Polynomials

Exercise 2.5 Part 6

Question: 5 – Factorise:

(i) `4x^2 + 9y^2 + 16z^2 + 12xy - 24yz – 16xz`

Answer: Given, `4x^2 + 9y^2 + 16z^2 + 12xy - 24yz – 16xz`

`= (2x)^2 + (3y)^2 + ( - 4z)^2`

` + 2(2x)xx(3y) + 2(3y)xx(-4z) + 2(2x)xx( - 4z)`


If a = 2x, b = 3y and c = - 4z
Then, using the identity `(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac`, we get;

`4x^2 + 9y^2 + 16z^2 + 12xy - 24yz – 16xz`

`= (2x + 3y – 4z)(2x + 3y – 4z)`

`= (2x + 3y – 4z)^2`


(ii) `2x^2 + y^2 + 8z^2 - 2sqrt2xy + 4sqrt2yz – 8xz`

Answer: Given, `2x^2 + y^2 + 8z^2 - 2sqrt2xy + 4sqrt2yz – 8xz`
This expression can be written as follows:
`(-sqrt2x)^2 + y^2 + (2sqrt2z)^2`

`+2( - sqrt2x)xx(z) + 2(y)xx(2sqrt2z) + 2(sqrt2x)xx(2sqrt2z)`

If, `a = -sqrt2x, b = y` and `c = 2sqrt2z`
Then, using the identity `(a + b + c)^2`

`= a^2 + b^2 + c^2 + 2ab + 2bc + 2ac`

We get;
`2x^2 + y^2 + 8z^2 - 2sqrt2xy + 4sqrt2yz – 8xz`

`= (- sqrt2x + y + 2sqrt2z)( - sqrt2x + y + 2sqrt2z)`

`= (- sqrt2x + y + 2sqrt2z)^2`



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