Question

24. Find the factors of the given 1 point
polynomial
2ycube + y2 – 2y-1​

Answer 1

Answer:

(y-1) (y+1) (2y+1) this is the answer

Answer 2

Step-by-step explanation:

p(1) = 2(1)³+1²-2(1)-1

= 2+1-2-1

= 0

By Factor theorem,

(y-1) is a factor of p(y).

2y³+y²-2y-1

=2y²(y-1)+3y(y-1)+1(y-1)

=(y-1)(2y²+3y+1)

=(y-1)(2y²+2y+1y+1)

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

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