12. Find the value of k so that $x^{3} - 3 x^{2} + 4 x + k$ is exactly divisible by x - 2.

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Solution

Dear Student,

$l e t p (x) = x^{3} - 3 x^{2} + 4 x + k \sin c e p (x) i s d i v i s i b l e b y x - 2 H e n c e p (x) = 0 a t x = 2 t h e r e f o r e p (2) = 0 p u t t i n g v a l u e o f x = 2 i n x^{3} - 3 x^{2} + 4 x + k w e g e t 2^{3} - 3 \times 2^{2} + 4 \times 2 + k = 0 \Rightarrow 8 - 3 \times 4 + 8 + k = 0 \Rightarrow 8 - 12 + 8 + k = 0 \Rightarrow 16 - 12 + k = 0 \Rightarrow k + 4 = 0 \Rightarrow k = - 4 H e n c e v a l u e o f k = - 4$
Regards,

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