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

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

1. [tex]\large\underline{\sf{Solution-}}[/tex]

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

   So, Let assume one degree polynomial as

   [tex]\bf :\longmapsto\:p(x) = ax + b – – – (1)[/tex]

   It is given that

   [tex]\bf :\longmapsto\:p(0) = 1[/tex]

   [tex]\rm :\implies\:a(0) + b = 1[/tex]

   [tex]\rm :\longmapsto\:0 + b = 1[/tex]

   [tex]\bf\implies \:b = 1 – – – (2)[/tex]

   Further,

   It is given that

   [tex]\bf :\longmapsto\:p(1) = 2[/tex]

   [tex]\rm :\implies\:a(1) + b = 2[/tex]

   [tex]\rm :\longmapsto\:a + 1 = 2[/tex]

   [tex]\rm :\longmapsto\:a = 2 – 1[/tex]

   [tex]\bf\implies \:a = 1 – – – (3)[/tex]

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

   [tex]\bf :\longmapsto\:p(x) = x + 1[/tex]

   Additional Information :-

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

   [tex]\rm :\longmapsto\:f(x) = {ax}^{2} + bx + c[/tex]

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

   [tex]\rm :\longmapsto\:f(x) = {ax}^{3} + b {x}^{2} + cx + d[/tex]

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

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

   ☆ The degree of constant polynomial is 0.

   ☆ The degree of 0 is not defined.

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