If (8 x^{3}+27 y^{3} ) \\div(2 x+3 y) = (4 x^{2}+B x y+C y^{2} )then the value of (5 {A}+4 {B}+3 {C}) is:

SSC CGL 7 06 2019 Shift 1 Paper

Question : 60 of 100

Marks: +1, -0

\ \ (8 x^{3}+27 y^{3}\ ) \÷(2 x+3 y)

=\ (4 x^{2}+B x y+C y^{2}\ )

then the value of

(5 \ {A}+4 \ {B}+3 \ {C})

is:

Solution:

\ {8 x^{3}+27 y^{3}}/{2 x+3 y}=4 x^{2}+9 y^{2}-6 x y

On comparing

A=4 ; B=-6, C=9

Now,

5 A+4 B+3 C

=20-24+27=23

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