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Factor the trinomial completely t^2 + (2/9)t + (1/81)
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(t + 1/9)^2
how did you do that?
See the middle term 2 * 1/9 * t the first term is just the t^2 the last term is the (1/9)^2 that's on the form a^2 + 2ab + b^2 = (a+b)^2
where does the 2 go in 2 * 1/9 * t? how did you plug it into that formula?
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We have 2ab = 2/9 t ab = 1/9 t a = t b = 1/9 The same as the product we found so we can apply the formula a^2 + 2ab + b^2 = (a+b)^2 = (t + 1/9)^2
thank you!
Any time.
k
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