Question

Geometrically expand the polynomial: $(t^2+u^2+2v^2{)}^2$

Answer 1

Step 1: Draw a square and cut into $9$ parts.
Step 2: There are $3$ squares (red, yellow, green) and $6$ rectangles ($2$ pink, $2$ purple, $2$ blue)
Step 3: Area of the full square $=$ $(t^2+u^2+2v^2{)}^2$
Step 4: Now we have to find the area of $3$ inside square(red, yellow, green) $=$ $(t^2{)}^2+(u^2{)}^2+(2v^2{)}^2$
Step 5: Consider the area of $2$ pink rectangle $=$ length $\times$ breadth $=$ $t^2+u^2+t^2+u^2=2t^2{u}^2$
Step 6: Area of $2$ purple rectangle $=$ $2t^2{v}^2+2t^2{v}^2=4t^2{v}^2$ and Area of $2$ blue rectangle $=$ $2v^2{u}^2+2v^2{u}^2=4v^2{u}^2$
Step 7: Area of full square $=$ area of $3$ inside square $+$ area of $2$ pink rectangle $+$ area of $2$ purple rectangle $+$ area of $2$ blue rectangle.
i.e., $(t^2+u^2+2v^2{)}^2=(t^2{)}^2+(u^2{)}^2+(2v^2{)}^2+2t^2{u}^2+4t^2{v}^2+4v^2{u}^2$
Hence, geometrically we expanded the identity $(t^2+u^2+2v^2{)}^2=(t^2{)}^2+(u^2{)}^2+(2v^2{)}^2+2t^2{u}^2+4t^2{v}^2+4v^2{u}^2$.