Question

plz help


important question​

Answer 1

▪IN THE FIRST QUESTION :

☯GIVEN :

• $\bf{p(x) =x^3-2x^2-4x-1}$

• $\bf{g(x) = x+1}$

☯TO FIND :

• Remainder.

➲SOLUTION :

• By Remainder Theorem :

Let us assume g(x) = 0.

$: \longrightarrow{ \tt{x+1= 0}}$

$: \longrightarrow{ \tt{x = 0-1}}$

$: \longrightarrow{ \bf{x = -1}}$

✞Substituting the value of x :

$:\longrightarrow{\tt{p(x) =x^3-2x^2-4x-1}}$

$: \longrightarrow{\tt{p( - 1) =(-1)^3-2(-1)^2 - 4\times(-1)-1}}$

$: \longrightarrow{\tt{p( - 1) = - 1-2\times 1 - 4\times( - 1)-1}}$

$: \longrightarrow{\tt{p( -1 ) = - 1 - 2 + 4 -1}}$

$: \longrightarrow{\tt{p( - 1) =4 - 1-2 -1}}$

$: \longrightarrow{\tt{p( - 1) = 4 - 4 }}$

$\dashrightarrow{\underline {\pink{\boxed{\bf{p(- 1) = 0}}}}}$

$\huge{\red{\therefore}}$Remainder = 0.

▪IN THE SECOND QUESTION :

☯GIVEN :

• $\bf{p(x) =x^3-3x^2+4x+50}$

• $\bf{g(x) = x-3}$

☯TO FIND :

• Remainder.

➲SOLUTION :

• By Remainder Theorem :

Let us assume g(x) = 0.

$: \longrightarrow{ \tt{x-3= 0}}$

$: \longrightarrow{ \tt{x = 0+3}}$

$: \longrightarrow{ \bf{x = 3}}$

✞Substituting the value of x :

$:\longrightarrow{\tt{p(x) =x^3-3x^2+4x+50}}$

$: \longrightarrow{\tt{p( 3) =(3)^3-3(3)^2 +4\times 3+50}}$

$: \longrightarrow{\tt{p( 3) = 27-3\times 9 + 4\times3+50}}$

$: \longrightarrow{\tt{p( 3 ) = 27 - 27 + 12 + 50}}$

$: \longrightarrow{\tt{p( 3) =27 + 12 + 50- 27}}$

$: \longrightarrow{\tt{p( 3) = 89 - 27}}$

$\dashrightarrow{\underline{ \purple{\boxed{\bf{p(3) = 62}}}}}$

$\huge{\red{\therefore}}$Remainder = 62.