Question

factorise x^3-3x^2-9x-5​

Answer 1

p(x) = x3 - 3x2 - 9x - 5 Factors of 5 = 1,5,-1,-5 p(1) = 13 - 3 x (1)2 - 9 x (1) - 5 = 1 -3 - 9 - 5 = 1 - 17 = - 16 ≠ 0 p(-1) = (-1)3 x 3 x (-1)2 - 9 x (-1) -5 = -1 - 3 + 9 - 5 = -9 + 9 = 0 p(-1) =0 ∴ (x+1) is a factor of p(x) p(x) ÷ (x+1) x3 - 3x2 - 9x - 5 ÷ x + 1 = x2 - 4x - 5 x2 - 4x - 5 x2 - 5x + 1x - 5 = x ( x - 5) + 1 ( x - 5) = ( x - 5) ( x + 1)

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Here we have :-

p(x) = x³ - 3x² - 9x - 5

Constant = 5

$\Large{\boxed{\sf\:{Numerical\;value\;of\;(-5) = 5}}}$

Factors of 5 = ±1 , ±5

Hence,

p(-1) = (-1)³ - 3(-1)² - 9(-1) - 5

= -1 - 3 + 9 - 5

= -4 + 4

= 0

Therefore,

$\Large{\boxed{\sf\:{(x + 1)\;is\;factor\;of\;p(x)}}}$

x³ - 3x² - 9x - 5 = x³ + x² - 4x² - 4x - 5x - 5

= (x³ + x²) - (4x² + 4x) - (5x + 5)

= x²(x + 1) - 4x(x + 1) - 5(x + 1)

= (x + 1)(x² - 4x - 5)

= (x + 1)(x² - 5x + x - 5)

= (x + 1){(x² - 5x) + (x - 5)}

= (x + 1){x(x - 5) + 1(x - 5)}

= (x + 1)(x + 1)(x - 5)