What is the product in simplest form? State any restrictions on the variable.

\frac{ y^2 }{ y-3 } times \frac{ y^2 - y - 6 }{ y^2 + 1y }

Question

What is the product in simplest form? State any restrictions on the variable.

 \frac{ y^2 }{ y-3 } times \frac{ y^2 - y - 6 }{ y^2 + 1y }

anonymous · Answer

$$ \frac{ y^2 }{ y-3 } \times \frac{ y^2 - y - 6 }{ y^2 + 1y }$$

ganeshie8 · Answer

factor numerator and denominator... on right side fraction

ganeshie8 · Answer

factor : $\large y^2-y-6$

find two numbers such that,
they multiply to $-6$
they add up to $-1$

whpalmer4 · Answer

Before you do any cancellation (assuming there are factors that cancel), take a look at your factored denominators and check to see what values of $y$ (if any) will cause the denominator to equal 0.  Those are your restricted variables, and they apply even for factors that may be canceled out of the final, simplest form.  It's easy to forget to check the ones that aren't present at the end!

anonymous · Answer

$$\frac{ y^2 }{ y - 3 } \times \frac{ (y + 2 )(y - 3) }{(y+1)(y-1) }$$

I think the numerator is correct but for some reason I'm stuck on the denominator because I think it's supposed to be two numbers that equal a positive one when you add and multiply it... I'm not sure >.<

anonymous · Answer

Oh wait i didn't you that you typed lol

anonymous · Answer

No... still confused /.\

ganeshie8 · Answer

yes numerator is correct. denominator is just  :
$y^2 + 1y  =  y(y+1)$

ganeshie8 · Answer

and as whpalmer is saying, before cancelling itself, u need to check the 'restricted values'

anonymous · Answer

Oooooh I always forget that that's an option >-\ 
So...
$$\frac{ y^2 }{ y-3 } \times  \frac{ (y + 2)(y-3) }{ y(y + 1) }$$

Who's Whpalmer? o. o

And I'm sorry but how do I locate restricted values as well as... cancel them out?

ganeshie8 · Answer

nope. cancelling is a different part of the story

ganeshie8 · Answer

to find restricted values of a fraction  :  look at denominator

ganeshie8 · Answer

wat values of y will make the denominator become 0 ?

ganeshie8 · Answer

if u put y = 3, denominator becomes 0 right ?

ganeshie8 · Answer

if u put y = -1, denominator becomes 0 right ?

ganeshie8 · Answer

if u put y = 0, denominator becomes 0 rihgt ?

ganeshie8 · Answer

so, $ 3, -1, 0$ are the 'restricted values' for variable $y$

ganeshie8 · Answer

if that makes any sense at all...

ganeshie8 · Answer

they're called restricted values cuz, having $0$ in the denominator is illegal in math.

ganeshie8 · Answer

basically, to find restricted values, u wud simply set the denominator to 0, and solve  for the variable.

anonymous · Answer

Why does it have to be a 3 or -1? And oops I see Whpalmer now... my fault. Sorries... I didn't read the name.

But...

the denominator on both sides or just one?

ganeshie8 · Answer

$\large    \frac{ y^2 }{\color{green}{ y-3} } \times  \frac{ (y + 2)(y-3) }{ \color{green}{y(y + 1)} } $

ganeshie8 · Answer

that whole thing is the denominator, so u need to set the whole thing to 0, and solve for y

ganeshie8 · Answer

$\large (y-3) y(y+1) = 0 $

ganeshie8 · Answer

solve y