Question

factorise 4x^2-9y^2+16z^2+12xy-24xy-16xz​

Answer 1

Answer:

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

Answer 2

Step-by-step explanation:

by using this identity

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

Now - 4x^2-9y^2+16z^2+12xy-24xy-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)

Hope it helps you

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