Factor the trinomial, if it's a perfect square. Or state that it's prime if it's prime. \[x ^{2}-13xy+169y ^{2}\]
Have you heard of the quadratic formula?
no
you cannot factor this, nor, i think, can you solve it using the quadratic formula if it was \[x^2-26xy+169y^2\] then it would be a perfect square, because \(136=13^2\) and \(2\times 13=26\)
You can certainly solve this in the complex plain
I have an online course and I was thinking this one was not solveable but wasn't sure
but since you have \(-13xy\) it is not a perfect square i can't land my complex plain
as nancy reagan used to say "just say no"
This is the first one I actually had right before putting on here to double check. :) Thank you!!!
She is doing how it is described in the textbook. She doesn't know why. Its all basic trinomials. Its like basic algebra.
I don't know how it's done in the textbook because I can't understand it. But yes, it's from an online course so I only have reading explanation and that doesn't quite work. Thank you!
What about these factors? \[(2x - 13y)(\frac{1}{2}x+13y)\]
Don't work either.
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