d Question 4. Factorize x 3 + 3 x y 2 – 2 x 2 y – 6y 3 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs
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Question 4. Factorize x 3 + 3 x y 2 – 2 x 2 y – 6y 3

2. ac+ad+bc+bd type expressions 0

Question 4. Factorize x 3 + 3 x y 2 – 2 x 2 y – 6y 3

factorization techniques —

Solution:

x^{3}+3x y^{2}-2 x^{2}y-6 y^{3}

Re-arranging terms;

= x^{3}-2 x^{2}y+3x y^{2}-6 y^{3}

we see that in first two terms x^{2} is common and in last two terms 3y^{2} is common – take out !

= x^{2}(x-2y)+3 y^{2}(x-2y)

Put brackets ‘[]’

= [x^{2}(x-2y)+3 y^{2}(x-2y)]

Here now (x-2y) is common – take out !

=(x-2y)[ x^{2}+3 y^{2}]

=(x-2y)( x^{2}+3 y^{2})

so,

x^{3}+3x y^{2}-2 x^{2}y-6 y^{3} = =(x-2y)( x^{2}+3 y^{2})



 


 


 


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