Question

5. Factorise:
(i) 4x² +9y2 + 1622 + 12xy - 24yz - 16xz
(ii) 2x2 + y2 +872 – 272 xy + 4V2 yz - 8xz​

Answer 1

Answer:

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We know that

(x

2

+y

2

+z

2

+2xy+2yz+zx)=(x+y+z)

2

(i) 4x

2

+9y

2

+16z

2

+12xy−24yz−16xz

=(2x)

2

+(3y)

2

+(−4z)

2

+2(2x)(3y)+2(3y)(−4z)+2(−4z)(2x)

=(2x+3y−4z)

2

=(2x+3y−4z)(2x+3y−4z)

(ii) 2x

2

+y

2

+8z

2

−2

2

xy+4

2

yz−8zx

=(−

2

x)+(y)

2

+(

2

z)+2(−

2

x)(y)+2(y)(2

2

z)+2(2

2

z)(y)

=(−

2

x+y+2

2

z)

2

=(−

2

x+y+2

2

z)(−

2

x+y+2

2

z)