Question

A) divide the polynomial 3x⁴+4x³+3x+1 by x-1 and find the remainder and quotient ,B) write any two solutions for the following equation: 3x+4y=12 ​

Answer 1

$\huge\bf\red{Answer ☃}</p><p>$

3x + 4y = 12 ----( 1 )

i ) Substitute x = 0 , in equation ( 1 ) ,

3 × 0 + 4y = 12

=> 4y = 12

=> y = 12/4

y = 3

Therefore ,

( x , y ) = ( 0 , 3 )

ii ) Substitute y = 0 in equation ( 1 ) ,

3x + 4 × 0 = 12

=> 3x = 12

=> x = 12/3

x = 4

Therefore ,

( x , y ) = ( 4 , 0 )

Required two solutions are ( 0 , 3 ) ,

( 4 , 0 )

••••

I hope it will help you army✌️☺️

Answer 2

Answer:

Hope it helpful to you army