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Factorise completely by using Remainder Theorem $2 x 3 − x 2 y − 36 x y 2 − 45 y 3$ .

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x + 3 y x − 5 y 2 x + 3 y

x + 3 y x + 5 y 2 x + 3 y

x + 3 y x − 5 y 2 x − 3 y

x − 3 y x − 5 y 2 x − 3 y

answer is A.

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Detailed Solution

Given,

2 x 3 − x 2 y − 36 x y 2 − 45 y 3

.
If a polynomial of x is divided by a linear expression of x, the remainder is zero, according to the Remainder theorem.
By hit and trial method substituting x = -3y to

2 x 3 − x 2 y − 36 x y 2 − 45 y 3

we get that x=-3y {x+3y=0} is a factor of

2 x 3 − x 2 y − 36 x y 2 − 45 y 3

.
Therefore, dividing

2 x 3 − x 2 y − 36 x y 2 − 45 y 3

by x = -3y, the quotient is

2 x 2 − 7 x y − 15 y 2

.
So,

2 x 3 − x 2 y − 36 x y 2 − 45 y 3

can be written as,

⇒ x + 3 y 2 x 2 − 7 x y − 15 y 2 ⇒ x + 3 y 2 x 2 − 10 x y + 3 x y − 15 y 2 ⇒ x + 3 y 2 x x − 5 y + 3 y x − 5 y ⇒ x + 3 y x − 5 y 2 x + 3 y

The factors of given equation are

x + 3 y x − 5 y 2 x + 3 y

.
The correct option is 1).

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