Polynomials II

 Value Based Questions: Polynomials

Ques 1: A charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth.
(i) Find the length and the breadth of the hall.
(ii) Which mathematical concept is used in the above problem?
(iii) By building a prayer hall, whose value is depicted by the trust?

Sol: 
(i) Let the breadth of the hall = x metres
∴ Length of the hall
= 2 [Breadth of the hall] + 1 metre
= (2x + 1) metre

∴ Carpet area of the hall= Length × Breadth
= (2x + 1) × x
= 2x2 + x
But the carpet area of the hall = 300 sq. metre
∴ 2x2 + x = 300
⇒ 2x2 + x - 300 = 0
⇒ 2x2 + 25x - 24x - 300 = 0
⇒ x(2x + 25) -12(2x + 25) = 0
⇒ (x - 12) (2x + 25) = 0
⇒ x - 12 = 0
or 2x + 25 = 0
⇒x = 12
or x = -  252 

But x = -  252   is not required as the length
or breadth cannot be negative.
∴ x = 12 m and 2x + 1 = (2 × 12) + 1 = 25 m
∴ Breadth of the hall = 12 m
Length of the hall = 25 m
(ii) Quadratic equations
(iii) Enhancing prayer and God fearing feelings.

Q2. Match the following, 

Sol: 

(P)
f(x) = x³ - 6x² + 11x - 6.
Find f(-1):
f(-1) = (-1)³ - 6(-1)² + 11(-1) - 6
= -1 - 6 - 11 - 6
= -24.
So, P → -24 (matches (iii)).

(Q)

f(x) = 2x³ - 13x² + 17x + 12.
Find f(-3):
f(-3) = 2(-3)³ - 13(-3)² + 17(-3) + 12
= 2(-27) - 13(9) - 51 + 12
= -54 - 117 - 51 + 12
= -210.
So, Q → -210 (matches (i))

(R)

Given x = 4/3 is a root of f(x) = 6x³ - 11x² + kx - 20.
A "root" means f(4/3) = 0. Substitute x = 4/3:
6(4/3)³ - 11(4/3)² + k(4/3) - 20 = 0
6·64/27 - 11·16/9 + 4k/3 - 20 = 0
= 128/9 - 176/9 + 4k/3 - 20 = 0
= (-48/9) + 4k/3 - 20 = 0
= (-16/3) + 4k/3 - 20 = 0.
Multiply everything by 3 to clear denominators:
-16 + 4k - 60 = 0 ⇒ 4k - 76 = 0 ⇒ k = 19.
So, R → 19 (matches (iv))
(S)

Given x = -1 is a root of f(x) = x¹⁰⁰ + 2x⁹⁹ + k.
So f(-1) = 0:

(-1)¹⁰⁰ + 2(-1)⁹⁹ + k = 0
1 - 2 + k = 0 ⇒ k = 1.
So, S → 1 (matches (ii)).

Q3. Santosh has ₹ (x³ - 3x² + 4x + 50). He want to buy chocolates each of cost ₹ (x - 3). After buying maximum number of chocolates with his money, how much money is left with him?

Sol: 
Step 1: Understand the question
Total money Santosh has: M = x³ - 3x² + 4x + 50
Cost of 1 chocolate: C = x - 3
We need to find remainder money after buying maximum chocolates. This means we have to divide the total money by the cost of one chocolate and see what is left (the remainder).

Step 2: Use polynomial division
We divide x³ - 3x² + 4x + 50 by x - 3.
After using long division method , we get
So, x³ - 3x² + 4x + 50 = (x - 3)(x² + 4) + 62

Step 3: Interpret remainder
The remainder 62 is the money left after buying the maximum number of chocolates.

Q4. Area of a rectangular field is (2x³ - 11x² - 4x + 5) sq. units and side of a square field is (2x² + 4) units. Find the difference between their areas (in sq. units).

Sol: Area of the square = (2x² + 4)² = 4x⁴ + 16x² + 16.
Area of the rectangle = 2x³ - 11x² - 4x + 5.
Difference = square area - rectangle area
= (4x⁴ + 16x² + 16) - (2x³ - 11x² - 4x + 5)
= 4x⁴ - 2x³ + 27x² + 4x + 11

Q5.  Length, breadth and height of a cuboidal tank are (x - 3y) m, (x + 3y) m and (x² + 9y²) m respectively. Find the volume of the tank.

