Math, asked by abhijeet0011, 10 months ago

factorise the following.....
2y²-y-1

Answers

Answered by SwaggerGabru
1

Answer:

2y³ + y² - 2y - 1

= 2y³ + 2y² -y² - y - y -1

= 2y²(y + 1) -y(y + 1) - 1(y + 1)

= (y +1 )(2y² - y - 1)

= (y + 1){2y² - 2y + y - 1}

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

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

hence, factors of 2y³ + y² -2y -1 are (y +1 ) , (2y + 1) and (y - 1)

Answered by aksingh57
9

2y^2-y-1

2y^2-(2-1)y-1

2y^2-2y+y-1

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

(y-1)(2y+1)

Hope it helps you....

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