For-what-value-of-k-can-be-expression-x-3-kx-2-7x-6-be-resolved-into-three-linear-factors-a-0-b-1-c-2-d-3-

Question Number 205018 by BaliramKumar last updated on 06/Mar/24

For what value of^{'} k^{'} can be expression x^{3} + {k x}^{2} - 7 x + 6

be resolved into three linear factors ?

(a) 0 (b) 1 (c) 2 (d) 3

Commented by Rasheed.Sindhi last updated on 06/Mar/24

(a) k = 0

Commented by mr W last updated on 06/Mar/24

q u e s t i o n i s n o t c l e a r . i n c o m p l e x

n u m b e r s w i t h a n y r e a l v a l u e o f k

t h e e x p r e s s i o n c a n a l w a y s b e r e s o l v e d

i n t o t h r e e l i n e a r f a c t o r s . i n r e a l

n u m b e r s t h e e x p r e s s i o n c a n b e

r e s o l v e d i n t o t h r e e l i n e a r f a c t o r s

i f k_{1} < k < k_{2} .

Answered by Rasheed.Sindhi last updated on 06/Mar/24

(a) k = 0

x^{3} + {k x}^{2} - 7 x + 6 \Rightarrow x^{3} - 7 x + 6

x^{3} - 1 - 7 x + 7

(x - 1) (x^{2} + x + 1) - 7 (x - 1)

(x - 1) (x^{2} + x - 6)

(x - 1) (x + 3) (x - 2)

F o r k = 1, 2, 3 t h e e x p r e s s i o n h a s

n o l i n e a r f a c t o r .

Commented by mr W last updated on 06/Mar/24

l i n e a r f a c t o r s (x + a) (x + b) (x + c)

{d o n}^{'} t r e q u e s t t h a t a, b, c s h o u l d b e

i n t e g e r n u m b e r s . t h e y m a y b e r e a l

n u m b e r s o r e v e n c o m p l e x n u m b e r s .

Commented by Rasheed.Sindhi last updated on 06/Mar/24

\boldsymbol Sir, I considered only numbers given

as options :

(a) k = 0 (b) k = 1 (c) k = 2 (d) k = 3

Commented by mr W last updated on 06/Mar/24

y e s . f r o m t h e s e 4 g i v e n o p t i o n s o n l y

k = 0 i s s u i t a b l e .

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