Question

factorise using remainder theorem:x^3-13x-12

Answer 1

Hola User,

→ p( x ) = x³ - 13x - 12

=> p( -1 ) = ( -1 )³ - 13( -1 ) -12

=> p( -1 ) = -1 + 13 - 12 = 0

=> By Remainder / Factor theorem, ( x - [ -1 ] ) is a factor

=> p( x ) = ( x + 1 )( x² - x - 12 ) 

♦ p( -3 ) = ( -3 + 1 )( 9 - ( -3 ) - 12 ) = ( -2 )( 0 ) = 0

♦ p( 4 ) = ( 4 + 1 )( 16 - 4 - 12 ) = 5 x 0 = 0

=> p( x ) = ( x + 1 )( x + 3 )( x - 4 )

Answer 2

$\huge\mathfrak\blue{Answer:-}$

$<font color = blue >$

→ p( x ) = x³ - 13x - 12

=> p( -1 ) = ( -1 )³ - 13( -1 ) -12

=> p( -1 ) = -1 + 13 - 12 = 0

=> By Remainder / Factor theorem, ( x - [ -1 ] ) is a factor

=> p( x ) = ( x + 1 )( x² - x - 12 )

• p( -3 ) = ( -3 + 1 )( 9 - ( -3 ) - 12 ) = ( -2 )( 0 ) = 0

• p( 4 ) = ( 4 + 1 )( 16 - 4 - 12 ) = 5 x 0 = 0

=> p( x ) = ( x + 1 )( x + 3 )( x - 4 )

____________________♡