Solve (1/2x-1)^2-(1/2x+2)(x+3)

$$( \frac { 1 } { 2 } x - 1 ) ^ { 2 } - ( \frac { 1 } { 2 } x + 2 ) ( x + 3 )$$

Evaluate

$$-\frac{x^{2}}{4}-\frac{9x}{2}-5$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{2}x-1\right)^{2}.

$$\frac{1}{4}x^{2}-x+1-\left(\frac{1}{2}x+2\right)\left(x+3\right)$$

Use the distributive property to multiply \frac{1}{2}x+2 by x+3 and combine like terms.

$$\frac{1}{4}x^{2}-x+1-\left(\frac{1}{2}x^{2}+\frac{7}{2}x+6\right)$$

To find the opposite of \frac{1}{2}x^{2}+\frac{7}{2}x+6, find the opposite of each term.

$$\frac{1}{4}x^{2}-x+1-\frac{1}{2}x^{2}-\frac{7}{2}x-6$$

Combine \frac{1}{4}x^{2} and -\frac{1}{2}x^{2} to get -\frac{1}{4}x^{2}.

$$-\frac{1}{4}x^{2}-x+1-\frac{7}{2}x-6$$

Combine -x and -\frac{7}{2}x to get -\frac{9}{2}x.

$$-\frac{1}{4}x^{2}-\frac{9}{2}x+1-6$$

Subtract 6 from 1 to get -5.

$$-\frac{1}{4}x^{2}-\frac{9}{2}x-5$$

Expand

$$-\frac{x^{2}}{4}-\frac{9x}{2}-5$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{2}x-1\right)^{2}.

$$\frac{1}{4}x^{2}-x+1-\left(\frac{1}{2}x+2\right)\left(x+3\right)$$

Use the distributive property to multiply \frac{1}{2}x+2 by x+3 and combine like terms.

$$\frac{1}{4}x^{2}-x+1-\left(\frac{1}{2}x^{2}+\frac{7}{2}x+6\right)$$

To find the opposite of \frac{1}{2}x^{2}+\frac{7}{2}x+6, find the opposite of each term.

$$\frac{1}{4}x^{2}-x+1-\frac{1}{2}x^{2}-\frac{7}{2}x-6$$

Combine \frac{1}{4}x^{2} and -\frac{1}{2}x^{2} to get -\frac{1}{4}x^{2}.

$$-\frac{1}{4}x^{2}-x+1-\frac{7}{2}x-6$$

Combine -x and -\frac{7}{2}x to get -\frac{9}{2}x.

$$-\frac{1}{4}x^{2}-\frac{9}{2}x+1-6$$

Subtract 6 from 1 to get -5.

$$-\frac{1}{4}x^{2}-\frac{9}{2}x-5$$