 - 4x2 + 5x2 - 10x - 3x + 6 = 2x2*(x - 2) + 5x*(x - 2) - 3*(x - 2) = (x - 2)*(2x2 + 5x - 3) Therefore, (2x3 + x2 - 13x + 6)/(x - 2) = (x - 2)*(2x2 + 5x - 3)/(x - 2) = 2x2 + 5x - 3 If x = 2 so that x - 2 = 0, then the expression is not defined.

2x4 - 7x3 + x2 + 7x - 3 = (x + 1)(2x3 - 9x2 + 10x - 3) = (x + 1)(x - 1)(2x2 -7x + 3) = (x + 1)(x - 1)(x - 3)(2x -1)

Related questions

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

-2x(x + 3)(x - 2)

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

-2x(x + 3)(x + 2)

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

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

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

whats the answer for -2x^3+2^2+12x

-2x(x + 3)(x + 2)

-2(x^3 + x^2 + 6)

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