How do you multiply #(x-9)^2#?

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Answer 1 · 2016-02-13T06:35:28

Alternate method.
#=x^2-18x+81#

We know that
#(a-b)^2=a^2-2ab+b^2#
In the given question #(x-9)^2#
#a=x and b=9#. Substituting the values we obtain
#(x-9)^2=x^2-2*x*9+9^2#

#=>x^2-18x+81#

Answer 2 · 2016-02-13T07:29:07

#x^2-18x+81#

#(x-9)^2#

Remember the formula #(a+b)^2=a^2+2ab+b^2#

In this case #a=x,b=-9#

And remember #-9^2=-9*-9=81#

Substitute the values:

We get:

#rArrx^2+2(x)(-9)+81#

#rArr=x^2-18x+81#