Question

Please factorise (2y^3-5y^2-19y+42)

Answer 1

Answer:

Step-by-step explanation:

The following is the process of solving the solution in a perfect manner.Shown all the steps...No Shortcut method used

Answer 2

Therefore the factorisation of the given cubic polynomial  2y³ - 5y² - 19y + 42 = 0 is ( y + 3 )( 2y -7 )( y - 2 ) = 0

Given:

2y³ - 5y² - 19y + 42 = 0

To Find:

Factorise the given polynomial 2y³ - 5y² - 19y + 42 = 0

Solution:

We can simply solve this numerical problem by using the following process.

Given that f ( y ) = 2y³ - 5y² - 19y + 42

First of all let us check any one root of the polynomial.

Suppose y = -3

Then f ( -3 ) = 2 ( -3 )³ - 5 ( -3 )² - 19 ( -3 ) + 42

⇒ f ( -3 ) = -2*27 - 5*9 + 57 + 42

⇒ f ( -3 ) = -54 - 45 + 57 + 42

⇒ f ( -3 ) = -99 + 99

⇒ f ( -3 ) = 0

Hence, ( y + 3 ) is a factor of 2y³ - 5y² - 19y + 42 = 0

On dividing 2y³ - 5y² - 19y + 42 = 0 by its factor ( y + 3 ) we will get a quadratic equation 2y² - 11y + 14

On Factorising, 2y² - 11y + 14 = 0

⇒ $2y^2 - 11y + 14 = 0$

⇒ $2y^2 - 4y - 7y + 14 = 0$

⇒ $2y ( y - 2 ) - 7 ( y - 2 ) =0$

⇒ $( 2y -7 )( y -2 ) = 0$

Therefore the factorisation of the given cubic polynomial  2y³ - 5y² - 19y + 42 = 0 is ( y + 3 )( 2y -7 )( y - 2 ) = 0.

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