Question

Let P(x) be a monic cubic equation such that P(1) = 1,
P(2) = 2, P(3) = 3, then find P(4).​

Answer 1

Given :  P(x) be a monic cubic equation such that P(1) = 1, P(2) = 2, P(3) = 3

To find : P(4)

Solution:

Let P(x) be a monic cubic equation  = x³  + bx² + cx  + d

P(1) = 1

=> 1 + b  + c +  d  = 1

=> b + c + d  = 0     Eq1

p(2) = 2³  + b*2² + c*2  + d  = 2

=> 4b + 2c  +  d  =  - 6     Eq2

p(3) = 3³  + b*3² + c*3  + d  = 3

=> 9b  +  3c  + d  =  - 24     Eq3

Eq2  - Eq1  

=> 3b  + c  = -6

Eq3 - Eq2

=> 5b + c  = -18

=> 2b = -12

=> b = -6

=> c = 12

b + c + d  = 0  =>  d  = - 6

p(x) = x³  -6x² + 12x  -6

p(4) = 4³  -6*4² + 12*4  -6

= 64  - 96  + 48  -  6

= 112 - 102

= 10

P(4) = 10

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Answer 2

Given:

P(1) = 1

P(2) = 2

P(3) = 3

To find:

P(4)=?

Solution:

Let P(x) =$x^3+ax^2+bx+c......(x)$

And let,

$\to P(1)= 1....(i)\\\to P(2)=2....(ii)\\\to P(3)=3...(iii)\\\to P(4)= 4.....(iv)$

Solve value for P(x):

When,

$\to p(1) =1 \\\\\to 1^3+ a1^2+b1+c =1\\\to 1+ a+b+c =1\\\to a+b+c = 0 .....(a)\\ \\\to p(2) =2 \\\\\to 2^3+ a2^2+b2+c =2\\\to 8+ 4a+2b+c =2\\\to 4a+2b+c = -6 .....(b)\\\\\to p(3) =3 \\\\\to 3^3+ a3^2+b3+c =3\\\to 27+ 9a+3b+c =3\\\to 9a+3b+c = -24 .....(c)$

by solving the equation a, b, and c we calculate the value of a, b, and c:

subtract equation b from the equation a, we get:

$\to 3a+b+6 =0....(d)$

subtract equation c from the equation b, we get:

$\to 5a+b+18....(e)$

subtract equation e from the equation d, we get the value of a that is:

$\to 5a+b+18-3a-b-6=0\\\to 2a+12=0\\\to a= -6...(f)$

put the value of equation f into the equation e:

$\to 5(-6)+b+18\\\to -30+b+18\\\to b= 12....(g)$

put the value of a, b in equation c:

$\to 9(-6)+3(12)+c=-24\\\to -54+36+c=-24\\\to -18+c=-24\\\to c=-24+18\\\to c = -6$

put the value of a, b, and c in equation (x):

$\to P(4) = 4^3+(-6)4^2+(12)4+(-6)\\\to P(4) = 64-6 \times 16+12\times 4-6\\\to P(4) = 64-96+48-6\\\to P(4) = 10$

The answer is "10".