(2x ^ { 3 } -4x ^ { 2 } -6x+8) \div (x+1)

asked by guest
on Mar 26, 2025 at 8:24 pm



You asked:

Evaluate the expression: 2x3−4x2−6x+8x+1\frac{2 {x}^{3} - 4 {x}^{2} - 6 x + 8}{x + 1}

MathBot Answer:

Evaluated



2x3−4x2−6x+8x+1=2x3−4x2−6x+8x+1\displaystyle \frac{2 {x}^{3} - 4 {x}^{2} - 6 x + 8}{x + 1} = \frac{2 x^{3} - 4 x^{2} - 6 x + 8}{x + 1}

Expanded

2x3−4x2−6x+8x+1=2x3x+1−4x2x+1−6xx+1+8x+1\frac{2 {x}^{3} - 4 {x}^{2} - 6 x + 8}{x + 1} = \frac{2 x^{3}}{x + 1} - \frac{4 x^{2}}{x + 1} - \frac{6 x}{x + 1} + \frac{8}{x + 1}

Factored

2x3−4x2−6x+8x+1=2(x−1)(x2−x−4)x+1\frac{2 {x}^{3} - 4 {x}^{2} - 6 x + 8}{x + 1} = \frac{2 \left(x - 1\right) \left(x^{2} - x - 4\right)}{x + 1}

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