Question

factorise the following question​

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given:

→ f(x) = x³ - 3x² - 9x - 5

⊕ We can factorize the expreession by using Factor Theorem.

Factor Theorem: If f(x) is a polynomial and α be a real number, then (x - α) is a factor of f(x) if f(α) = 0.

Put x = - 1:

→ f(-1) = (-1)³ - 3 × (1)² - 9 × (-1) - 5

→ f(-1) = -1 - 3 + 9 - 5

→ f(-1) = 9 - 9

→ f(-1) = 0

As f(-1) = 0, (x - (-1)) is a factor of f(x), i.e., (x + 1) is a factor of f(x).

Divide f(x) by (x + 1):

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

        x³ +   x²

----------------------------------------

              -4x² - 9x

              -4x² - 4x

----------------------------------------

                       -5x - 5

                       -5x - 5

----------------------------------------

                             0

Therefore:

→ f(x) = (x + 1)(x² - 4x - 5)

        = (x + 1){x² - 5x + x - 5}

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

        = (x + 1)(x + 1)(x - 5)

        = (x + 1)²(x - 5) (Answer)

$\texttt{\textsf{\large{\underline{Answer}:}}}$

• Factorised form - (x + 1)²(x - 5).

Answer 2

Let p(x) = x³ - 3x² - 9x - 5

All the factors of 5 have to be considered.

These are +1, + 5.

By trial method,

p(-1)=(-1)³ - 3(-1)²-9(-1)-5 = -1 -3 +9-5

= 0

Therefore, x + 1 is a factor of this polynomial.

Let us find the quotient on dividing x³ + 3x²9x5 by x + 1.

By long division: GIVEN IN ATTACHMENT.