Solve -2x^2/3-8x/3+10/3

$$-2 \frac{ { x }^{ 2 } }{ 3 } -8 \frac{ x }{ 3 } + \frac{ 10 }{ 3 }$$

Evaluate

$$\frac{2\left(1-x\right)\left(x+5\right)}{3}$$

Short Solution Steps

Express -2\times \frac{x^{2}}{3} as a single fraction.

$$\frac{-2x^{2}}{3}-8\times \frac{x}{3}+\frac{10}{3}$$

Express 8\times \frac{x}{3} as a single fraction.

$$\frac{-2x^{2}}{3}-\frac{8x}{3}+\frac{10}{3}$$

Since \frac{-2x^{2}}{3} and \frac{8x}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-2x^{2}-8x}{3}+\frac{10}{3}$$

Since \frac{-2x^{2}-8x}{3} and \frac{10}{3} have the same denominator, add them by adding their numerators.

$$\frac{-2x^{2}-8x+10}{3}$$

Factor

$$\frac{2\left(1-x\right)\left(x+5\right)}{3}$$

Solution Steps

Factor out \frac{2}{3}.

$$\frac{2\left(-x^{2}-4x+5\right)}{3}$$

Consider -x^{2}-4x+5. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-4 ab=-5=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-5$$

Rewrite -x^{2}-4x+5 as \left(-x^{2}+x\right)+\left(-5x+5\right).

$$\left(-x^{2}+x\right)+\left(-5x+5\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(-x+1\right)+5\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+5\right)$$

Rewrite the complete factored expression.

$$\frac{2\left(-x+1\right)\left(x+5\right)}{3}$$