Question

find the first degree polynomial p(x) with p(0)=1 and p(1)=2​

Answer 1

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

Since, it is asked to find one degree polynomial p(x),

So, Let assume one degree polynomial as

$\bf :\longmapsto\:p(x) = ax + b - - - (1)$

It is given that

$\bf :\longmapsto\:p(0) = 1$

$\rm :\implies\:a(0) + b = 1$

$\rm :\longmapsto\:0 + b = 1$

$\bf\implies \:b = 1 - - - (2)$

Further,

It is given that

$\bf :\longmapsto\:p(1) = 2$

$\rm :\implies\:a(1) + b = 2$

$\rm :\longmapsto\:a + 1 = 2$

$\rm :\longmapsto\:a = 2 - 1$

$\bf\implies \:a = 1 - - - (3)$

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

$\bf :\longmapsto\:p(x) = x + 1$

Additional Information :-

☆ The two degree polynomial is called Binomial and its algebraic expression is of the form

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

☆ The three degree polynomial is called Trinomial and its algebraic expression is of the form

$\rm :\longmapsto\:f(x) = {ax}^{3} + b {x}^{2} + cx + d$

☆ The four degree polynomial is called Bi - quadratic and its algebraic expression is of the form

$\rm :\longmapsto\:f(x) = {ax}^{4} + b {x}^{3} + c {x}^{2} + dx + e$

☆ The degree of constant polynomial is 0.

☆ The degree of 0 is not defined.