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Assuming Y=y, I see a 2nd degree polynomial in y. You're not asking a question, so I will just calculate some fun stuff associated with it... Factorize: (y-1)(y-4). Zeroes: (y-1)(y-4)=0, so y=1 or y=4. If f(y)=y^2-5y+4, then the graph of f is a parabola. There is a vertex when y is the mean of the zeroes: y=2.5. The value is f(2.5) = -2.25. The axis of symmetry of the parabola is x = 2.5. The derivative of f is f'(x)=2y-5. Maybe there is something of value for you in this all. You also could consider asking a question ;)
@ZeHanz so the answer is... (y-1)(y-4)?
@Harkirat can you help me/
are you trying to figure out how to factorize?
@DanL33 how to do this problem.
i am doing a study sheet that i can use on a test that starts in 15 minutes. and i need tto learn how to solve this problem
Yes but you can't just solve a problem... You first need to know if you need to factor it. Find the zeros etc..
okkk, does this problem need to be factorized?
\[Y^2-5y+4\] because it is Y^2 we know it must factor down into (Y plus or minus something) (Y plus or minus something) because it is a negative followed by a positive we know that they are both (Y - something) (Y - something) (if you need explanation just ask) we are now looking for numbers that multiply to make 4 and add together to make 5 1x4 = 4 1+4 = 5 2x2 = 4 2+2 = 4 so it must be 1 and 4 (Y - 1) (Y - 4)
ok... i can work with that
@DanL33 is correct
so (y-1)(y-4) is the answer?