How do you find the product of #(n-4)(n-6)#?

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Answer 1 · 2016-07-18T13:27:12

The product is #n^2 - 10n + 24#

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In order to multiply binomials we use the acronym FOIL

F - Firsts - Multiply the first terms
O - Outers - Multiply the outer terms
I - Inners - Multiply the Inner terms
L - Lasts - Multiply the Last terms

#(n)(n) + (n)(-6) + (-4)(n) + (-4)(-6)#

#n^2 - 6n -4n + 24#

Add like terms

#n^2 - 10n + 24#

Answer 2 · 2016-07-18T13:42:57

#n^2-10n+24#

Given:#" "color(blue)((n-4))color(brown)((n-6))#

Everything inside the brown bracket is multiplied by everything inside the blue bracket.

The 'minus' follows the blue 4

#color(brown)(color(blue)(n)(n-6)color(blue)(-4)(n-6) )#

#n^2-6n color(white)(...)-color(white)(...)4n+24#

#n^2-10n+24#