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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...
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 ;)
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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?
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)