For what value of k, (x + y + z)^2 + k(x^2 + y^2 + z^2 ) can be resolved into linear rational factors?


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Question Number 20308 by Tinkutara last updated on 25/Aug/17

$For what value of k, {(x + y + z)}^{2} +$ $k (x^{2} + y^{2} + z^{2}) can be resolved into$ $linear rational factors ?$
	Answered by Tinkutara last updated on 28/Aug/17

	${(x + y + z)}^{2} + k (x^{2} + y^{2} + z^{2})$ $= x^{2} + y^{2} + z^{2} + 2 x y + 2 y z + 2 z x + k (x^{2} + y^{2} + z^{2})$ $= z^{2} [{(\frac{x}{z})}^{2} + {(\frac{y}{z})}^{2} + 1 + \frac{2 x y}{z^{2}} + \frac{2 y}{z} + \frac{2 x}{z} + k {(\frac{x}{z})}^{2} + k {(\frac{y}{z})}^{2} + k]$ $= z^{2} [X^{2} + Y^{2} + 1 + 2 X Y + 2 Y + 2 X + {k X}^{2} + {k Y}^{2} + k]$ $= z^{2} [(k + 1) X^{2} + (k + 1) Y^{2} + 2 X Y + 2 Y + 2 X + (k + 1)]$ $Δ = (k + 1) (k + 1) (k + 1) + 2 (1) - (k + 1) - (k + 1) - (k + 1) = 0$ $= {(k + 1)}^{3} - 3 (k + 1) + 2 = 0$ $\Rightarrow k = 0, - 3$
	Commented by ajfour last updated on 28/Aug/17

	$F i n d t h e c o e f f i c i e n t s i f k = - 3 .$
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