Question

Find the value of k, if x-1 is a factor of p(x) in each of the following cases:
(i) p(x) =x^2+x+k​

Answer 1

$\implies\sf{x - 1 = 0} \\ \implies \sf{x = 0 + 1} \\ \boxed{ \sf \red{ \implies \: x = 1}}$

$\rule{200}2$

$\sf \purple{p(x) = {x}^{2} + x + k } \\ \implies \sf{p(1) = {1}^{2} + 1 + k = 0} \\ \implies \sf{ 1 + 1 + k = 0 } \\ \implies \sf{2 + k = 0} \\ \implies \sf{k = 0 - 2} \\ \boxed {\sf \red{\implies k = - 2}}$

Answer 2

Answer:

$\\ \huge \bigstar \tt \underbrace \red {information } \bigstar \\ \huge➬$

• Class 9
• Chapter ➲ Polynomials
• Excercise ➲ 2.4
• Question ➲ 3 (i)

$\\ \huge \bigstar \tt \underbrace \red {solution} \bigstar \\ \huge➬$

$\large \tt\blue➦ x - 1 = 0 \\ \large \tt\blue➦ x = 1 \\ \\ \large \tt\blue{(we \: got \: the \: value \: of \: x \:) } \\ \\ \large \tt \red{now \: using \: the \: remainder \: theorem} \\ \\ \large \tt{let \: p(x) = {x}^{2} + x + k} \\ \\ \large \tt \pink{ \: putting \: x = 1} \\ \\ \\ \large \tt\blue➦ p(1) = {1}^{2} + 1 + k = 0\\ \large \tt\blue➦ 1 + 1 + k = 0 \\ \large \tt\blue➦ 2 + k = 0 \\ \large \tt\blue➦ k = \bold{\fbox \red{\boxed{- 2}}}$