Question

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

Answer 1

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

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

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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)`