Question

Factorise x^3-3x^2-9x-5 using factor theorem

Answer 1

GIVEN :

Factorise $x^3-3x^2-9x-5$ using factor theorem

TO FIND :

The factors for the given polynomial

SOLUTION :

Given that the polynomial is $x^3-3x^2-9x-5$

Factor theorem states that   A polynomial f(x) has a factor x-a iff f(a)=0

By using Factor theorem we can find the factors :

Let $f(x)=x^3-3x^2-9x-5$

Put x=1 in f(x) we get

$f(1)=1^3-3(1)^2-9(1)-5$

=1-3-9-5

$=16\neq 0$

$f(1)\neq 0$

∴ x-1 is not a factor of f(x).

Put x=-1 in f(x) we get

$f(-1)=(-1)^3-3(-1)^2-9(-1)-5$

=-1-3+9-5

=0

∴ f(-1)=0

∴ x+1 is a factor of f(x).

By using the Synthetic division

 

-1 |   1       -3      -9     -5

      0       -1      4        5

    _______________

      1       -4      -5     0

∴ x+1 is a factor of f(x).

Now we have quadratic equation $x^2-4x-5=0$

Now factorise the equation $x^2-4x-5=0$

$x^2-4x-5=0$

(x+1)(x-5)=0

x+1=0 or x-5=0

x=-1 and x=5 are the zeroes.

∴ (x+1) and (x-5) are also the factors of f(x).

⇒ $x^3-3x^2-9x-5=(x+1)(x+1)(x-5)$

∴ the given polynomial is factorised by using Factor theorem is $x^3-3x^2-9x-5=(x+1)(x+1)(x-5)$

Answer 2

$= (x + 1) \: (x - 5) \: (x + 1)$