Factorize completely \(x^{2} + 2xy + y^{2} + 3x + 3y - 18\).
(x+y+6)(x+y-3)
(x-y-6)(x-y+3)
(x-y+6)(x-y-3)
(x+y-6)(x+y+3)
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x2 +2xy + y2 +3x + 3y -18
regrouping the terms above we have:
(x2 +2xy + y2) + [(3x+3y)-18]
(x + y)2 + 3(x + y) - 18
let z = x +y
z2 +3z -18
z2+6z-3z-18
z(z+6) -3(z+6)
(z+6) (z-3)
substituting x +y = z back into the expression:
(x+ y +6) (x +y -3)
nawa o..but if this question is not a multiple choice question..wat will u guys do to get the answer?
option A is the answer
if u expand option a which is (a)(x+y+6) (b)(x+y-3)by multiplyin both dat is bracket a to bracket b
u will get x^2+xy-3x+xy+y^2-3y+6x+6y-18
then collect like terms u get
x^2+2xy+y2+3x+3y-18 which was d question u were u given 2 factorize tanks
X^2 2xy y^2 3x 3y-18=(x y)^2 3(x y)-18
let (x y) be p,and subtitute to the equ.
= p^2 3p-18
=p^2 6p-3p-18
=p(p 6)-3(p 6)
=(p-3)(p 6)
recall: p=x y,sutitute to the answer
=(x y-3)(x y 6).
how about expanding each option to get the equation. consequently whichever option gives the given equation is the right answer. #capeesh
firstly you factorize it and we have
x^2 +(2y+3)x+(y^2 +3y)
a b c
Then we use quadratic formular
-b±√(b^2 -4ac)
2a
to get your answer
