Polynomial
Question: 1. Use suitable identities to find the following products:
(i) (x + 4) (x + 10)
Solution:
(x+4)(x+10)
= x^{2}+10x+4x+4 x 10
= x^{2}+14x+40
(ii) (x + 8) (x – 10)
Solution:
x^{2}10x+8x80
= x^{2}2x80
(iii) (3x + 4) (3x – 5)
Solution:
9x^{2}15x+12x20
= 9x^{2}3x20
(v) (3 – 2x) (3 + 2x)
Solution: This can be solved as the earlier question
(32x)(3+2x) = 94x^{2}
2. Evaluate the following products without multiplying directly:
(i) 103 × 107
Solution: 103 × 107
= (100+3)(100+7)
= 1002+7 × 100+3 × 100+7 × 3
= 10000+700+300+21
= 11021
(ii) 95 × 96
Solution: 95 × 96
= (1005)(1004)
= 1002400500+20
= 10000900+20
= 9120
(iii) 104 × 96
Solution: 104 × 96
= (100+4)(1004)
= 100242
= 1000016
= 9984
3. Factorise the following using appropriate identities:
(i)
9x^{2} + 6xy + y^{2}
Solution:
9x^{2}+6xy+y^{2}
= 3^{2}x^{2}+3xy+3xy+y^{2} …………………………… (1)
= 3x(3x+y)+y(3x+y)
= (3x+y)(3x+y)
= (3x+y)^{2}
Alternate way of solving this problem:
Equation 1 gives a hint that this can be solved through following formula:
(a+b)^{2} = a^{2}+2ab+b^{2}
(ii)
4y^{2} – 4y + 1
Solution:
2^{2}y^{2}2
x 2y+1^{2}
= (2y1)²
4. Expand each of the following, using suitable identities:
(i) (x
+ 2y + 4z)^{2}
Solution: As you know
(x + y + z)^{2}= x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx
Using this formula in the given equation,
(x+2y+4z)^{2}
= x^{2}+4y^{2}+16z^{2}+4xy+16yz+8zx
(ii)
(2x – y + z)^{2}
Solution:
(xy+z)^{2} = x^{2}+y^{2}+z^{2}2xy2yz+2zx
So,
(2xy+z)^{2}
= 4x^{2}+y^{2}+z^{2}4xy2yz+4zx
(iii) (–2x
+ 3y + 2z)^{2}
Solution:
(2x+3y+2z)^{2}
= 4x^{2}+9y^{2}+4z^{2}12xy+12yz8zx
(iv) (3a –
7b – c)^{2}
Solution:
(xyz)^{2}= x^{2}+y^{2}+z^{2}2xy2yz2zx
Hence,
(3a7bc)^{2}
= 9a^{2}+49b^{2}+c^{2}42ab14bc6ac
5. Factorise:
(i) 4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz
Solution: It is clear that this can be solved using
(x+yz)^{2} = x^{2}+y^{2}+z^{2}+2xy2yz2zx
Hence, 4x^{2}+9y^{2}+16z^{2}+12xy24yz2zx
=
(2x+3y4z)^{2}
6. Write the following cubes in expanded form:
(i)
(2x + 1)^{3}
Solution: As you know,
(x + y)^{3} = x^{3} + y^{3} + 3xy(x + y)
Hence, (2x+1)^{3}
= 8x^{3}+1+6xy(2x+1)
(ii) (2a – 3b)^{3}
Solution: As you know,
(x – y)^{3} = x^{3} – y^{3} – 3xy(x – y)
Hence,
(2a3b)^{3} = 8a^{3}27y^{3}18ab(2a3b)
7. Evaluate the following:
(i) 99^{3}
Solution:
99^{3} can be written as (1001)^{3}
(1001)^{3}
can be solved through using (xy)^{3}
Now, (1001)^{3}= 100^{3}1^{3}300(1001)
= 10000001300(99)
= 1000000129700
= 970299
(ii) 102^{3}
Solution: 102^{3} can be written as (100+2)^{3}
and can be solved using (x+y)^{3}
Hence,
(100+2)^{3}
= 100^{3}+2^{3}+600(100+2)
= 1000000+8+61200
= 1061208
8. Factorise each of the following:
(i) 8a^{3} + b^{3} + 12a^{2}b + 6ab^{2}
Solution:
= 8a^{3}+b^{3}+6ab(a+b)
=(2a+b)^{3}
(ii) 8a^{3} – b^{3} – 12a^{2}b + 6ab^{2}
Solution:
8a^{3}b^{3}12a^{2}b+6ab^{2}
=8a^{3}b^{3}6ab(a+b)
=(2ab)^{3}
(iii) 27 – 125a^{3} – 135a + 225a^{2}
Solution:
27 – 125a^{3} – 135a + 225a^{2}
= 3^{3}5^{3}a^{3}3^{3}5a+3^{2}5^{2}a^{2}
= 3^{3}5^{3}a^{3}3^{2}5a(35a)
if x=3 and y=5a
hence, (35a)^{3}= 3^{3}5^{3}a^{3}3^{2}5a(35a)
(iv) 64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}
Solution:
64a^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}
= 4^{3}a^{3}3^{3}b^{3}3 x 4a3b(4a3b)
= (4a3b)^{3}
Note: Try to identify the values of x and y by carefully analysing the first two terms of the equations. This will give you exact clue to the final answer.
