Factorize :
i)\(x^3 - 2x^2 - x + 2\)
ii)\(x^3 - 3x^2 -9x - 5\)
iii)\(x^3 + 13x^2 + 32x + 20\)
iv)\(2y^3 + y^2 - 2y - 1\)

i)Let \(p(x) = x^3 - 2x^2 - x + 2\)

Here, constant term is 2. So, finding the factors of +2, we get, ±1 and ±2.

Now, \(p(1) = (1)^3 - 2(1)^2 - (1) + 2\)
\(= 1 - 2 - 1 + 2\)
= 0

Therefore, by trial method, we find that,
p(1) = 0.

Hence, x - 1 is a factor of p(x).

\(\therefore x^3 - 2x^2 - x + 2 \)
\( = x^3 - x^2 - x^2 + x - 2x + 2\)
\(= x^2(x - 1) -x(x - 1) - 2(x - 1)\)
\(= (x - 1)(x^2 - x - 2)\)
now, splitting the middle term,
\(= (x - 1)(x^2 - 2x + x - 2)\)
\(= (x - 1)[x(x - 2) + 1(x - 2)]\)
\(= (x - 1)(x - 2)(x + 1)\)


ii)Let \(p(x) = x^3 - 3x^2 -9x - 5\)
Here, constant term is -5. So, finding the factors of -5, we get, 1 and -5 or vice versa.

Now, \(p(5) = (5)^3 - 3(5)^2 - 9(5) - 5\)
\(= 125 - 75 - 45 - 5\)
= 0

Therefore, by trial method, we find that, p(5) = 0.

Hence, x - 5 is a factor of p(x).

\(\therefore x^3 - 3x^2 -9x - 5 \)
\( = x^3 - 5x^2 + 2x^2 - 10x + x - 5\)
\(= x^2(x - 5) + 2x(x - 5) + 1(x - 5)\)
\(= (x - 5)(x^2 + 2x + 1)\)
now, splitting the middle term,
\(= (x - 5)(x^2 + x + x + 1)\)
\(= (x - 5)[x(x + 1) + 1(x + 1)]\)
\(= (x - 5)(x + 1)^2\)


iii)Let \(p(x) = x^3 + 13x^2 + 32x + 20\)

Here, constant term is +20. So, finding the factors of 20, we get, ±1, ±2, ±4 and ±5 or vice versa.

Now, \(p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20\)
\(= -1 + 13 - 32 + 20\)
= 0
Therefore, by trial method, we find that, p(-1) = 0.

Hence, x + 1 is a factor of p(x).

\(\therefore x^3 + 13x^2 + 32x + 20 \)
\( = x^3 + 2x^2 + 11x^2 + 22x + 10x + 20\)
\(= x^2(x + 2) + 11x(x + 2) + 10(x + 2)\)
\(= (x + 2)(x^2 + 11x + 10)\)
now, splitting the middle term,
\(= (x + 2)(x^2 + 10x + x + 10)\)
\(= (x + 2)[x(x + 10) + 1(x + 10)]\)
\(= (x + 2)(x + 1)(x + 10)\)


iv)Let \(p(x) = 2y^3 + y^2 - 2y - 1\)

Here, constant term is -1. So, finding the factors of -1, we get, -1.

Now, \(p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1\)
\(= -2 + 1 + 2 - 1\)
= 0

Therefore, by trial method, we find that, p(-1) = 0.

Hence, x + 1 is a factor of p(x).

\(\therefore 2y^3 + y^2 - 2y - 1 \)
\( = 2y^3 + 2y^2 - y^2 - y - y -1\)
\(= 2y^2(y + 1) - y(x + 1) - 1(y + 1)\)
\(= (y + 1)(2y^2 - y - 1)\)
now, splitting the middle term,
\(= (y + 1)(2y^2 - 2y + y - 1)\)
\(= (y + 1)[2y(y - 1) + 1(y - 1)]\)
\(= (y - 1)(y + 1)(2y + 1)\)