IMPORTANT! Find all the zeros of each function. State the multiplicity of any multiple zeros. y=(x-1)^2(2x-3)^3 = 1 has a multiplicity of 2 & 2/3 has a multiplicity of 3? This one I have no idea: y=(x+2)(x^2+3x-40)
y = (x+2)(x²+3x-40) (x+2)(x²+3x-40)=0 x+2=0 or x²+3x-40=0 then we have x = -3 +- srqrt169 / 2 x = -3 +- 13 / 2 x' = -16/2 and x'' = 10/2 finally... x = - 2, x = -8 and x = 5. is that what you mean?
so i got the first question i asked right? &
How did you get then we have x = -3 +- srqrt169 / 2?
Sorry, I don't know what you mean.
mmm lemme see
its Bhaskara
from x² + 3x - 40
Can you explain to me how you did that step please
remember this? http://3.bp.blogspot.com/_Pjkf6PNC_qk/TAQ8f271AlI/AAAAAAAAAAM/OHlBggN46JY/s320/bhaskara.jpg
quadratic formula
you do that with x² + 3x - 40 (ax² +bx + c)
just replace the letters
\[x^2+3x-40=(x+8)(x-5)\]
you can also factor this: x^2 +3x -40 = (x+8)(x-5)
No need to us the quadratic formula here.
oh that's right, easier indeed.
Can you explain to me how you factored that
@cubanito for your answer posted, the 2nd zero should be 3/2 not 2/3 2x-3 = 0 2x = 3 x = 3/2
Yeah, I meant to put 3/2. But those two answers are right?
cubanito, you can factor that like this: x² + 3x - 40 x 8 x -5
So, its x = - 2, x = -8 and x = 5?
Yep, it is. But make sure you got the whole process, not only the answers.
What about y= x^3+x^2-19x+5
You want to find the value of x?
no, find all the zeros & state the multiplicity
try zeros -1,-2,-3,-4,-5 and see if they factor out evenly i suggest using synthetic division
Yea, that's a good way, dumbcow... wait, dumbcow? wth is this name? LOL
i like the irony of it :) also if you have a graphing calculator, graph the function and see where it crosses x_axis
where does it cross the x axis
when y =0
I'm confused now :(
Melinda....dumbcow?????!!!???
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