Factories :
(iv) 2ycube + y square - 2y - 1
Answers
Answer:THE FACTORS ARE (y+1),(2y-1),(2y+1)
Step-by-step explanation:
Given - 2y³+y²-2y-1
Now by trial method
2(-1)³+(-1)²-2(-1)-1 (by putting y=-1)
= -2+1+2-1
= 0
Therefore, -1 is the zero of the polynomial
Since y+1 is a factor of the given polynomial
Now by long division method -
2y³+y²-2y-1/y+1
= 2y²-y-1 (sorry I'm not able to do full steps :(. )
Now by middle term spilit we get,
2y²+y²-1
= 2y²+(2-1)y-1
= 2y²+2y-1y-1
= 2y(y+1)-1(y+1)
= (2y-1) (2y+1)
Therefore the required factors are (y+1),(2y-1),(2y+1)
HOPE IT HELPS IF IT HELPS THAN PLZ GIVE 5 AND STAY HAPPY
Answer:
Step-by-step explanation:
Step-by-step explanation:
Given - 2y³+y²-2y-1
Now by trial method
2(-1)³+(-1)²-2(-1)-1 (by putting y=-1)
= -2+1+2-1
= 0
Therefore, -1 is the zero of the polynomial
Since y+1 is a factor of the given polynomial
Now by long division method -
2y³+y²-2y-1/y+1
= 2y²-y-1 (sorry I'm not able to do full steps :(. )
Now by middle term spilit we get,
2y²+y²-1
= 2y²+(2-1)y-1
= 2y²+2y-1y-1
= 2y(y+1)-1(y+1)
= (2y-1) (2y+1)
Therefore the required factors are (y+1),(2y-1),(2y+1)
HOPE IT HELPS IF IT HELPS THAN PLZ GIVE 5 AND STAY HAPPY