Question

Find a polynomial p of degree 3 so that

p(5) = 153,
p(−5) = −307,
p(3) = 13,
p(−1) = −3,

p(x)=
use the polynomial to find p(4)

p(4)=

Answer 1

do we know if the polynomial is quadratic or binomial or what type of polynomial P would be?

Answer 2

@DemolisionWolf I am not quite sure, but it says here that "You can find p by setting p(x) = ax3+bx2+cx+d, and substituting the given values of x and p(x) to obtain 4 equations in the 4 variables a...d."

Answer 3

p(x) = ax3+bx2+cx+d
p(5) = 5^3+5^2+5 = 155, so d would become -2 thus
p(5) = 5^3+5^2+5 -2 = 153

Answer 4

holy crap, is that so? All this time I would assume we put it into a matrix obtained from r+rx+rx^2

Answer 5

but how would you find p(x) @DemolisionWolf .

Answer 6

isn't p(x) given to you as
p(x) = ax3+bx2+cx+d ?

Answer 7

yes, but we have to find p(x) from those 4 equations given

Answer 8

so we need to find a, b, c, and d using a matrix then?

Answer 9

yeah i guess so

Answer 10

p(x) = ax3+bx2+cx+d

p(5) = A*5^3 + B*5^2 + C*5 + D = 153
p(-5) = A*5^3 + B*5^2 + C*5 + D = -307
p(3) = A*3^3 + B*3^2 + C*3 + D = 13
p(−1) = A*(-1)^3 + B*(-1)^2 + C*(-1) + D= −3

4 equations, 4 unknowns, so the 'x' of each term becomes the 'constant' so to speak, and we are needing to find A, B, C, D,
so I belive the first row would look like:
[ 5^3 5^2 5 1 153 ]
[ 5^3 5^2 5 1 -307 ]
[ 3^3 3^2 3 1 13 ]
[-1^3 -1^2 -1 1 -3 ]

get this matrix into identity form, and you can then solve for A, B, C, D

Answer 11

right on. that is exactly what i thought.

Answer 12

sorry it took me so long to understand what you were saying

Answer 13

yeah its no problem at all.