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Mathematics 7 Online
OpenStudy (anonymous):

someone help me please !!!!! :(

OpenStudy (anonymous):

what do you need help with

OpenStudy (anonymous):

writing polynomial functions of least degree with intergral coeffiecients that has the given zeros ?

OpenStudy (anonymous):

you gotta give all the info before someone can help you

OpenStudy (anonymous):

3 , 5 , -5

OpenStudy (anonymous):

that is all that is on my paper .i just typed what my directions said

OpenStudy (anonymous):

that is at least a degree 3 poly (x-3)(x-5)(x+5)

OpenStudy (anonymous):

(x-3)(x^2 -25) x^3 -3x^2 -25x +75

OpenStudy (anonymous):

ok so next what do i do ?

OpenStudy (anonymous):

Suppose you need to write a polynomial function of 6, 2i with intergral coeffiecients that has the given zeros then X-6 is the polynomial with least degree and 6 as zero. X=2i --> X^2=-4 --> X^2+4=0 --> X^2+4 is the polyn. with least degree and 2i as zero. ---> (X-6)(X^2+4)=X^3-6X^2+4X-24 is solution

OpenStudy (anonymous):

you write down your three roots like this: (x-3)(x-5)(x+5) ; then multiply them together

OpenStudy (anonymous):

im lost . i got \[x ^{2} + 15x\] thats all i have so far .

OpenStudy (anonymous):

thats not correct :)

OpenStudy (anonymous):

lol ok so what is correct then ?

OpenStudy (anonymous):

what do you have as the "root" factors of the problem?

OpenStudy (anonymous):

how do you take: x = 3,5,-5 and make root factors?

OpenStudy (anonymous):

x then the opposite of the number.. thats what my teacher said

OpenStudy (anonymous):

(3+a) = 0 (5+b) = 0 (-5+c) = 0 solve for a b and c and use them like this: (x+a)(x+b)(x+c)

OpenStudy (anonymous):

thats right, the opposite of the numbers...

OpenStudy (anonymous):

what do you get as your root factors then?

OpenStudy (anonymous):

uhm .... lol im sorry but i have ( NO ] clue

OpenStudy (anonymous):

root factors is a term im using that has no real mathmatical value except for what I use it for :) root factors, by me, are the (x....)(x......)(x......) things that polys factor down into; does that make sense? x^2 +7x +12 factors into: (x+4)(x+3) <---- root factors are these things

OpenStudy (anonymous):

ooooooooooooooooo yea . when you factor .. i know that .. like to factor \[x ^{2} + 5 +14\] wouldnt that be (x+7)(x+2)

OpenStudy (anonymous):

thats right :) except for the typos :)

OpenStudy (anonymous):

or i mean -14 and then (x+7)(x-2)

OpenStudy (anonymous):

:) you should really be my new tutor . lol i understand you more than the rest

OpenStudy (anonymous):

the root factor of x=-7 is: (x+7) the root factor of x=4 is: (x-4) the root factor of x=2 is: (x-2)

OpenStudy (anonymous):

the root factor of x=3 is: ?? the root factor of x=5 is: ?? the root factor of x=-5 is: ??

OpenStudy (anonymous):

thr root factor of x=3 is: (x-3) the root factor of x-5 is: (x+5) the root factor of x=-5 is: (x+5) :) right ?

OpenStudy (anonymous):

very good :) now lets multiply those root factors together :)

OpenStudy (anonymous):

(x-3)(x+5) = xx -3x +5x -15 = x^2 +2x -15 right?

OpenStudy (anonymous):

xx ? n

OpenStudy (anonymous):

i mean xx ?

OpenStudy (anonymous):

xx means x*x mean x^2

OpenStudy (anonymous):

ooooo sorry :) ok im following you again

OpenStudy (anonymous):

(x-5)(x^2 +2x -15) = xxx +2xx -15x -5xx -10x +75 : x^3 -3x^2 -25x +75 .... if im seeing it right :)

OpenStudy (anonymous):

and its good ;)

OpenStudy (anonymous):

lol how old are you ? you are a genius at this

OpenStudy (anonymous):

as long as you turn it into your root factors; and keep track of your multiplying, its simple enough :)

OpenStudy (anonymous):

....old :)

OpenStudy (anonymous):

lol you seem young too me . haha

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