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

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

Evaluate

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Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{4}x-\frac{1}{2}y\right)^{2}.

$$\frac{1}{16}x^{2}-\frac{1}{4}xy+\frac{1}{4}y^{2}-\left(\frac{1}{2}y-\frac{1}{4}x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{1}{2}y-\frac{1}{4}x\right)^{2}.

$$\frac{1}{16}x^{2}-\frac{1}{4}xy+\frac{1}{4}y^{2}-\left(\frac{1}{4}y^{2}-\frac{1}{4}yx+\frac{1}{16}x^{2}\right)$$

To find the opposite of \frac{1}{4}y^{2}-\frac{1}{4}yx+\frac{1}{16}x^{2}, find the opposite of each term.

$$\frac{1}{16}x^{2}-\frac{1}{4}xy+\frac{1}{4}y^{2}-\frac{1}{4}y^{2}+\frac{1}{4}yx-\frac{1}{16}x^{2}$$

Combine \frac{1}{4}y^{2} and -\frac{1}{4}y^{2} to get 0.

$$\frac{1}{16}x^{2}-\frac{1}{4}xy+\frac{1}{4}yx-\frac{1}{16}x^{2}$$

Combine -\frac{1}{4}xy and \frac{1}{4}yx to get 0.

$$\frac{1}{16}x^{2}-\frac{1}{16}x^{2}$$

Combine \frac{1}{16}x^{2} and -\frac{1}{16}x^{2} to get 0.

$$0$$

Factor

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Solution Steps

Factor out \frac{1}{16}.

$$\frac{\left(x-2y\right)^{2}-\left(2y-x\right)^{2}}{16}$$

Consider \left(x-2y\right)^{2}-\left(2y-x\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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Consider 2x-4y. Factor out 2.

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

Rewrite the complete factored expression.

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