Solve -frac{sqrt{3}}{4}x^2+frac{sqrt{3}}{2}x+2sqrt{3}

$$- \frac { \sqrt { 3 } } { 4 } x ^ { 2 } + \frac { \sqrt { 3 } } { 2 } x + 2 \sqrt { 3 }$$

Evaluate

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

Short Solution Steps

Express \left(-\frac{\sqrt{3}}{4}\right)x^{2} as a single fraction.

$$\frac{-\sqrt{3}x^{2}}{4}+\frac{\sqrt{3}}{2}x+2\sqrt{3}$$

Express \frac{\sqrt{3}}{2}x as a single fraction.

$$\frac{-\sqrt{3}x^{2}}{4}+\frac{\sqrt{3}x}{2}+2\sqrt{3}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2 is 4. Multiply \frac{\sqrt{3}x}{2} times \frac{2}{2}.

$$\frac{-\sqrt{3}x^{2}}{4}+\frac{2\sqrt{3}x}{4}+2\sqrt{3}$$

Since \frac{-\sqrt{3}x^{2}}{4} and \frac{2\sqrt{3}x}{4} have the same denominator, add them by adding their numerators.

$$\frac{-\sqrt{3}x^{2}+2\sqrt{3}x}{4}+2\sqrt{3}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2\sqrt{3} times \frac{4}{4}.

$$\frac{-\sqrt{3}x^{2}+2\sqrt{3}x}{4}+\frac{4\times 2\sqrt{3}}{4}$$

Since \frac{-\sqrt{3}x^{2}+2\sqrt{3}x}{4} and \frac{4\times 2\sqrt{3}}{4} have the same denominator, add them by adding their numerators.

$$\frac{-\sqrt{3}x^{2}+2\sqrt{3}x+4\times 2\sqrt{3}}{4}$$

Do the multiplications in -\sqrt{3}x^{2}+2\sqrt{3}x+4\times 2\sqrt{3}.

$$\frac{-\sqrt{3}x^{2}+2\sqrt{3}x+8\sqrt{3}}{4}$$

Factor

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

Solution Steps

Factor out \frac{1}{4}.

$$\frac{-\sqrt{3}x^{2}+2\sqrt{3}x+8\sqrt{3}}{4}$$

Consider -\sqrt{3}x^{2}+2\sqrt{3}x+8\sqrt{3}. Factor out \sqrt{3}.

$$\sqrt{3}\left(-x^{2}+2x+8\right)$$

Consider -x^{2}+2x+8. Factor the expression by grouping. First, the expression needs to be rewritten as -x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=2 ab=-8=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 2.

$$a=4 b=-2$$

Rewrite -x^{2}+2x+8 as \left(-x^{2}+4x\right)+\left(-2x+8\right).

$$\left(-x^{2}+4x\right)+\left(-2x+8\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(x-4\right)-2\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x-2\right)$$

Rewrite the complete factored expression.

$$\frac{\sqrt{3}\left(x-4\right)\left(-x-2\right)}{4}$$