what do you mean by "solve"? It's not an equation.
You can expand it out, but the only thing that drops out are the 4t^4 and v^2 terms.
It doesn't factor to anything simple.
Here is a doozy:
Solve: ((-2w^2+v)^2)+4t^4-(-1+v-2t^2)^2
3 answers
Oh sorry, meant simplify. It's a super strange one I guess but that's what they've given me and I was fairly confused.
( - 2 w² + v )² + 4 t⁴ - ( - 1 + v - 2 t² )² =
( - 2 w² )² + 2 ∙ ( - 2 w ) ∙ v + v ² + 4 t⁴ - ( - 1 + v - 2 t² ) ∙ ( - 1 + v - 2 t² ) =
4 w⁴ - 4 w ∙ v + v ² + 4 t⁴ - [ - 1 ∙ ( - 1 ) + ( - 1 ) ∙ v - ( - 1 ) ∙ ( 2 t² ) + v ∙ ( - 1 ) + v ∙ v + v ∙ ( - 2 t² ) - ( - 2 t² ) ∙ 1 + ( - 2 t² ) ∙ v - ( - 2 t² ) ∙ 2 t² ] =
4 w⁴ - 4 w ∙ v + v ² + 4 t⁴ - ( 1 - v + 2 t² - v + v² - 2 t² ∙ v + 2 t² - 2 t² ∙ v + 4 t⁴ ) =
4 w⁴ - 4 w ∙ v + v ² + 4 t⁴ - 1 + v - 2 t² + v - v² + 2 t² ∙ v - 2 t² + 2 t² ∙ v - 4 t⁴ =
4 w⁴ - 4 w ∙ v + 2 v - 4 t² + 2 t² ∙ v + 2 t² ∙ v - 1 =
2 ( 2 w⁴ - 2 w ∙ v + v - 2 t² + t² ∙ v + t² ∙ v ) - 1
( - 2 w² )² + 2 ∙ ( - 2 w ) ∙ v + v ² + 4 t⁴ - ( - 1 + v - 2 t² ) ∙ ( - 1 + v - 2 t² ) =
4 w⁴ - 4 w ∙ v + v ² + 4 t⁴ - [ - 1 ∙ ( - 1 ) + ( - 1 ) ∙ v - ( - 1 ) ∙ ( 2 t² ) + v ∙ ( - 1 ) + v ∙ v + v ∙ ( - 2 t² ) - ( - 2 t² ) ∙ 1 + ( - 2 t² ) ∙ v - ( - 2 t² ) ∙ 2 t² ] =
4 w⁴ - 4 w ∙ v + v ² + 4 t⁴ - ( 1 - v + 2 t² - v + v² - 2 t² ∙ v + 2 t² - 2 t² ∙ v + 4 t⁴ ) =
4 w⁴ - 4 w ∙ v + v ² + 4 t⁴ - 1 + v - 2 t² + v - v² + 2 t² ∙ v - 2 t² + 2 t² ∙ v - 4 t⁴ =
4 w⁴ - 4 w ∙ v + 2 v - 4 t² + 2 t² ∙ v + 2 t² ∙ v - 1 =
2 ( 2 w⁴ - 2 w ∙ v + v - 2 t² + t² ∙ v + t² ∙ v ) - 1
1 answer