Extra Curriculum Support

SecA

1. If p(x) = 5x - 4x^2 + 3 the p(-1) = ?

2. 3 is a polynomial of degree ?

(a) 0

(b)12

(c) 1

(d) 2

3. If x + 1 is a factor of the polynomial: 2x2+kx then k = ?

(a) -3

(b) 4

(c) -2

(d) 2

4. x + 1 is a factor of the polynomial:

(a) x3+2x2−x−2

(b) x3+2x2−x+2

(c) x3−2x2+x+2

(d) x3+2x2+x−2

Sol

1. Solution:

Substitute x=−1 into the polynomial:

p(-1) = 5−1−4−12+3 = −5−4+3 = -6

So, p(-1) = - 6

2. Solution: Option (a)

3. Solution: Option (d)

4. Solution: Option (a)

SecB

5. Factorise: x4−23x2+142x−120

6. Factorise: 8a3−b3−12a2b+6ab2

7. If x3+ax2+bx=6 has x−2 as a factor and leaves a remainder of 3 when divided by x−3, find a and b.

8. If x+2a is a factor of the polynomial: x5−4a2x3+2x+2a+3, find a.

Sol

5. Solution:

x4−23x2+142x−120 = x4−23x2+142x−120 (Group the terms and factorise)

= x2x2−23 + 271x−60

= x2x+23x−23 + 271x−60

Thus, x4−23x2+142x−120 = x2x+23x−23 + 271x−60

6. Solution: Group the terms in pairs

8a3−b3−12a2b+6ab2 = 8a3−b3+−12a2b+6ab2

= 2a−b4a2+2ab+b2−6ab2a−b

= 2a−b4a2+2ab+b2−6ab = 2a−b4a2−4ab+b2

Therefore, the factorized form of 8a3−b3−12a2b+6ab2 is 2a−b4a2−4ab+b2

7. Solution: Since x−2 is a factor of x3+ax2+bx=6, let's substitute x = 2.

p(2) = 23+a22+2b = 8+4a+2b = 6

2a+b=−1 (1)

Since the remainder when divided by (x−3) is 3, substituting x = 3 gives us:

p(3) = 33+a32+3b = 27+9a+3b = 3

9a+3b=−24

3a+b=−8 (2)

Solving (1) and (2), we get: a = -7 and b = 13

Therefore, the values of a and b are -7 and 13, respectively.

8. Solution:

To find the value of a such that x+2a is a factor of the polynomial x5−4a2x3+2x+2a+3, we can use the Factor Theorem. According to the Factor Theorem, if x+2a is a factor, then substituting x = −2a into the polynomial should result in zero.

P(-2a) = −2a5−4a2−2a3+2−2a+2a+3

= −32a5+32a5−4a+2a+3 = 3−2a

Since x+2a is a factor, P(−2a) should equal zero: 3−2a = 0

a = 32

Hence, the value of a = 32

SecC

9. Using factor theorem, factorize the polynomial: 2x4−7x3−13x2+63x−45

10. Verify whether x+1 is a factor of Q(x)= x3+3x2−4x−4. If yes, find the quotient polynomial.

Sol

9. Solution:

To factorize the polynomial 2x4−7x3−13x2+63x−45 using the Factor Theorem, we look for potential rational roots using the Rational Root Theorem.

Let's check these possible roots by substituting them into the polynomial to find the actual roots and thereby factorize the polynomial.

After testing some values, we find that x=3 is a root of the polynomial.

Thus, x−3 is a factor of 2x4−7x3−13x2+63x−45.

Doing synthetic division of 2x4−7x3−13x2+63x−45 by x−3 , we get the quotient:

2x3−x2−16x+15

After factorizing the quotient polynomial using synthetic divison, we get:

2x3−x2−16x+15 = (x−1)(2x−5)(x+3)

Thus, 2x4−7x3−13x2+63x−45 = x−12x−5x+3x−3

10. Solution:

To verify whether x+1 is a factor of x3+3x2−4x−4 and find the quotient polynomial if it is, we can use the Factor Theorem.

