Question

Find the Factor of y³+y²+y+1 by hit and Trial Method​

Answer 1

Answer:

Correct option is B)

The value of y

3

+y

2

−y−1 is

y

3

−y

2

+2y

2

−2y+y−1

=y

2

(y−1)+2y(y−1)+1(y−1)

=(y−1)(y

2

+2y+1)

=(y−1)[y

2

+y+y+1]

=(y−1)[y(y+1)+1(y+1)]

=(y−1)(y+1)

2

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Answer 2

Answer:

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Step-by-step explanation:

The value of y

3

+y

2

−y−1 is

y

3

−y

2

+2y

2

−2y+y−1

=y

2

(y−1)+2y(y−1)+1(y−1)

=(y−1)(y

2

+2y+1)

=(y−1)[y

2

+y+y+1]

=(y−1)[y(y+1)+1(y+1)]

=(y−1)(y+1)

2