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


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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

	$Sir, I considered only numbers given$ $as options :$ $(a) k = 0 (b) k = 1 (c) k = 2 (d) k = 3$
	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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