According to the Factor Theorem, x+1 is a factor of Q(x) if Q(−1)= 0

Let's check if Q(−1) = 0.

Q(−1) = −13+3−12−4−1−4 = −1+3+4−4 = 2

Since Q(−1) ≠ 0, x+1 is not a factor of Q(x).

Therefore, x+1 is not a factor of the polynomial x3+3x2−4x−4 and there is no need to find the quotient polynomial.

SecD

11. Verify that: x3+y3+z3−3xyz = 12x+y+zx2+y2+y2+z2+z2+x2

Sol

11. Solution: To verify, we need to manipulate and simplify both sides of the equation to check if they are indeed equal.

LHS

x3+y3+z3−3xyz

RHS

12x+y+zx2+y2+y2+z2+z2+x2 = 12x+y+z2x2+2y2+2z2

= 12x+y+z2x2+y2+z2 = x+y+zx2+y2+z2

= xx2+y2+z2+yx2+y2+z2+zx2+y2+z2

= x3+y3+z3+xy2+xz2+yx2+yz2+zx2+zy2

Now, observe the LHS:

x3+y3+z3−3xyz

For these two expressions to be equal, the extra terms in the RHS xy2,xz2,yx2,yz2,zx2,zy2 must cancel out with −3xyz.

Consider the symmetric form of the polynomial identity:

x3+y3+z3−3xyz = x+y+zx2+y2+z2−xy−yz−zx

Compare this with our simplified RHS: x+y+zx2+y2+z2

The simplified RHS x+y+zx2+y2+z2 can be split into:

= x+y+zx2+y2+z2−xy−yz−zx+x+y+zxy+yz+zx

= x+y+zxy+yz+zx

Hence, the polynomial identity holds true, verifying: x3+y3+z3−3xyz = 12x+y+zx2+y2+y2+z2+z2+x2

Problem 1

A health club is creating a new rectangular yoga area where the length is given by L(x)=2x+5 and the width by W(x)=x+3. The club wants to encourage healthy living through regular yoga sessions.

1. Calculate the polynomial for the area of the yoga area.

2. If x=4, what is the area of the yoga area?

3. Discuss the importance of physical fitness and how activities like yoga contribute to a healthy lifestyle.

Sol

Solution:

1. The area A of a rectangle is given by the formula: A = Length × Width

Here, the length L(x)=2x+5 and the width W(x)=x+3.

So, the area A(x) as a function of x will be: A(x) = L(x) × W(x) = (2x+5)(x+3)

= 2x(x) + (2x)(3) + (5)(x) + 5(3) = 2x2+6x+5x+15 = 2x2+11x+15

So, the polynomial for the area of the yoga area is A(x) = 2x2+11x+15

2. Substitute x = 4 into the polynomial A(x):

A(4) = 242+114+15 = 216+44+15 = 32+44+15 = 91

Therefore, when x = 4, the area of the yoga area is 91 square units.

3. Regular yoga practice, combined with other forms of exercise and a balanced diet, contributes to a healthy lifestyle by improving physical fitness, mental well-being, and overall quality of life. It promotes longevity and reduces the risk of various chronic diseases.

Problem 2

A team of students is working on a project to install solar panels on the school roof. The length of the roof is represented by L(x)=7x−1 and the width by W(x)=4x+2.

1. Find the polynomial that represents the area of the roof.

2. If x=3, calculate the area of the roof. How many solar panels can be installed if each panel requires 10 square units of space?

3. Discuss the benefits of using renewable energy sources and how solar panels can help reduce the carbon footprint of the school.

