Question

The factors of x cube -10x square -53x-42

Answer 1

Answer:

$x^3 - 10 x^2 -53x -42 = ( x + 1)(x + 3) (x - 14)$

Step-by-step explanation:

 $x^3 - 10x^2 -53x - 42$

The factors of 42 : ±1 , ±2 , ±3 , ±6, ±7, ±14, ±21 , ±42

Using trial and error method we will find the first root of the polynomial.

1 : ( 1 )³ - 10 ( 1 )² - 53 ( 1 ) - 42

     1 - 10 - 53 - 42  ≠ 0

   Therefore 1 is not a root

-1 : ( -1 )³ - 10 (- 1 )² - 53 (- 1 ) - 42

      - 1 - 10 + 53 - 42

       53 - 53 = 0

Therefore - 1 is a root of the polynomial.

Therefore ( x + 1) is a factor.

Now by long division or by using synthetic division we can find other factors.

Synthetic Division :

                             -1 | 1      -10       -53      - 42

                                | 0     - 1          11          42

                                |______________________

                                  1      - 11       - 42        0

Therefore ,

$x^3 - 10 x^2 -53x -42 = ( x + 1)(x^2 - 11x +42)$  ------ ( 1 )

Next further factorize x² - 11x - 42

   x² - 11x - 42

= x² -  14x + 3x - 42

=x(x - 14) + 3 (x - 14)

=(x + 3 )(x - 14)

Therefore ,

$x^3 - 10 x^2 -53x -42 = ( x + 1)(x + 3) (x - 14)$