Answer :

(i) Let, f (x) = x3+13x2+31x-45

Given that (x + 9) is a factor of f (x)


Let us divide f (x) by (x + 9) to get the other factors


By using long division method, we have


f (x) = x3+13x2+31x-45


= (x + 9) (x2 + 4x – 5)


Now,


x2 + 4x – 5 = x2 + 5x – x – 5


= x (x + 5) – 1 (x + 5)


= (x – 1) (x + 5)


f (x) = (x + 9) (x + 5) (x – 1)


Therefore, x3+13x2+31x-45 = (x + 9) (x + 5) (x – 1)


(ii) Let, f (x) = 4x3+20x2+33x+18


Given that (2x + 3) is a factor of f (x)


Let us divide f (x) by (2x + 3) to get the other factors


By long division method, we have


4x3+20x2+33x+18 = (2x + 3) (2x2 + 7x + 6)


2x2 + 7x + 6 = 2x2 + 4x + 3x + 6


= 2x (x + 2) + 3 (x + 2)


= (2x + 3) (x + 2)


4x3+20x2+33x+18 = (2x + 3) (2x + 3) (x + 2)


= (2x + 3)2 (x + 2)


Hence,


4x3+20x2+33x+18 = (2x + 3)2 (x + 2)


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