Question

please solve this question by remainder theorem

Answer 1

I am only doing 1st of 1s question 
1. p(x)= x^3+3x^2+3x+1
(i) g(x)=x+1
          zero of g(x)=x+1=0
                        is, x=-1
    p(-1)= -1^3+3*-1^2+3*-1+1
            =-1+3-3+1
            =0
 Remainder= 0    

2. p(x)=3x^3-ax^2+6x-a
    g(x)=x-a
         zero of g(x)=x-a=0
               is, x=a
p(a)=a^3-a*a^2+6a-a
      =5a

Answer 2

☛ Factor Theorem ;

If p ( x ) is a polynomial of degree ≥ one and a is any real number , then

( i ) ( x - a ) is a factor of p ( x ) , if p ( a ) = 0.

( ii ) p ( a ) = 0, if ( x - a ) is a factor of p (x).


☛ Remainder Theorem ;

Division and Algorithm for polynomials : If p(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find q ( x ) and r (x) and such that

p ( x ) = g ( x ) * q ( x ) + r ( x ), where
r (x) = 0 (or) deg r(x) < deg g(x)

Dividend = Divisor × quotient + remainder