i am having trouble expanding this expression.
$\displaystyle (y-2)(y-4)(y+1)$
can you please help?
Hello,
you could expand it as follows :
$\displaystyle (y - 2)(y - 4)(y + 1)$
Let's first expand $\displaystyle (y - 2)(y - 4)$. This is done by distributing :
$\displaystyle (y - 2)(y - 4) = y \times y + y \times (-4) + (-2) \times y + (-2) \times (-4) = y^2 - 4y - 2y + 8 = y^2 - 6y + 8$
Therefore :
$\displaystyle (y - 2)(y - 4)(y + 1) = (y^2 - 6y + 8)(y + 1)$
Now apply the distributive law again :
$\displaystyle (y^2 - 6y + 8)(y + 1) = y^2 \times y + y^2 \times 1 + (-6y) \times y + (-6y) \times 1 + 8 \times y + 8 \times 1$ $\displaystyle = y^3 + y^2 -6y^2 -6y + 8y + 8 = y^3 - 5y^2 + 2y + 8$
Finally :
$\displaystyle (y - 2)(y - 4)(y + 1) = y^3 - 5y^2 + 2y + 8$
Does this help ?
PS : since I'm in a good mood I'll give you a nice website to check your results (if you don't know it already). It is called Wolfram Alpha, and you can type in what you want it to do (like "expand/factor/simplify [expression]", "solve [equation]"), and it will most likely do it correctly.