Question

Find the area of the rectangle with a length of (x^2-2) and a width of (2x^2-x+2)

Answer 1

$A = L*w = (x^2-2)(2x^2-x+2)=2x^4-x^3+2x^2-4x^2+2x-4$
$A =2x^4-x^3-2x^2+2x-4$

So the area is : $A =2x^4-x^3-2x^2+2x-4$

Answer 2

Area=legnth times width

so multiply them together use distributive property
a(b+c)=ab+ac so
in this problem
(a+b)(c+d+e)=(a+b)(c)+(a+b)(d)+(a+b)(e)

so
x^2-2 times (2x^2-x+2)=(x^2)(2x^2-x+2)-(2)(2x^2-x+2)=(2x^4-x^3+2x^2)-(4x^2-2x+4)
add like terms
2x^4-x^3+(2x^2-4x^2)-2x+4
2x^4-x^3-2x^2-2x+4