Sol

Solution:

1. The area A(x) of the roof can be found by multiplying the polynomials representing the length and width:

L(x)=7x−1, W(x)=4x+2

The area A(x) is given by:

A(x) = L(x) × W(x)

A(x) = (7x−1) (4x+2)

Using the distributive property to expand the expression:

A(x) = 7x(4x) + (7x)(2) − (1)(4x) − (1)(2) = 28x2+14x−4x−2 = 28x2+10x−2

So, the polynomial representing the area of the roof is: 28x2+10x−2

2. Substitute x = 3 into the area polynomial A(x):

A(3) = 2832+103−2 = 289+103−2 = 252+30−2 = 280 unit2

So, the area of the roof when x = 3 is: 280 unit2

If each solar panel requires 10 square units of space, the number of solar panels that can be installed is:

Number of panels = TotalareaAreaperpanel = 28010 = 28

Therefore, the maximum number of solar panels that can be installed is: 28

3. Using renewable energy sources like solar panels offers numerous benefits, from reducing greenhouse gas emissions and pollution to promoting sustainability and providing economic and health advantages. By installing solar panels, the school can actively contribute to environmental protection and educate students about the importance of renewable energy and sustainable practices.

Problem 3

Rahul is designing a garden in his neighborhood park. The length of the garden is represented by the polynomial L(x) = 4x+3 and the width by W(x) = 2x+1. He wants to ensure that the garden area can accommodate various plants and trees to promote green cover in the locality.

1. Calculate the polynomial representing the area of the garden.

2. If x = 5, what is the area of the garden? How many different types of plants can be planted if each type requires 5 square units of space?

3. Discuss the importance of green spaces in urban areas and how they contribute to environmental sustainability.

Sol

Solution: The area A(x) of the garden can be found by multiplying the polynomials representing the length and width:

L(x)=4x+3

W(x)=2x+1

The area A(x) is given by:

A(x) = L(x)×W(x)

A(x) = (4x+3)(2x+1)

Using the distributive property to expand the expression:

A(x) = 4x(2x) + (4x)(1) + (3)(2x) + (3)(1) = 8x2+4x+6x+3 = 8x2+10x+3

So, the polynomial representing the area of the garden is: 8x2+10x+3

2. Substituting x=5 into the area polynomial A(x):

A(5) = 852+105+3 = 8(25) + 50 + 3 = 200 + 50 + 3 = 253 unit2

If each type of plant requires 5 square units of space, the number of different types of plants that can be planted is:

Number of plants = TotalareaAreaperplant = 2535 = 50.6 ≈ 50

Thus, a maximum of 50 plants can be planted, with 3 square units of space left over.

3. Urban green spaces are essential for promoting environmental sustainability, enhancing health and well-being, supporting biodiversity, and providing social, community, and economic benefits. Initiatives like Rahul's garden project play a crucial role in creating greener, healthier, and more sustainable urban environments.

Q1

1. A polynomial P(x) represents the volume of a cube in terms of its side length x. If P(x) = x3−6x2+12x, find the side length of the cube when the volume is maximum.

Sol

Solution: To find the side length of the cube when the volume P(x) = x3−6x2+12x is maximum, we need to follow these steps:

Find the critical points by taking the derivative and setting it to zero.

Determine which critical point corresponds to the maximum volume.

Find the critical points

First, let's find the first derivative of P(x):

The first derivative,

P(x)′ = dx3−6x2+12xdx = 3x2−12x+12

Set the first derivative equal to zero to find the critical points: 3x2−12x+12 = 0

x2−4x+4 = 0

x−22 = 0

x = 2

Determine if this critical point is a maximum To determine if this critical point is a maximum, we can use the second derivative test. Find the second derivative,

P(x)′′ = d3x2−12x+12dx = 6x−12

Evaluate the second derivative at x=2:

P(2)′′ = 6(2) − 12 = 0

Since P(2)′′ = 0, the second derivative test is inconclusive. We need to check the nature of the critical point by other means, such as evaluating the sign of the first derivative around the critical point or using the first derivative test.

Evaluate P(x)′just before and just after x = 2:

For x = 1.9:

P(1.9)′ = 31.92−121.9+12 = 10.83−22.8+12 = 0.03

For x = 2.1:

P(2.1)′ = 32.12−122.1+12 = 13.23−25.2+12 = 0.03

Both values are very close to zero, suggesting a possible maximum or minimum around x = 2. However, we see a pattern indicating that at x = 2, the function is at an extreme point.

