Given the polynomial p(x)= x3-11x2+17x-6 . If p(x)= (x-2) q(x) + r , Then what is the valur of r and q(x) . Write p(x) as the products of three first degree polynomials .

We have  p ( x ) =  x3 - 11x2 + 17x  - 6 , And

p( x ) = ( x -  2 ) q ( x ) +  r

From remainder theorem , Here we know  At , x  =  2 we get our remainder ( r ) So

p ( 2 )  = 23 - 11 ( 2 )2 + 17 ( 2 ) - 6

p( 2 ) =  8 -44 + 34 - 6  =  - 8

SO,

r  =  - 8  ( Ans )
Now we  get

p ( x ) = ( x -  2 ) q ( x ) +  r  , Now we substitute values of r and p ( x ) , and get

x3 - 11x2 + 17x  - 6 = ( x -  2 ) q ( x ) -  8

( x -  2 ) q ( x ) = x3 - 11x2 + 17x  + 2  , So

q ( x ) =   , No we solve this by long division method , As :

So,
q ( x ) = x2 - 9x - 1       ( Ans )

We solve it by quadratic formula
And

x  = , Here a  =  1  , b  =  - 9  and  c  =  -1  , So

x  = , So

=
And
x  =   , So we have

q ( x ) = x2 - 9x - 1 = ( x  - ) ( x  -

Then we can write p( x ) as multiple of three first  degree polynomial

p ( x ) = ( x -  2 ) ( x  - ) ( x  - ) - 8  ( Ans )

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p(2)= 8-44+34-6= -8

so r = -8

q(x)= x2-9x-1

therefore p(x)=(x-2)(x2-9x-1)+(-8)

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