Question

factorise : 4x^3-4x^2-21x-9​

Answer 1

Answer:

Dear Student,

where p = factor of constant (9) and q = factor of leading coefficient (4)

p = ±1, ±3, ±9

q = 1, 2, 4

Using trial and error, we find that 3 is a root

Now we use synthetic division to find quadratic factor:

3 | 4 -4 -21 -9

.. | .. 12 24. 9

.. ----------------

... 4 . 8 . 3 . 0

So dividing (4x^3 - 4x^2 - 21x - 9) by (x - 3), we get (4x^2 + 8x + 3)

(4x^3 - 4x^2 - 21x - 9) = 0

(x - 3) (4x^2 + 8x + 3) = 0

(x - 3) (4x^2 + 2x + 6x + 3) = 0

(x - 3) (2x (2x + 1) + 3 (2x + 1)) = 0

(x - 3) (2x + 1) (2x + 3) = 0

x = 3, -1/2, -3/2

Regards!