<P>ANSWER:

Answer:

(i)

Let the given POLYNOMIAL be,

p(x)=x3−2x2−x+2

By trial method,

p(1)=(1)3−2(1)2−1+2=1−2−1+2=0

So, (x−1) is the FACTOR of the given polynomial.

Divide the given polynomial by (x−1) by long division.

x−1x2−x−2x3−2x2−x+2x3−x2−−−−−−     −x2−x     −x2+x−−−−−−−−−            −2x+2            −2x+2−−−−−−−−−−−                      0−−−−−−−−

It is known that,

Dividend=Divisor×Quotient+Remainderx3−2x2−x+2=(x−1)(x2−x−2)+0=(x−1)(x2−2x+x−2)=(x−1)[x(x−2)+1(x−2)]

Further simplify,

x3−2x2−x+2=(x−1)(x+1)(x−2)

(ii)

Let the given polynomial be,

p(y)=2y3+y2−2y−1

By trial method,

p(1)=2(1)3+(1)2−2(1)−1=2+1−2−1=0

So, (y−1) is the factor of the given polynomial.

Divide the given polynomial by (y−1) by long division.

y−12y2+3y+12y3+y2−2y−12y3−2y2−−−−−−−−         3y2−2y         3y2−3y−−−−−−−−−−                   y−1                   y−1−−−−−−−−−−                        0−−−−−−−−

It is known that,

Dividend=Divisor×Quotient+Remainder2y3+y2−2y−1=(y−1)(2y2+3y+1)+0=(y−1)(2y2+2y+y+1)=(y−1)[2y(y+1)+1(y+1)]

Further simplify,

2y3+y2−2y−1=(y−1)(y+1)(2y+1)