expanding this

**johnsy123** · March 3rd 2010, 10:59 PM

i am having trouble expanding this expression.

$(y-2)(y-4)(y+1)$

can you please help?

**Bacterius** · March 3rd 2010, 11:35 PM

Hello,
you could expand it as follows :

$(y - 2)(y - 4)(y + 1)$

Let's first expand $(y - 2)(y - 4)$ . This is done by distributing :

$(y - 2)(y - 4) = y \times y + y \times (-4) + (-2) \times y + (-2) \times (-4) = y^2 - 4y - 2y + 8 = y^2 - 6y + 8$

Therefore :

$(y - 2)(y - 4)(y + 1) = (y^2 - 6y + 8)(y + 1)$

Now apply the distributive law again :

$(y^2 - 6y + 8)(y + 1) = y^2 \times y + y^2 \times 1 + (-6y) \times y + (-6y) \times 1 + 8 \times y + 8 \times 1$ $= y^3 + y^2 -6y^2 -6y + 8y + 8 = y^3 - 5y^2 + 2y + 8$

Finally :

$(y - 2)(y - 4)(y + 1) = y^3 - 5y^2 + 2y + 8$

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.

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