Q: How do you solve -x2 plus 2x-3?

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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 + 40Râˆ´ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)

2x3 + 16 = 2(x3 + 8) = 2(x3 + 23) = 2(x + 2)(x2 - 2x + 22) = 2(x + 2)(x2 - 2x + 4)

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

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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 + 40Râˆ´ x4 - 2x3 + 2x2 + x + 4 = (x2 + x + 1)(x2 - 3x + 4)

x2 • (5x2 + x + 8)

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

x2+7x+12

(3x4 + 2x3 - x2 - x - 6)/(x2 + 1)= 3x2 + 2x - 4 + (-3x - 2)/(x2 + 1)= 3x2 + 2x - 4 - (3x + 2)/(x2 + 1)where the quotient is 3x2 + 2x - 4 and the remainder is -(3x + 2).

2x3 + 16 = 2(x3 + 8) = 2(x3 + 23) = 2(x + 2)(x2 - 2x + 22) = 2(x + 2)(x2 - 2x + 4)

If that's 2x2, the answer is (x + 2)(x2 + 4)

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

x2 × 2x = 2x3

-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)