Question

using factor theorem factorise the following (a) x³-3x²-9x-5

Answer 1

$\textbf{Answer:}$

➡ x³ - 3x² -9x - 5

= x³ + x² - 4x² - 4x - 5x - 5

= x²(x + 1) - 4x(x + 1) - 5(x - 1)

= (x + 1)(x² - 4x - 5) ----【Taking (x + 1) common】

= (x + 1)(x² - 5x + x - 5)

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

= (x + 1)(x + 5)

Therefore,(x + 1) & (x + 5) are the factors of the above mentioned polynomial.
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Answer 2

Answer:

Answer: x³ - 3x² -9x - 5

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

Answer: x³ - 3x² -9x - 5= x³ + x² - 4x² - 4x - 5x - 5= x²(x + 1) - 4x(x + 1) - 5(x - 1)

Answer: x³ - 3x² -9x - 5= x³ + x² - 4x² - 4x - 5x - 5= x²(x + 1) - 4x(x + 1) - 5(x - 1)= (x + 1)(x² - 4x - 5) ----【Taking (x + 1) common】

Answer: x³ - 3x² -9x - 5= x³ + x² - 4x² - 4x - 5x - 5= x²(x + 1) - 4x(x + 1) - 5(x - 1)= (x + 1)(x² - 4x - 5) ----【Taking (x + 1) common】= (x + 1)(x² - 5x + x - 5)

Answer: x³ - 3x² -9x - 5= x³ + x² - 4x² - 4x - 5x - 5= x²(x + 1) - 4x(x + 1) - 5(x - 1)= (x + 1)(x² - 4x - 5) ----【Taking (x + 1) common】= (x + 1)(x² - 5x + x - 5)= (x + 1)(x + 5)(x + 1)

Answer: x³ - 3x² -9x - 5= x³ + x² - 4x² - 4x - 5x - 5= x²(x + 1) - 4x(x + 1) - 5(x - 1)= (x + 1)(x² - 4x - 5) ----【Taking (x + 1) common】= (x + 1)(x² - 5x + x - 5)= (x + 1)(x + 5)(x + 1)= (x + 1)(x + 5)

Answer: x³ - 3x² -9x - 5= x³ + x² - 4x² - 4x - 5x - 5= x²(x + 1) - 4x(x + 1) - 5(x - 1)= (x + 1)(x² - 4x - 5) ----【Taking (x + 1) common】= (x + 1)(x² - 5x + x - 5)= (x + 1)(x + 5)(x + 1)= (x + 1)(x + 5)Therefore,(x + 1) & (x + 5) are the factors of the above mentioned polynomial.