Question

Divide The Following Polynomial 4y³+4y²-y-1 , 2y+1​

Answer 1

Given, p(y)=4y

3

+4y

2

−y−1 is a multiply of (2y+1).

To check that 2y+1 is a multiple, the value of 2y+1=0 ⟹ y=−

2

1

.

Now, put y=−

2

1

in expression, we get,

p(−

2

1

)=4(−

2

1

)

3

+4(−

2

1

)

2

−(−

2

1

)−1

⇒p(−

2

1

)=(4×−

8

1

)+(4×

4

1

)+

2

1

−1

⇒p(−

2

1

)=−

2

1

+1+

2

1

−1

⇒p(−

2

1

)=0.

Here, the remainder is zero (0), when 2y+1 is divided the polynomial 4y

3

+4y

2

−y−1 .

Then 2y+1 is factor of the polynomial 4y

3

+4y

2

−y−1 .

Hence, 4y

3

+4y

2

−y−1 is a multiple of 2y+1 .