Find the integral value of $x$ such that $(x^{3} + 7 x^{2} + 14 x + 8) = 0$ .

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Solution

$W e h a v e$

$x^{3} + 7 x^{2} + 14 x + 8 = 0$

$p u t t i n g t h e v a l u e o f x a s - 1$

$t h e n,$

${(- 1)}^{3} + 7 \times {(- 1)}^{2} + 14 \times (- 1) + 8$

$= - 1 + 7 - 14 + 8$

$= - 15 + 15$

$= 0$

$w h e n w e p u t t h e v a l u e o f x = - 1,$

$T h e w h o l e e q u a t i o n g e t s 0, t h e n (x + 1) i s a f a c t o r o f t h i s e q u a t i o n$

$N o w, o n d i v i d i n g t h e e q u a t i o n b y (x + 1)$

$T h e n, w e g e t a q u a d r a t i c e q u a t i o n x^{2} + 6 x + 8.$

$t h e n a f t e r f a c t o r i z e i t w e g e t (x + 2) (x + 4) .$

$t h e n t h e v a l u e o f x = - 2 a n d - 4$

$H e n c e, t h e i n t e r g r a l v a l u e o f x i s - 1.$

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