Factorize:

(i)


(ii)


(iii)


(iv)


(i) Let p (x) = x3 – 2x2 – x + 2

Factors of 2 are ±1 and ±2


By trial method, we find that


p (1) = 0


So, (x + 1) is a factor of p (x)


Now,


p (x) = x3 – 2x2 – x + 2


p (-1) = (-1)3 – 2 (-1)2 – (-1) + 2


= -1 – 2 + 1 + 2 = 0


Therefore, (x + 1) is the factor of p (x)



Now, Dividend = Divisor * Quotient + Remainder


(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


Factors of 5 are ±1 and ±5


By trial method, we find that


p (5) = 0


So, (x – 5) is a factor of p(x)


Now,


p (x) = x3 – 2x2 – x + 2


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, Dividend = Divisor * Quotient + Remainder


(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


Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20


By trial method, we find that


p (-1) = 0


So, (x + 1) is factor of p (x)


Now,


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


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


= -1 + 13 – 32 + 20 = 0


Therefore, (x + 1) is the factor of p (x)



Now, Dividend = Divisor * Quotient + Remainder


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


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


= (x – 5) {x (x + 2) + 10 (x + 2)}


= (x – 5) (x + 2) (x + 10)


(iv) Let, p(y) = 2y3 + y2 – 2y – 1


Factors of ab = 2 * (-1) = -2 are ±1 and ±2


By trial method, we find that


p (1) = 0


So, (y – 1) is a factor of p (y)


Now,


p (y) = 2y3 + y2 – 2y – 1


p (1) = 2 (1)3 + (1)2 – 2 (1) – 1


= 2 + 1 -1 -1 = 0


Therefore, (y – 1) is the factor if p (y)



Now, Dividend = Divisor * Quotient + Remainder


(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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