Question

factorise:-2 y cube + y square - 2y - 1 ​

Answer 1

Answer:

prefer to it !!!!!!!!!!

Answer 2

2y³+y²-2y-1

let's take factor (y+1)

therefore,

p(y)=2y³+y²-2y-1

p(y)= y+1 =0

i.e y= -1

therefore,

p(-1) = 2(-1)³+(-1)²-2(-1)-1

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

= -2+3-1

= -3+3

p(-1)= 0

therefore (y+1) is the factor

now,

let's divide the sum

i.e 2y³+y²-2y-1 ÷ (y+1)

y+1/2y³+y²-2y-1 ..........(pretend the slash[/] as divide sign)

y+1/2y³+y²-2y-1 ........(so first divide y from 2y³)

.........[2y³/y=2y²] now multiply it by the factor

.........[ 2y²(y+1)=2y³+2y²]

therefore,

2y²-y-1

_________

y+1/2y³+y²-2y-1

2y³+2y²

- -

_________

0-y²-2y

-y²-y ...............(-y²/y=-y)

+ + ........[ -y(y+1)=y²-y]

___________

0 -y-1 ....................... -y/y=-1

-y-1

+ + ...........-1(y+1)= -y-1

________

0

now we got Q=2y²-y-1

and R=0

so now we can factorise

2y²-y-1 ......... 2 x 1 =2

/ \

2 1

therefore

2y²-2y+1y -1

now making pairs and taking common

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

so, (2y+1)(y-1)

so the factors of 2y³+y²-2y-1 = (y+1)(2y+1)(y-1)