# Polynomial

Question: 1. Use suitable identities to find the following products:

(i) (x + 4) (x + 10)

Solution:- (x+4)(x+10)

= x2+10x+4x+4 x 10

= x2+14x+40

(ii) (x + 8) (x – 10)

Solution: x2-10x+8x-80

= x2-2x-80

(iii) (3x + 4) (3x – 5)

Solution: 9x2-15x+12x-20

= 9x2-3x-20  (v) (3 – 2x) (3 + 2x)

Solution: This can be solved as the earlier question

(3-2x)(3+2x) = 9-4x2

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

= (100-5)(100-4)

= 1002-400-500+20

= 10000-900+20

= 9120

(iii) 104 �— 96

Solution: 104 �— 96

= (100+4)(100-4)

= 1002-42

= 10000-16

= 9984

3. Factorise the following using appropriate identities:

(i) 9x2 + 6xy + y2

Solution: 9x2+6xy+y2

= 32x2+3xy+3xy+y2 …………………………… (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 = a2+2ab+b2

(ii) 4y2 – 4y + 1

Solution: 22y2-2 x 2y+12

= (2y-1)² 4. Expand each of the following, using suitable identities:

(i) (x + 2y + 4z)2

Solution: As you know (x + y + z)2= x2 + y2 + z2 + 2xy + 2yz + 2zx

Using this formula in the given equation,

(x+2y+4z)2

= x2+4y2+16z2+4xy+16yz+8zx

(ii) (2x – y + z)2

Solution: (x-y+z)2 = x2+y2+z2-2xy-2yz+2zx

So, (2x-y+z)2

= 4x2+y2+z2-4xy-2yz+4zx

(iii) (–2x + 3y + 2z)2

Solution: (-2x+3y+2z)2

= 4x2+9y2+4z2-12xy+12yz-8zx

(iv) (3a – 7b – c)2

Solution: (x-y-z)2= x2+y2+z2-2xy-2yz-2zx

Hence, (3a-7b-c)2

= 9a2+49b2+c2-42ab-14bc-6ac

5. Factorise:

(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz

Solution: It is clear that this can be solved using

(x+y-z)2 = x2+y2+z2+2xy-2yz-2zx

Hence, 4x2+9y2+16z2+12xy-24yz-2zx

= (2x+3y-4z)2 6. Write the following cubes in expanded form:

(i) (2x + 1)3

Solution: As you know, (x + y)3 = x3 + y3 + 3xy(x + y)

Hence, (2x+1)3

= 8x3+1+6xy(2x+1)

(ii) (2a – 3b)3

Solution: As you know, (x – y)3 = x3 – y3 – 3xy(x – y)

Hence, (2a-3b)3 = 8a3-27y3-18ab(2a-3b) 7. Evaluate the following:

(i) 993

Solution: 993 can be written as (100-1)3

(100-1)3 can be solved through using (x-y)3

Now, (100-1)3= 1003-13-300(100-1)

= 1000000-1-300(99)

= 1000000-1-29700

= 970299

(ii) 1023

Solution: 1023 can be written as (100+2)3 and can be solved using (x+y)3

Hence, (100+2)3

= 1003+23+600(100+2)

= 1000000+8+61200

= 1061208

8. Factorise each of the following:

(i) 8a3 + b3 + 12a2b + 6ab2

Solution:

= 8a3+b3+6ab(a+b)

=(2a+b)3

(ii) 8a3 – b3 – 12a2b + 6ab2

Solution:

8a3-b3-12a2b+6ab2

=8a3-b3-6ab(a+b)

=(2a-b)3

(iii) 27 – 125a3 – 135a + 225a2

Solution: 27 – 125a3 – 135a + 225a2

= 33-53a3-335a+3252a2

= 33-53a3-325a(3-5a)

if x=3 and y=5a

hence, (3-5a)3= 33-53a3-325a(3-5a)

(iv) 64a3 – 27b3 – 144a2b + 108ab2

Solution: 64a3 – 27b3 – 144a2b + 108ab2

= 43a3-33b3-3 x 4a3b(4a-3b)

= (4a-3b)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) x3 + y3 = (x + y) (x2 – xy + y2)

Solution: RHS (x+y)(x2-xy+y2)

= x3-x2y+xy2+x2y-xy2+y3

=x3+y3 LHS proved

(ii) x3 – y3 = (x – y) (x2 + xy + y2)

Solution: RHS (x – y) (x2 + xy + y2)

= x3+x2y+xy2-x2y-xy2-y3

= x3-y3 LHS proved

10. Factorise each of the following:

(i) 27y3 + 125z3

Solution: From the previous question you can recall

x3+y3=(x+y)(x2+xy+y2)

Hence, 33y3+53z3 can be written as follows: (27=33 and 125=53)

(3y+5z)(9y2+15yz+25z2

(ii) 64m3 – 343n3

Solution: As you know, x3-y3=(x-y)(x2+xy+y2)

Hence, 43m3-73n3 can be written as follows: (64=43 and 343=73)

4m-7n)(16n2+28mn+49n2)

11. Factorise : 27x3 + y3 + z3 – 9xyz

Solution: As you know,

x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

Hence, 27x3 + y3 + z3 – 9xyz

= (3x+y+z) (9x2+y2+z2-3xy-yz-3zx)

12. Verify that 13. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.

Solution: As you know,

x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

Now, as per question x+y+z=0,

Putting value of x+y+z=0 in the equation we get

x3 + y3 + z3 – 3xyz = 0 (x2 + y2 + z2 – xy – yz – zx)

Or, x3 + y3 + z3 – 3xyz = 0

Or, x3 + y3 + z3 = 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,

x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

Or, x3 + y3 + z3 =(x + y + z) (x2 + y2 + z2 – xy – yz – zx)+3xyz

Hence, (–12)3 + (7)3 + (5)3

=(-12+7+5) [-122+72+52-(-12 x 7)-(7 x 5)-(-12 x5) ]+3(-12 x 7 x 5)

=0 -122+72+52[-(-12 x 7)-(7 x 5)-(-12 x 5)] -1260

= 0-1260 = -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) = (28-15-13) = 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.

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