Question

factorise 2y³+y²-2y-1 in trial method,division method and remainder theorem

Answer 1

Step-by-step explanation:

Given that 2y³ + y² – 2y – 1

Now Factorizing

2y³ + y² – 2y – 1

= y² (2y + 1) -1 (2y + 1)

= (2y + 1)(y² – 1)

= (2y + 1)(y² – 1²)

By algebraic identity:

a² – b² = (a + b) (a – b)

(y² – 1²) = (y + 1)(y – 1)

(y² – 1²) can be written as (y + 1) (y – 1)

∴ (2y + 1)(y² – 1²)= (2y + 1) (y + 1) (y – 1)

Therefore,

2y³ + y² – 2y – 1 = (2y + 1) (y + 1) (y-1)