Solve (x+1)(x-1)-(x-2)^2+4x-1/4

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

Evaluate

$$8x-\frac{21}{4}$$

Solution Steps

Consider \left(x+1\right)\left(x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

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

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

To find the opposite of x^{2}-4x+4, find the opposite of each term.

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

Combine x^{2} and -x^{2} to get 0.

$$-1+4x-4+4x-\frac{1}{4}$$

Subtract 4 from -1 to get -5.

$$-5+4x+4x-\frac{1}{4}$$

Combine 4x and 4x to get 8x.

$$-5+8x-\frac{1}{4}$$

Subtract \frac{1}{4} from -5 to get -\frac{21}{4}.

$$-\frac{21}{4}+8x$$

Expand

$$8x-\frac{21}{4}$$

Solution Steps

Consider \left(x+1\right)\left(x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

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

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

To find the opposite of x^{2}-4x+4, find the opposite of each term.

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

Combine x^{2} and -x^{2} to get 0.

$$-1+4x-4+4x-\frac{1}{4}$$

Subtract 4 from -1 to get -5.

$$-5+4x+4x-\frac{1}{4}$$

Combine 4x and 4x to get 8x.

$$-5+8x-\frac{1}{4}$$

Subtract \frac{1}{4} from -5 to get -\frac{21}{4}.

$$-\frac{21}{4}+8x$$