How do you simplify #(x-3)^2-4(x-3)+3#?

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Answer 1 · 2018-04-13T10:57:09

#(x-6)(x-4)#

#(x-3)^2-4(x-3)+3#

#((x-3)-3)((x-3)-1)#

#(x-6)(x-4)#

Answer 2 · 2018-04-16T07:47:49

Simplified gives #x^2 -10x+24#

and factorised give: #(x-6)(x-4)#

To simplify we need to multiply out the brackets:

#(x-3)^2 -4(x-3) +3#

#=x^2-6x+9-4x+12+3#

#=x^2 -10x+24#

However, I suspect that the question was meant to be to factorise..

We could multiply out as done above and then factorise from this, but let's regard #(x-3)# as a variable, #p#.

#(x-3)^2 -4(x-3) +3#

#rarr p^2 -4p +3#

#=(p-3)(p-1)#

#rarr(x-3-3)(x-3-1)#

#=(x-6)(x-4)#