Question

The function f(x)=x^3-p^2x^2+2x-p has a remainder of -5 when divided by (x+1).find the possible values f constant p

Answer 1

Answer:

p = - 2 or p = 1

Step-by-step explanation:

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{p(x) = {x}^{3} - {p}^{2} {x}^{2} + 2x - p } \\ &\sf{remainder \: = - 5}\\ &\sf{divisor \: x \: + \: 1} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{value \: of \: p}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

We know that

☆ If polynomial p(x) is divided by x + 1, then remainder is given by p(- 1).

$\boxed{ \purple{ \bf \: p( - 1) = \tt \: - \: 5}}$

$\tt\implies \: {( - 1)}^{3} - {p}^{2} {( - 1)}^{2} + 2( - 1) - p = - 5$

$\tt\implies \: - 1 - {p}^{2} - 2 - p = - 5$

$\tt\implies \: { - p}^{2} - p - 3 = - 5$

$\tt\implies \: {p}^{2} + p - 2 = 0$

$\tt\implies \: {p}^{2} + 2p - p - 2 = 0$

$\tt\implies \:p( p + 2) - 1(p + 2) = 0$

$\tt\implies \:(p + 2)(p - 1) = 0$

$\tt\implies \: \boxed{ \purple{\bf \:  p \: = \: - \: 2 \: \: \: or \: \: \: p \: = \: 1}}$