A rectangle has an area represented by x^2-4x-12 square feet. If the length is x+2 feet, what is the width of the rectangle?

Question

anonymous · Answer

The area of a rectangle is defined as the product of its length and width, $A=lw$. It should make sense that we can factor our area polynomial into the length and width, just as you can factor a rectangle with area five into its dimensions 5x1.

To factor $A=x^2-4x-12$ we find factors of -12 which sum to -4... i.e. -6 and 2. Our linear factors must turn to 0 for these values of $x$. we can factor A into $(x+6)(-2)$. If $x-2$ is the length then the width must be $x+6$.

anonymous · Answer

(x+6)(x-2)... sorry, I'm on an iPad.

anonymous · Answer

Omg thanks ;-;

anonymous · Answer

But wouldn't that be the other way around since 2 times -6 is -4?   x+2  and x-6

anonymous · Answer

No; to make $(x-2)(x+6)=0$, we need to make either $x-2=0$ or $x+6=0$. Solve both for $x$ to yield $x=2$ and $x=-6$.