Exercise - 2.5 : Polynomials - Class 9 PDF Download

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.