Question

is (2y-1) is a factor of 2y³+y²-2y-1?​

Answer 1

Step-by-step explanation:

$answer$

Given :-

$\binom{3}{5} \times \binom{3}{5}$

= 3 × 3 / 5 × 5

= 6 / 25

= $\frac{6}{25}$

Hence, it is proved !!!!!!

Answer 2

Step-by-step explanation:

Given that 2y³+y²-2y-1

=(2y+1)(y+1)(y-1)

Now Factorizing

2y³+y²-2y-1

= y² (2y + 1) -1 (2y + 1)

= (2y + 1)(y² – 1)

= (2y + 1)(y² – 1²)

= (2y + 1) (y + 1) (y – 1)

By algebraic identity:

a² – b² = (a + b) (a – b)

Therefore,

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