Sol: Volume = length × breadth × height
= (x - 3y)(x + 3y)(x² + 9y²)
= (x² - 9y²)(x² + 9y²) [using (a - b)(a + b) = a² - b²]
= x⁴ - 81y⁴.​​​​

 

Mind Map: Polynomials

Case Based Questions: Polynomials

Q1: Read the source below and answer the questions that follow:

Ramesh was asked by one of his friends Anirudh to find the polynomial whose zeroes are -2/√3 and √3/4.
He obtained the polynomial by following steps which are as shown below:
Let α = -2/√3 and β = √3/4
Then,
α + β = (-2/√3) + (√3/4) = (-8 + 1) / (4√3) = -7 / (4√3)
and
αβ = (-2/√3) × (√3/4) = -1/2
∴ Required polynomial = x² - (α + β)x + αβ
= x² - (-7/4√3)x + (-1/2)
= x² + (7x/4√3) - 1/2
= 4√3x² + 7x - 2√3
His another friend Kavita pointed out that the polynomial obtained is not correct.

i. Is the claim of Kavita correct? (1 mark)
ii. If given polynomial is incorrect, then find the correct quadratic polynomial. (1 mark)
iii. Find the value of α² + β². (2 mark)
Or
iii. If correct polynomial p(x) is a factor of (x - 2), then find f(2). (2 mark)

Ans:
i. Given, α = -2/√3 and β = √3/4
∴ α + β = (-2/√3) + (√3/4) = (-8 + 3) / 4√3 = -5 / 4√3
and αβ = (-2/√3) × (√3/4) = -1/2
Yes, because value of (α + β) calculated by Anirudh is incorrect.
ii. Required polynomial = k(x² - (α + β)x + αβ)
k(x² + (5x/4√3) - 1/2)
= (k/4√3)(4√3x² + 5x - 2√3)
= (4√3x² + 5x - 2√3) where k = 4√3
iii. α² + β² = (α + β)² - 2αβ
= (-5/4√3)² - 2 × (-1/2)
= 25/48 + 1 = 73/48
Alternate method:
α² + β² = (-2/√3)² + (√3/4)²
= 4/3 + 3/16 = 64/48 + 9/48 = 73/48
Or
iii. We have, p(x) = 4√3x² + 5x - 2√3
Since, p(x) is a factor of (x - 2), then
p(2) = 4√3(2)² + 5(2) - 2√3
= 16√3 + 10 - 2√3 = 14√3 + 10
Hence, remainder is 14√3 + 10.

Get additional INR 200 off today with EDUREV200 coupon.

Avail Offer

Q2: Read the source below and answer the questions that follow:

A group of school friends went on an expedition to see caves. One person remarked that the entrance of the caves resembles a parabola and can be represented by a quadratic polynomial f(x)= ax2 + bx + c, a# 0, where a, b and c are real numbers.
i. Draw a neat labelled figure to show above situation diagrammatically. (1 mark)
ii. If one of the zeroes of the quadratic polynomial (p-1)x² + px + 1 is 4. Find the value of p. (1 mark)
iii. Find the quadratic polynomial whose zeroes are 5 and -12. (1 mark)
iv. If one zero of the polynomial f(x) = 5x² + 13x + m is reciprocal of the other, then find the value of m. (1 mark)

Ans:
i. We have, f(x) = ax² + bx + c, a<0 It means a figure is a shape of parabola which open downwards.
ii. Since, x = 4 is one of the zeros of the polynomial (p - 1)x² + px + 1.
∴ (p - 1)(4)² + p(4) + 1 = 0
⇒ 16p - 16 + 4p + 1 = 0
⇒ 20p = 15
⇒ p = 3/4
iii. Required quadratic polynomial = k [x² - (Sum of zeroes)x + (Product of zeroes)] = k(x²-(-12+5)x+(-12)(5)) = k(x²+7x-60), where k is any arbitrary constant.  
iv. Let the zeroes of the quadratic polynomial be α. Since, one zero of the polynomial f(x) is reciprocal of the other. 
∴ Product of zeroes = (-1)² × (Constant term / Coefficient of x²)
⇒ α × (1/α) = 1 × (m/5) ⇒ m = 5

Also read: Mind Map: Polynomials

Q3: Read the source below and answer the questions that follow:

