Question

"Question 5 Factorise:
(i) 4x^2 + 9y^2 +16z^2 + 12xy - 24yz - 16xz
(ii) 2x^2 + y^2 + 8z^2 - 2.2^(1/2)xy + 4.2^(1/2)yz - 8xz

Class 9 - Math - Polynomials Page 49"

Answer 1

Hlo friend
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Here is ur answer :

(i) 4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca 
4x2 + 9y2 + 16z2 + 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) 2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz
Using identity, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
2x2 + y2 + 8z2 - 2√2 xy + 4√2 yz - 8xz 
= (-√2x)2 + (y)2 + (2√2z)2 + (2×-√2x×y) + (2×y×2√2z) + (2×2√2z×-√2x)
= (-√2x + y + 2√2z)2
=  (-√2x + y + 2√2z) (-√2x + y + 2√2z)

HOPE IT HELPS YOU.. PLS MARK ME AS BRAINLIEST...

REGARDS
@SUNITA

Answer 2

Identity:

An identity is an equality which is true for all values of a variable in the equality.

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

In an identity the right hand side expression is called expanded form of the left hand side expression.

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Solution:

(i) 4x²+9y²+16z²+12xy–24yz–16xz

 We can we write the given expression as

 

=(2x)²+(3y)²+(−4z)²+(2×2x×3y)+(2×3y×−4z)+(2×−4z×2x)

 

=(2x+3y–4z)²

 Identity used:

 [a²+b²+c²+2ab+2bc+2ca = (a²+b²+c²)]

 

=(2x+3y–4z) (2x+3y–4z)

 

(ii) 2x²+y²+8z²–2√2xy+4√2yz–8xz

 

We can rewrite the given expression as

=(−√2x)²+(y)²+(2√2z)²+(2×(−√2x)×y)+(2×y×2√2z)+(2×2√2z ×(−√2x))

 

=(−√2x+y+2√2z)²

 

Identity used:

 [a²+b²+c²+2ab+2bc+2ca = (a²+b²+c²)]

 

=(−√2x+y+2√2z) (−2√x+y+2√2z)

 

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