Thus, checking the values directly:

P(1.9) = 1.93−61.92+121.9 = 6.859−21.66+22.8 = 7.999

P(2.1) = 2.13−62.12+122.1 = 9.261−26.46+25.2 = 8.001

These values suggest a slight peak at x = 2.

Therefore, the side length of the cube when the volume is at a maximum is x = 2.

Q2

2. If P(x)= x3−6x2+11x−6, find the zeros of the polynomial and verify the relationship between the zeros and coefficients using Vieta's formulas.

Sol

Solution: Find the zeros of the polynomial We will use the fact that the polynomial can often be factored by trying possible rational roots (factors of the constant term divided by factors of the leading coefficient). For the polynomial x3−6x2+11x−6, the possible rational roots are the factors of −6 (constant term) divided by the factors of 1 (leading coefficient).

Taking x = 1:

P(1) = 13 − 612 + 111− 6 = 1 − 6 + 11 − 6 = 0

So, x = 1 is a root.

Using polynomial division or synthetic division to find the quadratic factor, we get: x2−5x+6

So, P(x) = x−1x2−5x+6

Factorising the quadratic equation: x2−5x+6 = x−2x−3

Thus, P(x) = x−1x−2x−3

So, the zeros of the polynomial are: x = 1,2,3

Verify Vieta's formulas

Sum of the roots α + β + γ = −ba

Sum of the product of the roots αβ + βγ + γα = ca

Product of the roots αβγ = −da

​For the polynomial P(x) we have: a = 1, b = −6, c = 11, d = −6 with the roots are α = 1, β = 2 and γ = 3. ​ Sum of the roots: α + β + γ = 1 + 2 + 3 = 6

According to Vieta's formula: −ba = −−61 = 6

Sum of the product of the roots: αβ + βγ + γα = 1(2) + 2(3) +3(1) = 2 + 6 + 3 = 11

According to Vieta's formula: ca = 111 = 11

Product of the roots: αβγ = 6

According to Vieta's formula: −da = −−61 = 6

The zeros 1,2,3 satisfy Vieta's formulas, confirming the correctness of the factorization and the relationship between the zeros and coefficients.

Questions

1. Which of the following is a factor of x+y3−x3+y3 ?

(A) x2+y2+2xy

(B) x2+y2−xy

(C) xy2

(D) 3xy

2. If x51+51 is divided by x + 1, the remainder is:

(A) 0

(B) 1

(C) 49

(D) 50

3. Which one of the following is a polynomial?

(A) x22−2x2

(B) 2x−1

(C) x2+3x32x

(D) x−1x+1

Sol

1. Solution: Option (d)

2. Solution: Option (d)

3. Solution: Option (c)

Questions

4. Prove that a+b+c3−a3−b3−c3=3a+bb+cc+a.

5. If a, b, c are all non-zero and a + b + c = 0, prove that: a2bc + b2ca + c2ab = 3

6. Give possible expressions for the length and breadth of the rectangle whose area is given by 4a2+4a−3.

7. Find the value of:

(a) x3+y3−12xy+64, when x + y = – 4

(b) x3−8y3−36xy−216, when x = 2y + 6

8. Factorise :

(i) a3−8b3−64c3−24abc

(ii) 22a3+8b3−27c3+182abc

(iii) 1−64a3−12a+48a2

(iv) 8p3+125p2+625p+1125

9. Show that p – 1 is a factor of p10−1 and also of p11−1.

10. (i) Without actually calculating the cubes, find the value of 483−303−183.

(ii) Without finding the cubes, factorise x−y3+y−z3+z−x3

Sol

4. Solution: Start by expanding both sides and then show that they are equal.

LHS

a+b+c3 = (a+b+c)(a+b+c)(a+b+c)

Expanding a+b+ca+b+c

a+b+ca+b+c = a2+ab+ac+ab+b2+bc+ac+bc+c2

= a2+b2+c2+2ab+2ac+2bc

Now,

a+b+ca+b+ca+b+c = a+b+ca2+b2+c2+2ab+2ac+2bc

= a3+b3+c3+3a2b+a2c+ab2+ac2+bc2+ac2+6abc

Then,

a+b+c3−a3−b3−c3 = 3a2b+a2c+ab2+ac2+bc2+ac2+6abc (1)

RHS

3a+bb+cc+a = 3a2b+a2c+ab2+b2c+bc2+c2a+2abc

= 3a2b+a2c+ab2+b2c+bc2+c2a+6ab (2)

Since, both (1) and (2) are equal.

a+b+c3−a3−b3−c3=3a+bb+cc+a

Hence, proved.

