Polynomial ch 2 class 9th exercise 2.5 question 4 part 5

1.	Polynomial ch 2 class 9th exercise 2.5 question 4 part 5
Answer» Question no :- 05 SOLUTION- (i) 4x2+9y2+16z2+12xy–24yz–16xz Using identity, (X+y+z)2 = x2+y2+z2+2xy+2yz+2zx We can SAY that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2 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√2xy+4√2yz–8xz Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2 2x2+y2+8z2–2√2xy+4√2yz–8xz = (-√2x)2+(y)2+(2√2z)2+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x) = (−√2x+y+2√2z)2 = (−√2x+y+2√2z)(−√2x+y+2√2z)

Answer»

Question no :- 05 SOLUTION-

(i) 4x2+9y2+16z2+12xy–24yz–16xz

Using identity, (X+y+z)2 = x2+y2+z2+2xy+2yz+2zx

We can SAY that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2

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√2xy+4√2yz–8xz

Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx

We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2

2x2+y2+8z2–2√2xy+4√2yz–8xz

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

= (−√2x+y+2√2z)2

= (−√2x+y+2√2z)(−√2x+y+2√2z)

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