Cubic Polynomial

Algebra Level 3

A cubic polynomial p(x) is such that p(1)=1 , p(2)=2 , p(3)=3 , p(4)=5. find the value of P(6) ????

The answer is 16.

3 solutions

Deepanshu Gupta
Aug 20, 2014

let $f(x)=p(x) - x$ .

where f(x) has roots 1,2,3

$f(x)=p(x) - x = a(x-1)(x-2)(x-3)$ .

$p(4)=5$ .

$a=1/6$ .

$p(6)=16$ .

I used the method of differences .

mathh mathh - 6 years, 9 months ago

I used the method, but got some problem. Can you please post the solution?

Swapnil Das - 5 years, 11 months ago

$\begin{array}{l|c|r}x& p(x)& D_1& D_2& D_3\\\hline 1& 1& 1& 0& 1\\\hline 2& 2& 1& 1\\\hline 3& 3& 2\\\hline 4& 5\\\hline 5\\\hline 6\end{array}$

Since $p(x)$ is a cubic polynomial, $D_3$ is constant.

$\begin{array}{l|c|r}x& p(x)& D_1& D_2& D_3\\\hline 1& 1& 1& 0& 1\\\hline 2& 2& 1& 1& 1\\\hline 3& 3& 2& & 1\\\hline 4& 5\\\hline 5\\\hline 6\end{array}$

Knowing this, you find the other differences, and eventually find $p(5),p(6)$ .

$\begin{array}{l|c|r}x& p(x)& D_1& D_2& D_3\\\hline 1& 1& 1& 0& 1\\\hline 2& 2& 1& 1& 1\\\hline 3& 3& 2& 2& 1\\\hline 4& 5& & 3\\\hline 5\\\hline 6\end{array}$

$\begin{array}{l|c|r}x& p(x)& D_1& D_2& D_3\\\hline 1& 1& 1& 0& 1\\\hline 2& 2& 1& 1& 1\\\hline 3& 3& 2& 2& 1\\\hline 4& 5& 4& 3\\\hline 5& & 7\\\hline 6\end{array}$

$\begin{array}{l|c|r}x& p(x)& D_1& D_2& D_3\\\hline 1& 1& 1& 0& 1\\\hline 2& 2& 1& 1& 1\\\hline 3& 3& 2& 2& 1\\\hline 4& 5& 4& 3\\\hline 5& 9& 7\\\hline 6& 16\end{array}$

mathh mathh - 5 years, 11 months ago

@Mathh Mathh – I am extremely indebted for your kind help!

Swapnil Das - 5 years, 11 months ago

Sai Ram
Aug 18, 2015

Let the polynomial be $ax^3+bx^2+cx+d.$

Now,

$p(x)=ax^3+bx^2+cx+d.$

It is given that $p(1)=1 , p(2)=2 ,p(3)=3,p(4)=5.$

$p(1)=a+b+c+d =1 .....................\dots \boxed{1}$

$p(2)=8a+4b+2c+d =2 ..................\dots \boxed{2}$

$p(3)=27a+9b+3c+d = 3 .................\dots \boxed{3}$

$p(4)=64a+16b+4c+d=4 .................\dots \boxed{4}$

Solving these four equations ,we get

$a=\dfrac{1}{6}$

$b=-1$

$c=\dfrac{17}{6}$

$d=-1.$

Now,the polynomial becomes

$\dfrac{x^3}{6}-x^2+\dfrac{17x}{6}-1.$

Substituting $6$ we get $16.$

Divyansh Saxena
Oct 8, 2014

let a cubical polynomial equation be

P(x)= ax^{3} + bx^{2} + cx + d . So now,

P(1)=a+b+c+d=1

p(2)=8a+4b+2c+d=2

P(3)=27a+9b+3c+d=3

P(4)=64a+16b+4c+d=5

use CRAMER'S RULE to solve these equations and get the values of (a,b,c,d).

you will get the values as a=0.1666666 , b=-1 , c=2.8333333 , d=-1

so now put these values in above equation,

P(x)=0.166666x^{3} - x^{2} + 2.833333x - 1

put x=6

you will get P(6)=15.98

approx. (16).

That's exactly how i did it!

Stewart Feasby - 6 years, 8 months ago

nice try. gr8 explanation

Piyush Joshi - 6 years, 6 months ago

Cubic Polynomial

The answer is 16.

3 solutions

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