Solve m^2+2m(m+sqrt{2m^2})-(m+sqrt{2m^2})^2

$$m ^ { 2 } + 2 m ( m + \sqrt { 2 m ^ { 2 } } ) - ( m + \sqrt { 2 m ^ { 2 } } ) ^ { 2 }$$

Evaluate

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Solution Steps

Use the distributive property to multiply 2m by m+\sqrt{2m^{2}}.

$$m^{2}+2m^{2}+2m\sqrt{2m^{2}}-\left(m+\sqrt{2m^{2}}\right)^{2}$$

Combine m^{2} and 2m^{2} to get 3m^{2}.

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

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

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

Calculate \sqrt{2m^{2}} to the power of 2 and get 2m^{2}.

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

Combine m^{2} and 2m^{2} to get 3m^{2}.

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

Combine 3m^{2}+2m\sqrt{2m^{2}} and -\left(3m^{2}+2m\sqrt{2m^{2}}\right) to get 0.

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Factor

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