If $x^{3} - 3 x^{2} - 9 x - 5 = (x + a)^{b} (x - c)$ .Find $a + b + c$

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Solution

$x^{3} - 3 x^{2} - 9 x - 5 c l e a r l y x = - 1 i s o n e o f t h e z e r o e s o f t h e g i v e n p o l y n o m i a l = (x + 1) (x^{2} - 4 x - 5) = (x + 1) [x^{2} - 5 x + x - 5] = (x + 1) [x (x - 5) + 1 (x - 5)] = (x + 1) (x - 5) (x + 1) = (x + 1)^{2} (x - 5) c o m p a r e i t w i t h (x + a)^{b} (x - c) w e g e t a = 1, b = 2 a n d c = 5 ∴ a + b + c = 8$

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Similar questions

Solve: $x^{3} - 3 x^{2} - 9 x - 5$

x^{3} + 3 x^{2} - 9 x + 5

(x - 1) (x^{2} + a x - 5)

a

x^{3} + 3 x^{2} - 9 x + c = 0

c .

x^{3} + 3 x^{2} - 9 x + c

(x - α)^{2} (x - β)

c

x^{3} - 3 x^{2} + 3 x - 7 = (x + 1) (a x^{2} + b x + c)

a + b + c

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