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anonymous
 4 years ago
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.
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anonymous
 4 years ago
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?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay got the asnwers. thank you very much!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait no i got a=2 b=19 :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you show me your calculation? i want to see where i did wrong

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1well, 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why is the second equation is equal to 0 and not 36 since its the remainder?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0shouldnt it be a(1)^4  (1)^3  14(1)^2 + b(1)  6 = 36 ?

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1opsss.. 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but 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,

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1factor ? i think we didnt see and neednt factor there but elimination and substitution method

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1oh 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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?

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1how come x^2x2 ? what is it

anonymous
 4 years ago
Best 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?

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1if 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh right i see...so i should try to find the root by myself?

RadEn
 4 years ago
Best ResponseYou've already chosen the best response.1yea, u might use trial and error or use synthetic division to find the other factors

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay i get. thank you so much, again! you've been a great help :)

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