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- Polynomials : Polynomial : Exercise

**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+8x-80*

*= x ^{2}-2x-80*

**(iii) (3x + 4) (3x – 5)**

**Solution:** *9x ^{2}-15x+12x-20*

*= 9x ^{2}-3x-20*

*(v) (3 – 2x) (3 + 2x)*

*Solution:* This can be solved as the earlier question

* (3-2x)(3+2x) = 9-4x*^{2}

**2. Evaluate the following products without multiplying directly:**

**(i) 103 Ã— 107 **

**Solution:** 103 Ã— 107

= (100+3)(100+7)

= 1002+7 Ã— 100+3 Ã— 100+7 Ã— 3

= 10000+700+300+21

= 11021

**(ii) 95 Ã— 96 **

**Solution:** 95 Ã— 96

= (100-5)(100-4)

= 1002-400-500+20

= 10000-900+20

= 9120

**(iii) 104 Ã— 96**

**Solution:** 104 Ã— 96

= (100+4)(100-4)

= 1002-42

= 10000-16

= 9984

**3. Factorise the following using appropriate identities:**

**(i) ** **9x ^{2} + 6xy + y**

**Solution:** *9x ^{2}+6xy+y^{2}*

*= 3 ^{2}*x

*= 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}

= (2y-1)Â²

**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:** *(x-y+z) ^{2} = x^{2}+y^{2}+z^{2}-2xy-2yz+2zx*

So, * (2x-y+z) ^{2}*

*= 4x ^{2}+y^{2}+z^{2}-4xy-2yz+4zx*

**(iii) (–2x + 3y + 2z) ^{2}**

**Solution:** *(-2x+3y+2z) ^{2}*

*= 4x ^{2}+9y^{2}+4z^{2}-12xy+12yz-8zx*

**(iv) (3a – 7b – c) ^{2}**

**Solution:** *(x-y-z) ^{2}= x^{2}+y^{2}+z^{2}-2xy-2yz-2zx*

Hence, * (3a-7b-c) ^{2}*

*= 9a ^{2}+49b^{2}+c^{2}-42ab-14bc-6ac*

**5. Factorise:**

**(i) 4x ^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz**

**Solution:** It is clear that this can be solved using

(x+y-z)^{2} = x^{2}+y^{2}+z^{2}+2xy-2yz-2zx

Hence, *4x ^{2}+9y^{2}+16z^{2}+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} = 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, *(2a-3b) ^{3} = 8a^{3}-27y^{3}-18ab(2a-3b)*

7. Evaluate the following:

**(i) 99 ^{3}**

**Solution:** 99^{3} can be written as (100-1)^{3}

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

Now, *(100-1) ^{3}= 100^{3}-1^{3}-300(100-1)*

*= 1000000-1-300(99)*

*= 1000000-1-29700*

*= 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)*

*=(2a-b)*^{3}

**(iii) 27 – 125a ^{3} – 135a + 225a^{2} **

**Solution:** *27 – 125a ^{3} – 135a + 225a^{2} *

*= 3 ^{3}-5^{3}*a

*= 3 ^{3}-5^{3}*a

if x=3 and y=5a

hence, *(3-5a) ^{3}= 3^{3}-5^{3}*a

**(iv) 64a ^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}**

**Solution:** *64a ^{3} – 27b^{3} – 144a^{2}b + 108ab^{2}*

*= 4 ^{3}*a

*= (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: 27pÂ³ can be written as 3Â³pÂ³

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}y-xy^{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}y-xy^{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

*(3y+5z)(9y ^{2}+15yz+25z*

**(ii) 64m ^{3} – 343n^{3}**

**Solution:** As you know, *x ^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2}*)

Hence, *4 ^{3}*m

4m-7n)(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}-3xy-yz-3zx)*

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)Â³ + (7)Â³ + (5)Â³**

**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*

*= 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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