# Is there a cubic polynomial that takes...

• Dec 30th 2009, 12:06 AM
zorro
Is there a cubic polynomial that takes...
Question : Is there a cubic polynomial that takes these values

$\begin{array}{|c|c|c|c|c|c|c|}x & 1 & -2 & 0 & 3 & -1 & 7 \\ \hline
y & -2 & -56 & -2 & 4 & 16 & 376 \end{array}$
• Dec 30th 2009, 12:34 AM
Moo
Hello,

Just try it out !

A cubic polynomial is in the form $ax^3+bx^2+cx+d$
• Dec 30th 2009, 12:50 AM
zorro
but what should i try to do .Do i stubstitute the values in the table, but where should i substitute
• Dec 30th 2009, 12:52 AM
Moo
You substitute the x with the values your given.
Then you get an equation with each of the first 4 values, which gives you a system of 4 equations with 4 unknowns (a,b,c,d).
Then see if the remaining values are such that the equations are satisfied.

I suggest you first consider 0,1,-1 because they're nice values to deal with ;)
• Dec 30th 2009, 12:56 AM
zorro
but what about the y values what are those for???
• Dec 30th 2009, 12:57 AM
Moo
Quote:

Originally Posted by zorro
but what about the y values what are those for???

$y=ax^3+bx^2+cx+d
$

They're the values of the polynomial taken at a given x.
• Dec 31st 2009, 05:30 PM
zorro
Quote:

Originally Posted by Moo
You substitute the x with the values your given.
Then you get an equation with each of the first 4 values, which gives you a system of 4 equations with 4 unknowns (a,b,c,d).
Then see if the remaining values are such that the equations are satisfied.

I suggest you first consider 0,1,-1 because they're nice values to deal with ;)

Putting x=1 in the eq
a+b+c+d=0

Putting x=0 in the eq
d=0

Putting x=-1 in the eq
-a+b-c+d=0

Is this what u meant ..........but i dont knw what are we doing
• Dec 31st 2009, 06:50 PM
Soroban
Hello, zorro!

You don't understand a thing we're saying, do you?

Quote:

Is there a cubic polynomial that takes these values?

. . $\begin{array}{|c||c|c|c|c|c|c|}x & \text{-}2 & \text{-}1 & 0 & 1 & 3 & 7\\ \hline
y & \text{-}56 & 16 & \text{-}2 & \text{-}2 & 4 & 376 \end{array}$

The general cubic polynomial is: . $f(x) \:=\:ax^3 + bx^2 + cx + d$
. . and we must determine the four coefficients: $a,b,c,d.$

We can use the first four values of the function and create a system of equations.

. . $\begin{array}{ccccccccc}
f(\text{-}2) = 56: & \text{-}8a + 4b - 2c + d &=& \text{-}56 \\
f(\text{-}1) = 16: & \text{-}a + b - c + d &=& 16 \\
f(0) = 0: & 0 + 0 + 0 + d &=& \text{-}2 \\
f(1) = \text{-}2: & a + b + c + d &=& \text{-}2 \end{array}$

Solve the system of equations: . $a = 18,\;b = 9,\;c = -27,\;d = -2$

Hence, the cubic is: . $f(x) \;=\;18x^3 + 9x^2 - 27x - 2$
. . which goes through the first four points.

But: . $f(3) \:=\:18(3^3) + 9(3^2) - 27(3) - 2 \;=\;322$
. . It does not go through the fifth point.

Therefore, there is no cubic polynomial through the given six points.

• Dec 31st 2009, 09:39 PM
simplependulum
Quote:

Originally Posted by zorro
Question : Is there a cubic polynomial that takes these values

$\begin{array}{|c|c|c|c|c|c|c|}x & 1 & -2 & 0 & 3 & -1 & 7 \\ \hline
y & -2 & -56 & -2 & 4 & 16 & 376 \end{array}$

we only need four points to constuct a cubic polynomial so

I choose those points which are located at $x = -2 , -1 , 0 , 1$

To find the cubic polynomial which passes through those points

we need to find the first , second and the third difference ,

$-56 , 16 , -2 , -2$

$\delta = 16 - (-56) , -2 - 16 , -2 - (-2) = 72 , -18 , 0$

$\delta^2 = - 90 , 18$

$\delta^3 = 108$

Therefore , by using Newton's Interpolation Formula ,

the cubic polynomial is :

$-56 + 72 \binom{x+2}{1} - 90\binom{x+2}{2} + 108\binom{x+2}{3}$

To check whether there is a cubic polynomial passing through the other points either , we just sub. $x = 3 , x = 7$
and see if the values the polynomial gives us equal to $4$ and $376$ respectively .
• Jan 1st 2010, 02:26 AM
zorro
thanks moo and simplependulum for ur post.........