How do you factor #x^2- (4/3)x+ (4/9)#?

2 Answers
MathIsMyWaifu
Sep 2, 2016

#(x-2/3)^2#

Explanation:

Because this polynomial has a coefficient of #1# for the squared term, #x^2#, you can easily find out that the factored form will look some like this:
#(x +- _ )(x +- _ )#

Now you just look at the constant term, which is #4/3#, and see what numbers you can multiply by itself to get that number.
Now since the constant is a fraction this might be a little difficult but
that number is #2/3#.
And now this problem is being extremely nice to you because #2/3+2/3=4/3#
Because the #4/3# is negative and the #4/9# is positive we get:
#(x-2/3)(x-2/3)# or #(x-2/3)^2#

Hope this helped! :3

CW
Sep 2, 2016

#(x-2/3)(x-2/3)#

Explanation:

#-(2/3)xx-(2/3) = 4/9#

#-(2/3)-(2/3) = -(4/3)#

#(x-2/3)(x-2/3) = x^2-(2/3)x -(2/3)x +(4/9) #

#= x^2-(4/3)x+(4/9)#

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