Question #23cd8

What is shown is the same as the FOIL method but just looks different.

Let do this in stages.

First consider the two left most brackets

#color(blue)((n-3)) color(green)( (n-4) )#

Multiply everything in the right brackets by everything in the left

#color(green)(color(blue)(n)(n-4) color(white)("ddd")color(blue)(-3)(n-4))" "# Notice the minus followed the 3

#n^2-4ncolor(white)("ddd") -3n+12#

#n^2-7n+12#

Putting it all back together with the last bracket we now have:

#color(blue)((n^2-7n+12))color(green)( (n-5) )#

As per the method used before:

#color(green)( color(blue)(n^2) (n-5) color(white)("dddd")color(blue)(-7n) (n-5) color(white)("dddd") color(blue)(+12)(n-5))#

#n^3color(white)("d") -5n^2 color(white)("ddd")-7n^2color(white)("dd")+35n color(white)("d") +12ncolor(white)("d")-60#

#n^3color(white)("dddd")-12n^2color(white)(ddddddddd"d")+47ncolor(white)("dddd")-60#

#color(white)()#

#n^3-12n^2+47n-60#

#"here are 2 approaches to expanding the factors"#

#color(blue)"Approach 1"#

#"expand any pair of factors then multiply by the third factor"#

#"expanding "(n-3)(n-4)" using the FOIL method"#

#(n-3)(n-4)=n^2-4n-3n+12=n^2-7n+12#

#rArr(n-3)(n-4)(n-5)#

#=(color(red)(n-5))(n^2-7n+12)#

#=color(red)(n)(n^2-7n+12)color(red)(-5)(n^2-7n+12)#

#"distribute both sets of brackets"#

#=n^3-7n^2+12n-5n^2+35n-60#

#"collect like terms"#

#=n^3-12n^2+47n-60#

#color(blue)"Approach 2"#

#"given "(n+a)(n+b)(n+c)" then expansion is of form"#

#n^3+(a+b+c)n^2+(ab+bc+ac)n+abc#

#rArr(n-3)(n-4)(n-5)#

#=n^3+(-3-4-5)n^2+(12+20+15)n#

#color(white)(=)+(-3)(-4)(-5)#

#=n^3-12n^2+47n-60#