# Factorize:</p

(i) Let p(x) = x3 – 2x2 – x + 2

By trial method, we find that

p (-1) = (-1)3 – 2 (-1)2 – (-1) + 2

= -1 – 2 + 1 + 2 = 0

Therefore,  x= -1  satisfies the equation and (x + 1) is the factor of p (x)
Now, we divide  p(x) =  x3  - 2x - x + 2  by  (x + 1) Now, Dividend = (Divisor × Quotient) + Remainder

p (x) = (x + 1) (x2 – 3x + 2)

= (x + 1) (x2 – x – 2x + 2)

= (x + 1) {x (x – 1) – 2 (x – 1)}

= (x + 1) (x – 1) (x - 2)

(ii) Let p (x) = x3 – 3x2 – 9x – 5

By trial method,

p (5) = (5)3 – 3 (5)2 – 9 (5) – 5

= 125 – 75 – 45 – 5 = 0

Therefore, (x – 5) is the factor of p (x)

Now, we divide p(x) = x3 - 3x2 - 9x - 5  by (x - 5) Now, Dividend = (Divisor × Quotient) + Remainder

p(x) = (x – 5) (x2 + 2x + 1)

= (x – 5) (x2 + x + x + 1)

= (x – 5) {x (x + 1) + 1 (x + 1)}

= (x – 5) (x + 1) (x + 1)

(iii) Let p (x) = x3 + 13x2 + 32x + 20

By trial method,

p (-1) = (-1)3 + 13 (-1)2 + 32 (-1) + 20

= -1 + 13 – 32 + 20 = 0

Therefore,  x = -1 satisfies the equation p(x) = 0  so, (x + 1) is the factor of p (x)

Now we divide p (x) = x3 + 13x2 + 32x + 20  by (x + 1) Now, Dividend = (Divisor × Quotient) + Remainder

p(x) = (x + 1) (x2 + 12x + 20)

= (x + 1) (x2 + 2x + 10x + 20)

= (x + 1) {x (x + 2) + 10 (x + 2)}

= (x + 1) (x + 2) (x + 10)

(iv) Let, p(y) = 2y3 + y2 – 2y – 1

By trial method,

p (1) = 2 (1)3 + (1)2 – 2 (1) – 1

= 2 + 1 -1 -1 = 0

Therefore, y = 1 satisfies the equation p(y) = 0   so, (y – 1) is a factor of p(y)
Now, we divide  p(y) = 2y3 + y2 – 2y – 1  by (y - 1) Now, Dividend = (Divisor × Quotient) + Remainder

p(y) = (y – 1) (2y2 - 3y + 1)

= (y – 1) (2y2 - 2y - y + 1)

= (y – 1) {2y (y - 1) - 1 (y - 1)}

= (y – 1) (2y - 1) (y - 1)

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