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shaqadry
Group Title
Consider the polynomial P(x) = ax^4  x^3  14x^2 + bx  6 where a and b are constant. If one of the zeros of P(x) is 2 and the remainder is 36 when P(x) is devided by (x + 1). Find the values of a and b.
How do I solve this?
 one year ago
 one year ago
shaqadry Group Title
Consider the polynomial P(x) = ax^4  x^3  14x^2 + bx  6 where a and b are constant. If one of the zeros of P(x) is 2 and the remainder is 36 when P(x) is devided by (x + 1). Find the values of a and b. How do I solve this?
 one year ago
 one year ago

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RadEn Group TitleBest ResponseYou've already chosen the best response.1
it means P(2) = 0 and P(1) = 36 now, aplly of them to P(x)
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
okay got the asnwers. thank you very much!
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
a=6, b=13, right ?
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
wait no i got a=2 b=19 :/
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
can you show me your calculation? i want to see where i did wrong
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
well, look this: P(x) = ax^4  x^3  14x^2 + bx  6 P(2) = 0, it means a(2)^4  (2)^3  14(2)^2 + b(2)  6 = 0 16a  8  56 + 2b  6 = 0 16a + 2b  70 = 0 16a + 2b = 70 8a + b = 35 .......... (1st equation) then P(1) = 36, means a(1)^4  (1)^3  14(1)^2 + b(1)  6 = 0 a + 1  14  b  6 = 0 a  b  19 = 0 a  b = 19 ............ (2nd equation) solve for x and y, by using elimination and substitution method (1st) + (2nd), gives 9a = 54 (cancel out for b) a = 6 now, substitute a=6 to one of equation, let the (2nd eq) ab=19 6  b = 19 b = 6  19 = 13
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
why is the second equation is equal to 0 and not 36 since its the remainder?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
i used remainder theorem, if P(x) divided by (x+a) so its remainder is P(a)
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
shouldnt it be a(1)^4  (1)^3  14(1)^2 + b(1)  6 = 36 ?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
opsss.. u are correct, sorry my bad because copy and paste from the1st equ... yeah, it should a(1)^4  (1)^3  14(1)^2 + b(1)  6 = 36 sorry, ur answer is right and mine was wrong
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
dont angry to me, yea :)
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
but when i cant seem to factorise it completely :/ there's a remainder. is it correct? or is my calculation wrong? bcs the 2nd questions ask me to factorise it completely,
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
not at all :)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
factor ? i think we didnt see and neednt factor there but elimination and substitution method
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
i dont understand
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
oh sorry, miss comunication (because my english soo bad :(, sorry again ) if u have pounded the value of a and b, just plug of them to P(x), from here we can find the factors of P(x)
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
so what i should do is devide 2x^4  x^3  14x^2 + 19x  6 with x^2  x 2 rite? i tried calculating it but i end up with remainders. what should i do?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
how come x^2x2 ? what is it
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
(x2) and (x+1) is two of the four factors right? (x+1) is given. and x=2 is also given therefore x2=0 correct? or am i wrong?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
if the problem told u one of the zeros or one of the factor, yea right but wrong for (x+1), it's not a root because it has remainder
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
oh right i see...so i should try to find the root by myself?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
yea, u might use trial and error or use synthetic division to find the other factors
 one year ago

shaqadry Group TitleBest ResponseYou've already chosen the best response.0
okay i get. thank you so much, again! you've been a great help :)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
i have asked to wolfram, it said to me here : http://www.wolframalpha.com/input/?i=2x^4x^314x^2%2B19x6
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
you're welcome :)
 one year ago
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