Question

If (x^2020 + 2x^2019 + k) is divisible by (x + 1) than the value of k is​

Answer 1

Step-by-step explanation:

let ,

$f(x) = {x}^{2020} + 2 {x}^{2019} + k$

$f(x) \: is \: divisible \: by \: (x + 1) \: then$

$f( - 1) = {( - 1)}^{2020} + 2 {( - 1)}^{2019} + k = 0$

$1 - 2 + k = 0$

$k = 1$