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moongazer
 3 years ago
1.) Find a polynomial of degree that has zeros 1, 1 and 1/2 and whose leading coefficient is 3.
2.) Find the value of k such that x+3 is a product of P(x) = x^2  10kx + 6
moongazer
 3 years ago
1.) Find a polynomial of degree that has zeros 1, 1 and 1/2 and whose leading coefficient is 3. 2.) Find the value of k such that x+3 is a product of P(x) = x^2  10kx + 6

This Question is Closed

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1AT FIRST. do u know what a degree of polynomial means.

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1\[3x^3+(3x^2/2)−(3x/2)+(3/2)\] is this correct?

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1Is my answer correct?

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1343207441484:dw

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1how about for no. 2 could you help me?

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1can u simplify this plz

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1I DONT THINK YOUR ANSWER IS CORRECT.

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1FOR NUMBER 2) do u know factor theorem

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1could I make \[(x1/2) > (2x1)\] ????????

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1343207772794:dw

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1NOW do u know factor theorem? and did u get answer for 1)

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1\[3x^3(3x^2/2)−3x−(3/2)\]

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1BUT U R VERY CLOSE

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1u just made the sign mistake

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1could I make (x1/2) into (2x1) because my teacher said there should be no fractions?

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1343208282102:dw

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1is the correct answer

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1could I make (x1/2) into (2x1) ??????

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1NOPE becausedw:1343208355724:dw

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1but if dw:1343208386811:dw

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1that's what I mean for the second part

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1becausedw:1343208443188:dw

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1isn't that it is zero so that it is possible?

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1343205903760:dw

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1IF it is equal to zero then yes u can because dw:1343208534245:dw

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1343205945423:dw

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1I got the correct answer :)

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1it mean that there zeroes are both 1/2

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1so you can use either of the two?

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1yeah. BUT only in that case. U cannot use this in simplification.

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1AND NOW SHALL WE BEGIN NO. 2

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1do u know factor theorem.

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1wait I am trying to solve 1 with (2x1) :)

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1I got the same answer :)

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1I know factor theorem. If the remainder when dividing P(x) by D(x) is zero, then D(x) is a factor of P(x). P(x) is any polynomial D(x) is the divisor being a factor means it is included in multiplying.

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1is that correct? could you improve my definition of factor ?

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1No u r defination is correct. And can u use this to solve this problem

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1how I am confused with the question?

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1I am thinking of distributing 3 to x

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1but I can't explain why it is correct :)

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1your 2 number question simply means that if u divide P(x) by (x+3) then remainder is zero. dw:1343209619170:dw

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1ohh, I think it also has remainder theorem? :)

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1343207190855:dw

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1I added =0 so that the remainder will be equal to zero making (x+3) a product ?

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1am I correct with my reason?

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.1@sauravshakya thank you very much for your help :)

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1U r WELCOME. AND best of luck
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