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Higher Order Thinking Skills (HOTS) - Polynomials Class 9 Notes | EduRev

Class 9 : Higher Order Thinking Skills (HOTS) - Polynomials Class 9 Notes | EduRev

 Page 1


  
Question 1. If  p(x) = x
2
 – 2 2 x + 1, then find  p
()
22
.
Solution: Since, p(x) = x
2
 – 2 2 x + 1
Then p
()
22
=
()
2
22
??
??
–
()
22 22 1
??
+
??
= 4 (2) – 4 (2) + 1 = 8 – 8 + 1 = 1
Question 2. If  a + b + c = 9, and ab + bc + ca = 26, find a
2
 + b
2
 + c
2
Solution: (a + b + c)
2
=(a
2
 + b
2
 + c
2
) + 2 (ab + bc + ca)
? [9]
2
=(a
2
 + b
2
 + c
2
) + 2 (26)  = (a
2
 + b
2
 + c
2
) + 52
? a
2
 + b
2
 + c
2
=9
2
– 52 = 81– 52 = 29
Question 3. Factorise : 8p
3
 + 
12
5
p
2 
 + 
6
25
 p
 
 + 
1
125
Solution: ? 8p
3
 = (2p)
3
, and 
1
125
 = 
3
1
5
??
??
??
? 8p
3
 + 
12
5
p
2 
 + 
6
25
p
 
 + 
1
125
  =  (2p)
3 
+ 3 (2p)
2
 
1
5
??
??
??
 + 3 (2p)
 
23
11
55
?? ??
+
?? ??
?? ??
= 
3
1
2p +
5
??
??
??
[? a
3
 + 3a
2
b + 3ab
2 
+
  
b
3 
= (a + b)
3
]
= 
1
2p +
5
??
??
??
 
1
2p +
5
??
??
??
 
1
2p +
5
??
??
??
Question 4. If a, b, c are all non-zero and a + b + c = 0, prove that  
22 2
ab c
3
bc ca ab
++ =
Solution: ? a + b + c = 0  ?  a
3
 + b
3
 + c
3 
– 3abc = 0
or a
3
 + b
3
 + c
3 
= 3abc 
 
?  
33 3
abc
ab c 3
aa a a bc bc bc bc
++ = ? 
22 2
ab c
3
bc ca ab
++ =
Question 5. Prove that (a + b + c)
3
 – a
3
 – b
3
 – c
3 
=  3(a + b) (b + c) (c + a)
Solution: L.H.S. = (a + b + c)
3
 – a
3
 – b
3
 – c
3 
=  [(a + b + c)
3
– a
3
] – [b
3
 + c
3
] ...(1)
(a + b + c)
3
 – a
3
 = (b + c) [3a
2
 + b
2 
+ c
2 
+ 3ab + 2bc + 3ca] ...(2)
[using x
3
 – y
3 
= (x – y) (x
2
 + y
2 
+ xy)]
and b
3
 + c
3 
= (b + c) [b
2
 + c
2
 – bc]            [using x
3
 + y
3 
= (x + y) (x
2
 + y
2 
– xy)]...(3)
From (1), (2) and (3), we get
L.H.S. = (b + c) (3a
2 
+ b
2 
+ c
2 
+ 3ab + 2bc + 3ca) – (b + c) (b
2 
+ c
2 
– bc)
= (b + c) [3a
2 
+ b
2 
+ c
2 
+ 3ab + 2bc + 3ca – b
2 
– c
2 
+ bc]
= (b + c) [3a
2 
+ (b
2 
– b
2
) + (c
2 
– c
2
) + 3ab + (2bc + bc) + 3ca]
= (b + c) [3a
2 
+ 0
 
+ 0 + 3ab + 3bc + 3ca]
= (b + c) [3
 
(a
2 
+ ab
 
+ bc + ca)]
= 3 (b + c) [(a + b) (c + a)]
= 3 (a + b) (b + c) (c + a) = RHS

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Author

Indu Gupta

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533

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