Solution: 27p³ can be written as 3³p³
Hence, x= 3p
Note: This step is to help you develop the problem solving skills. In exam situation you have to write all steps to get full marks.
9. Verify :
(i) x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})
Solution: RHS
(x+y)(x^{2}xy+y^{2})
= x^{3}x^{2}y+xy^{2}+x^{2}yxy^{2}+y^{3}
=x^{3}+y^{3} LHS proved
(ii) x^{3} – y^{3} = (x – y) (x^{2} + xy + y^{2})
Solution: RHS
(x – y) (x^{2} + xy + y^{2})
= x^{3}+x^{2}y+xy^{2}x^{2}yxy^{2}y^{3}
= x^{3}y^{3} LHS proved
10. Factorise each of the following:
(i) 27y^{3} + 125z^{3}
Solution: From the previous question you can recall
x^{3}+y^{3}=(x+y)(x^{2}+xy+y^{2})
Hence, 3^{3}y^{3}+5^{3}z^{3} can be written as follows: (27=3^{3} and
125=5^{3})
(3y+5z)(9y^{2}+15yz+25z^{2}
(ii) 64m^{3} – 343n^{3}
Solution: As you know,
x^{3}y^{3}=(xy)(x^{2}+xy+y^{2})
Hence, 4^{3}m^{3}7^{3}n^{3} can be written as follows: (64=4^{3} and 343=7^{3})
4m7n)(16n^{2}+28mn+49n^{2})
11. Factorise :
27x^{3} + y^{3} + z^{3} – 9xyz
Solution: As you know,
x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)
Hence, 27x^{3} + y^{3} + z^{3} – 9xyz
= (3x+y+z) (9x^{2}+y^{2}+z^{2}3xyyz3zx)
12. Verify that
13. If x + y + z = 0, show that x^{3} + y^{3} + z^{3} = 3xyz.
Solution: As you know,
x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)
Now, as per question x+y+z=0,
Putting value of x+y+z=0 in the equation we get
x^{3} + y^{3} + z^{3} – 3xyz = 0 (x^{2} + y^{2} + z^{2} – xy – yz – zx)
Or, x^{3} + y^{3} + z^{3} – 3xyz = 0
Or, x^{3} + y^{3} + z^{3} = 3xyz proved
14. Without actually calculating the cubes, find the value of each of the following:
(i) (–12)³ + (7)³ + (5)³
Solution: As you know,
x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)
Or, x^{3} + y^{3} + z^{3} =(x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx)+3xyz
Hence, (–12)^{3}
+ (7)^{3} + (5)^{3}
=(12+7+5)
[12^{2}+7^{2}+5^{2}(12 x 7)(7 x 5)(12 x5) ]+3(12
x 7 x 5)
=0 12^{2}+7^{2}+5^{2}[(12
x 7)(7 x 5)(12 x 5)] 1260
= 01260 = 1260 answer
(ii) (28)^{3} + (–15)^{3}
+ (–13)^{3}
Solution: This question can be solved in the same way as above.
Here, value of (x+y+z) = (281513) = 0
Hence, you need to calculate the value of 3xyz
3 x 28 x 15 x 13 = 16380
Hence, the required answer = 16380
But, while practicing at home try following every step for better learning.
