Question

If f(x) is a polynomial of degree three with leading coefficient 1 such that f(1)=2, f(2)=5, f(3)=10 then
the value off (4) is:​

Answer 1

$\large\underline{\sf{Solution-}}$

Since it is given that f(x) is a polynomial of degree 3 with leading coefficient 1.

So, it means the required polynomial is

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

Now, it is given that

$\red{\bf :\longmapsto\:f(1) = 2}$

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

$\rm :\longmapsto\:a + b + c = 1 - - - (2)$

Again, Given that

$\purple{\bf :\longmapsto\:f(2) = 5}$

$\rm :\longmapsto\:{2}^{3} + {a(2)}^{2} + b(2)+ c = 5$

$\rm :\longmapsto\:8 + 4a + 2b + c = 5$

$\rm :\longmapsto\:4a + 2b + c = - 3 - - - (3)$

$\pink{\bf :\longmapsto\:f(3) = 10}$

$\rm :\longmapsto\:{3}^{3} + {a(3)}^{2} + b(3)+ c = 10$

$\rm :\longmapsto\:27 + 9a + 3b + c = 10$

$\rm :\longmapsto\: 9a + 3b + c = - 17 - - - (4)$

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

$\rm :\longmapsto\: 3a + b= - 4 - - - (5)$

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

$\rm :\longmapsto\: 5a + b= - 14- - - (6)$

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

$\rm :\longmapsto\: 2a= - 10$

$\bf\implies \:a \: = \: - \: 5$

On substituting value of 'a' in equation (5), we get

$\rm :\longmapsto\: - 15 + b = - 4$

$\rm :\longmapsto\: b \: = 15 - 4$

$\bf\implies \:\: b \: = \: 11$

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

$\rm :\longmapsto\: - 5 + 11 + c = 1$

$\rm :\longmapsto\: 6 + c = 1$

$\rm :\longmapsto\:c = 1 - 6$

$\bf\implies \:c \: = \: - \: 5$

Hence,

$\bf :\longmapsto\:f(x) = {x}^{3} - 5 {x}^{2} + 11 x - 5$

Therefore,

$\bf :\longmapsto\:f(4) = {4}^{3} - 5 ({4)}^{2} + 11 (4) - 5$

$\bf :\longmapsto\:f(4) = 64 - 80 + 44 - 5$

$\bf :\longmapsto\:f(4) = 23$