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 .

Answer :

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 ) = $\frac{x^3 - 11x^2 + 17x + 2 }{\left( x - 2 \right)}$  , No we solve this by long division method , As :

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

We solve it by quadratic formula
And
We Know quadratic formula 

x  = $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , Here a  =  1  , b  =  - 9  and  c  =  -1  , So

x  = $\frac{-\left( - 9\right) \pm \sqrt{{\left( - 9\right)}^2 - 4\times 1 \times \left( - 1\right)}}{2 \times 1} = \frac{9\pm \sqrt{81+ 4}}{2 } = \frac{9\pm \sqrt{85}}{2 }$ , So

x  = $\frac{9+ \sqrt{85}}{2 }$
And
x  = $\frac{9 - \sqrt{85}}{2 }$  , So we have


q ( x ) = x² - 9x - 1 = ( x  - $\frac{9+ \sqrt{85}}{2 }$ ) ( x  - $\frac{9 - \sqrt{85}}{2 }$ ) 

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

p ( x ) = ( x -  2 ) ( x  - $\frac{9+ \sqrt{85}}{2 }$ ) ( x  - $\frac{9 - \sqrt{85}}{2 }$ ) - 8  ( Ans )