5. Solution:

Square both sides: a+b+c2 = 0

a2+b2+c2+2ab+2bc+2ac = 0

a2+b2+c2 = −2ab+bc+ac

Now, a2bc + b2ca + c2ab, multiply each fraction with abcabc.

a2bc + b2ca + c2ab = a3+b3+c3abc (1)

Now we need a3+b3+c3.

a3+b3+c3−3abc = a+b+ca2+b2+c2−ab−bc−ca

Since, a + b + c = 0: a3+b3+c3−3abc = 0

a3+b3+c3 = 3abc (2)

Putting (2) in (1),

a2bc + b2ca + c2ab = a3+b3+c3abc = 3abcabc = 3

Hence, proved.

6. Solution: To find possible expressions for the length and breadth of the rectangle whose area is given by 4a2+4a−3 , we need to factorize it.

To factorize, we look for two numbers that multiply to 4×−3=−12 and add up to 4.

The two numbers are 6 and −2 because:

6×−2=−12

6+−2=4

So, we can write: 4a2+4a−3 = 4a2+6a−2a−3

Now, we can group the terms and factor by grouping:

4a2+6a−2a−3 = 4a2+6a+−2a−3 = 2a2a+3−12a+3 = 2a+32a−1

Therefore, the possible expressions for the length and breadth of the rectangle are: 2a+3 and 2a−1.

So, the length and breadth of the rectangle could be 2a+3 and 2a−1, respectively.

7. Solution:

(a) Given: x + y = −4

We need to find the value of: x3+y3−12xy+64

Using the identity for the sum of cubes:

x3+y3 = x+yx2−xy+y2

We also need to express x2−xy+y2 in terms of x+y and xy. We can use:

x+y2 = x2+2xy+y2

Thus, x2+y2 = x+y2−2xy

Substituting x+y=−4: x2+y2 = 16−2xy

Now, let's use this to find x2−xy+y2:

x2−xy+y2 = x2+y2−xy = 16−2xy−xy = 16−3xy

Now we have:x3+y3 = x+yx2−xy+y2 = −416−3xy = −64+12xy

So, x3+y3−12xy+64 = −64+12xy−12xy+64 = 0

Therefore, the value is 0.

(b) Given : x=2y+6

Putting this in x3−8y3−36xy−216 = 2y+63−8y3−362y+6y−216

= 8y3+72y2+216y+216−8y3−362y+6y−216

= 8y3+72y2+216y+216−8y3−72y2+216y−216 = 0

Therefore, the value is 0.

8. Solution:

(i) To factorize the expression a3−8b3−64c3−24abc, we recognize it as a sum of cubes. The given expression can be rewritten using the identity for the difference of cubes:

a3−8b3 = a−2ba2+2ab+4b2

Similarly, 64c3 can be expressed as 4c3 and 24abc involves a common factor of 4abc:

64c3 = 4c3

Now, let's factorize the entire expression:

a3−8b3−64c3−24abc = a3−8b3−64c3+24abc = a−2ba2+2ab+4b2−44c3+6abc

So, the factorized form of a3−8b3−64c3−24abc is a−2ba2+2ab+4b2−44c3+6abc.

