Question

factorise x3-3x3-9x-5 into linear factors note 2nd term is 3 x square​

Answer 1

Answer:  x3 – 3x2 – 9x – 5

               We have p(x) = x3 – 3x2 – 9x – 5

               By trial, let us find:

               p(1) = (1)3 – 3(1)2 – 9(1) –5

               = 3 – 3 – 9 – 5

               = –14 ≠ 0

               Now p(–1) = (–1)3 – 3(–1)2 – 9(–1) –5

               = –1 – 3(1) + 9 – 5

               = –1 – 3 + 9 – 5 = 0

               ∴ By factor theorem, [x – (–1)] is a factor of p(x).

               Now,  

              x3-3x2-9x-5/x-(-)=x2-4x-5

               ∴ x3 – 3x2 – 9x – 5 = (x + 1)(x2 – 4x – 5)

               = (x + 1)[x2 – 5x + x – 5]

               [Splitting –4 into –5 and +1]

               = (x + 1)[x(x – 5) +1(x – 5]

               = (x + 1)[(x – 5) (x + 1)]

               = (x + 1)(x – 5)(x + 1)

Please mark as the Brainliest Answer...................

Answer 2

Answer :

= ${\sf{{x}^{3}-{3x}{3}-9x-5}}$

= ${\sf{{1x}^{3}-{3x}{3}-9x-5}}$

= ${\sf{{1x - 3x}{3}-9x-5}}$

= ${\sf{{-2x}{3}-9x-5}}$

That's the answer !!

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