How do you find the product #(y+4)(y-8)#?

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Answer 1 · 2016-12-01T12:09:00

#y=-4 or 8#

#(y+4)(y-8)#

#y^2-8y+4y-32#

This is a quadratic equation (because of the power of 2), so we will have to use the quadratic formula which is:

#ax+by+c=0#

So

#y^2-4y-32 = 0#

So basically, now we know the quadratic equation, we can say:

#(y+4)(y-8) = 0#

Possible Answers are:

#y=-4#

#y=8#

because if one or the other equals to 0, then multiplying it with any number makes it 0.

Answer 2 · 2016-12-01T14:03:33

The question only askes for the product which is:

#y^2-4y-32#

#color(blue)((y+4))color(green)((y-8))#

Multiply everything in the right hand side bracket by everything in the left one.

#color(blue)(ycolor(green)((y-8))" "+" "4color(green)((y-8))#

#y^2-8y" "+" "4y-32#

#y^2-4y-32#