(ii) Factor out 2 from the first and last terms:

22a3+8b3−27c3+182abc = 22a3+92abc+8b3−27c3

Factorize 22a3+92abc

22a3+92abc = 2aa2+9bc

Factorize 8b3−27c3

8b3−27c3 = 2b3−3c3 = 2b−3c4b2+6bc+9c2

Substituting back into the expression:

22a3+8b3−27c3+182abc = 2aa2+9bc + 2b−3c4b2+6bc+9c2

Thus, the factorized form is 2aa2+9bc + 2b−3c4b2+6bc+9c2

(iii) Group the terms in pairs:

1−64a3−12a+48a2 = 1−64a3+48a2−12a

1−64a3 = 1−4a1+4a+16a2 (difference of cubes)

For the second group 48a2−12a,

48a2−12a = 12a(4a−1)

Now, rewrite the polynomial with the factored forms:

1−64a3−12a+48a2 = 1−4a1+4a+16a2−12a1−4a = 1−4a1+4a+4a2

Thus, the factorized form of 1−64a3−12a+48a2 is 1−4a1+4a+4a2.

(iv) 8p3+125p2+625p+1125

We will convert these fractions to have a common denominator to see if we recognize a cube pattern. Now we can see that the terms fit the pattern of the cube of a binomial a+b3=a3+3a2b+3ab2+b3, where: a=2p and b = 15

Thus, the given expression can be factored as follows: 8p3+125p2+625p+1125 = 2p+153

Therefore, the factorized form is 2p+153.

9. Solution:

Showing that p – 1 is a factor of p10−1

p10−1 = p52−12 = p5+1p5−1

To check if p−1 is a factor , substituting p = 1:

p10−1 = 110−1 = 1 - 1 = 0

Thus, p−1 is a factor of p10−1

Showing that p – 1 is a factor of p11−1

p11−1 = p−1p10+p9+p8+p7+p6+p5+p4+p3+p2+p+1 (This factorization comes from the sum of a geometric series)

To check if p−1 is a factor , substituting p = 1:

p11−1 = 111−1 = 1 - 1 = 0

Thus, p−1 is a factor of p11−1

10. Solution:

(i) Using a3−b3 = a−ba2+ab+b2 we get:

483−303 = 48−30482+4830+302 = 182304+1440+900 = 184644 = 83592

So, 483−303−183 = 83592 - 5832 = 77760

(ii) To factorize the expression x−y3+y−z3+z−x3, we can use the identity for the sum of cubes, which states:

a3+b3+c3−3abc = a+b+ca2+b2+c2−ab−bc−ca

Here, let a=x−y, b=y−z, and c=z−x. Applying these substitutions:

x−y3+y−z3+z−x3 = x−y+y−z+z−xx−y2+y−z2+z−x2−x−yy−z−y−zz−x−z−xx−y = 0

Therefore, the factorized form of x−y3+y−z3+z−x3 is 0.

Q1

Rahul is working on a polynomial project for his mathematics class. He decides to explore polynomials through practical examples and real-life applications. One day, he noticed a pattern in his garden. The length of the garden is represented by the polynomial L(x)=3x+2, and the width of the garden is represented by the polynomial W(x)=x−1. Rahul decides to calculate various aspects of his garden using these polynomials.

1. Rahul wants to find the polynomial that represents the area of his garden. The area of a rectangle is given by the product of its length and width. Calculate the polynomial representing the area of the garden.

2. Rahul also wants to find the polynomial representing the perimeter of his garden. The perimeter of a rectangle is given by 2×(Length+Width). Calculate the polynomial representing the perimeter of the garden.

3. If x = 4, find the area and perimeter of the garden.

4. Find the roots of the polynomial representing the width W(x) = x−1.

5. Determine the degree of the polynomial representing the area of the garden.

Sol

Solution:

1. Area of the Garden:

A(x) = L(x) × W(x) = (3x+2)(x−1) = 3x2−3x+2x−2

A(x) = 3x2−x−2

2. Perimeter of the Garden: P(x) = 2 × (L(x)+W(x)) = 2×(3x + 2 + x −1) = 2×(4x+1) = 8x+2

3. Value at x = 4:

L(4) = 3(4)+2 = 12+2 = 14

W(4) = 4−1 = 3

Area = L(4)×W(4) = 14×3 = 42

Perimeter = 8(4)+2 = 32+2 =

4. Roots of the Polynomial W(x) = x−1:

x−1 = 0

x =

5. Degree of the Polynomial Representing the Area:

The polynomial representing the area is 3x2−x−2. The degree of this polynomial is 2.