Algebra 2 Help please!! which of the following represents the zeros of f(x)=x^3-11x^2+38x-40 a) 5, -4, 2 b) 5, -4, -2 c) 5, 4, 2 d) 5, 4, -2
Zeros of a problem is when you want f(X)=0. Right?
This problem is very simple. Finding the zeros of a function, we want to find any 'x' value(s) that will make the function = 0.
I think so. I'm not sure >.< I really have no idea how to do this.
In this case, you are given MC answers that you can put into your problem. I'll do the first on for you. :-)
Oh okay that makes sense!
thank you
Glad it does :-)
Without MC answers to help, you would factor to find the zeros.
So I'm basically plugging in the 3 numbers for x and finding which of them all make it f(x)=0?
f(x)=x^3-11x^2+38x-40 a) 5, -4, 2 b) 5, -4, -2 c) 5, 4, 2 d) 5, 4, -2 5^3-11(5)^2+38(5)-40=0. This satisfies what we are looking for, so 5 is a zero to the function.
f(-4)= -4^3-11(-4)^2+38(-4)-40=-432 this doesn't satisfy, so -4 doesn't work.
Brb, I have to do something.
well that narrows it down, I can figure out if it's 2 or -2 and I'll let you know.
nvmd I guess I did something wrong cause neither of them came out as 0
I am back @notamathgenuis
yay!
Actually, 2 produces a zero.
f(x)=x^3-11x^2+38x-40 a) 5, -4, 2 (not zero function) b) 5, -4, -2(not zero function) c) 5, 4, 2 d) 5, 4, -2 Now its 50/50. Since we know its both 5 and 4, lets try -2 and 2. 2 first. 2^3-11(2)^2+38(2)-40=0 when f(x)=f(2)
-2^3-11(-2)^2+38(-2)-40=-168. Not z ero.
c) is the only one to satisfy.
oh okay I see what I did wrong!! thank you so much.
We learn math through mistakes. I am in Pre-Calc/Trig, and have made many, many mistakes but learned from them! That's how we gotta learn :-) If you need any more help, mention me. I LOVE Algebra.
I will definitely mention you on anything else I need help with. I HATE algebra! lol
Fantastic, Algebra II was my favorite course. You sound like my brother, he doesn't care for it too much, haha :)
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