Question

2 y cube + y square - 2 Y - 1 factorise

Answer 1

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)

Answer 2

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)