Algebra
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Q.

(x^3-6x^2+px+q) is exactly divisible by (x^2-3x+2) then,

 A.

$p+q > 0$ and $pq > 0$

 B.

$p+q > 0$ and $pq < 0$

 C.

$p+q < 0$ and $pq < 0$

 D.

$p+q < 0$ and $pq > 0$

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Solution:
Option(B) is correct

$x^2−3x+2=(x−2)(x−1)$

$\Rightarrow$ The factors of the cubic equation is $x =2$ and $x= 1$

On putting $x=2$, we get, $8−24+2p+q=0$

$\Rightarrow 2p+q=16$ -------- (i)

On putting $x=1$, we get, $1−6+p+q=0$ 

$\Rightarrow p+q=5$ -------- (ii)

Thus on solving these two equation we get: 

$p =11$ and $q = -6$

So p+q > 0 and pq < 0


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