Question

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?

Answer 1

it means P(2) = 0 and P(-1) = -36
now, aplly of them to P(x)

Answer 2

okay got the asnwers. thank you very much!

Answer 3

a=6, b=-13, right ?

Answer 4

wait no i got a=2 b=19 :-/

Answer 5

can you show me your calculation? i want to see where i did wrong

Answer 6

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)
a-b=19
6 - b = 19
b = 6 - 19 = -13

Answer 7

why is the second equation is equal to 0 and not -36 since its the remainder?

Answer 8

i used remainder theorem, if P(x) divided by (x+a) so its remainder is P(-a)

Answer 9

shouldnt it be a(-1)^4 - (-1)^3 - 14(-1)^2 + b(-1) - 6 = -36 ?

Answer 10

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

Answer 11

dont angry to me, yea :)

Answer 12

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,

Answer 13

not at all :)

Answer 14

factor ? i think we didnt see and neednt factor there but elimination and substitution method

Answer 15

i dont understand

Answer 16

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)

Answer 17

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?

Answer 18

how come x^2-x-2 ? what is it

Answer 19

(x-2) and (x+1) is two of the four factors right? (x+1) is given. and x=2 is also given therefore x-2=0 correct? or am i wrong?

Answer 20

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

Answer 21

oh right i see...so i should try to find the root by myself?

Answer 22

yea, u might use trial and error or use synthetic division to find the other factors

Answer 23

okay i get. thank you so much, again! you've been a great help :)

Answer 24

i have asked to wolfram, it said to me here :
http://www.wolframalpha.com/input/?i=2x^4-x^3-14x^2%2B19x-6

Answer 25

you're welcome :)