Question

P(x)= x³+ ax²+ bx+ c be a polynomial such that P(1)=1, P(2)=2, P(3)=3, then find the value of P(4).​

Answer 1

Given that

$\rm :\longmapsto\:P(x) = {x}^{3} + {ax}^{2} + bx + c - - - - (1)$

Now,

$\rm :\longmapsto\:P(1) = 1$

$\rm :\longmapsto\: {(1)}^{3} + a {(1)}^{2} + b(1) + c = 1$

$\rm :\longmapsto\:1 + a + b + c = 1$

$\bf\implies \:a + b + c = 0 - - - (2)$

Again,

$\rm :\longmapsto\:P(2) = 2$

$\rm :\longmapsto\: {(2)}^{3} + a {(2)}^{2} + b(2) + c = 2$

$\rm :\longmapsto\:8 + 4a + 2b + c = 2$

$\bf\implies \:4a + 2b + c = - 6 - - - (3)$

Also,

$\rm :\longmapsto\:P(3) = 3$

$\rm :\longmapsto\: {(3)}^{3} + a {(3)}^{2} + b(3) + c = 3$

$\rm :\longmapsto\:27 + 9a + 3b + c = 3$

$\rm :\longmapsto\:9a + 3b + c = - 24 - - - (4)$

On Subtracting equation (2) from equation (3), we get

$\rm :\longmapsto\:3a + b = - 6 - - - (5)$

On Subtracting equation (3) from equation (4), we get

$\rm :\longmapsto\:5a + b = - 18 - - - - (6)$

Now,

On Subtracting equation (5) from equation (6), we get

$\rm :\longmapsto\:2a = - 12$

$\bf\implies \:a = - 6 - - - (7)$

Put a = - 6 in equation (6), we get

$\rm :\longmapsto\:5( - 6) + b = - 18$

$\rm :\longmapsto\: - 30 + b = - 18$

$\bf\implies \:b = 12 - - - (8)$

Substituting the value of a and b in equation (2), we get

$\rm :\longmapsto\: - 6 + 12 + c = 0$

$\rm :\longmapsto\: 6+ c = 0$

$\bf\implies \:c \: = \: - \: 6$

So,

On substituting the values of a, b and c in equation (1), we get

$\rm :\longmapsto\:P(x) = {x}^{3} - 6 {x}^{2} + 12x - 6$

So,

$\rm :\longmapsto\:P(4) = {(4)}^{3} - 6 {(4)}^{2} + 12 \times 4- 6$

$\rm :\longmapsto\:P(4) = 64 - 96 + 48 - 6$

$\rm :\longmapsto\:P(4) = 10$