A student was given a task to prepare a graph of quadratic polynomial p(x)=-8-2x+x². To draw this graph, he take seven values of y corresponding to different values of x. After plotting the points on the graph paper with suitable values, he obtain the graph as shown below.
i. What is the shape of graph of a quadratic polynomial? (1 mark)
ii. Find the zeroes of given quadratic polynomial. (1 mark)
iii. The graph of the given quadratic polynomial cut at which points on the X-axis? (1 mark)
iv. The graph of the given quadratic polynomial cut at which point on Y-axis? (1 mark)

Ans:
i. The graph of a quadratic polynomial is a parabola which open upwards. 
ii. The zeroes of the quadratic polynomial p(x) = -8 -2x + x² are x-coordinates of the points where the graph intersects the X-axis. From the given graph, -2 and 4 are the x-coordinates of the points where the graph of p(x)=-8-2x+x² intersects the X-axis. Hence, -2 and 4 are zeroes of p(x)=-8-2x+x². 
iii. The graph of the given quadratic polynomial cut X-axis at points (-2, 0) and (4,0). 
iv. The graph of the given quadratic polynomial cut Y-axis at point (0, -8). 

Attention! Sale expiring soon, act now & get EduRev Infinity at 40% off!

Claim Offer

Q4: Read the source below and answer the questions that follow:

Pooja was asked by her friend Arjun to find a polynomial whose zeroes are 3/√2 and -√2/5. She performed the following steps: Let α = 3/√2 and β = -√2/5. Then, α + β = 3/√2 + (-√2/5) = (15 - 2)/(5√2) = 13/(5√2) and αβ = 3/√2 × (-√2/5) = -3/5. Thus, the polynomial obtained is x² - (α + β)x + αβ = x² - 13x/(5√2) - 3/5 = 5√2x² - 13x - 3√2. However, her friend Neha pointed out an error in her solution.

i. Is Neha correct in her claim? (1 mark) 
ii. If the polynomial is incorrect, find the correct polynomial. (1 mark) 
iii. Find the value of α² + β². (1 marks)
iv. If the correct polynomial q(x) is a factor of (x+3), find q(-3). (1 marks)

Ans:
i. Yes, Neha is correct. 
The sum of roots should be α + β = 3/√2 + (-√2/5) = (15 - 2)/(5√2) = 13/(5√2), 
which is correct, but the final polynomial calculation by Pooja is incorrect.
ii. Correct Polynomial:
p(x) = k(x² - (α + β)x + αβ)
= k(x² - 13x/(5√2) - 3/5)
Multiplying by 5√2 to remove the denominator:
= 5√2x² - 13x - 3√2
where k = 5√2.
iii. Value of α² + β²:
α² + β² = (α + β)² - 2αβ
= (13/(5√2))² - 2(-3/5)
= 169/50 + 6/5
= 169+60/50 = 229/50.
iv. If q(x) = 5√2x² - 13x - 3√2 is a factor of (x + 3):
q(-3) = 5√2(-3)² - 13(-3) - 3√2
= 5√2(9) + 39 - 3√2
= 45√2 + 39 - 3√2
= 42√2 + 39.

Long Questions: Polynomials

Q1: Find the value of "p" from the polynomial x2 + 3x + p, if one of the zeroes of the polynomial is 2.
Ans: As 2 is the zero of the polynomial.
We know that if α is a zero of the polynomial p(x), then p(α) = 0
Substituting x = 2 in x2 + 3x + p,
⇒ 22 + 3(2) + p = 0
⇒ 4 + 6 + p = 0
⇒ 10 + p = 0
⇒ p = -10

Q2: Compute the zeroes of the polynomial 4x2 - 4x - 8. Also, establish a relationship between the zeroes and coefficients.
Ans: Let the given polynomial be p(x) = 4x2 - 4x - 8
To find the zeroes, take p(x) = 0
Now, factorise the equation 4x2 - 4x - 8 = 0
4x2 - 4x - 8 = 0
4(x2 - x - 2) = 0
x2 - x - 2 = 0
x2 - 2x + x - 2 = 0
x(x - 2) + 1(x - 2) = 0
(x - 2)(x + 1) = 0
x = 2, x = -1
So, the roots of 4x2 - 4x - 8 are -1 and 2.
Relation between the sum of zeroes and coefficients:
-1 + 2 = 1 = -(-4)/4 i.e. (- coefficient of x/ coefficient of x2)
Relation between the product of zeroes and coefficients:
(-1) × 2 = -2 = -8/4 i.e (constant/coefficient of x2)

Q3: Find the quadratic polynomial if its zeroes are 0, √5.
Ans: A quadratic polynomial can be written using the sum and product of its zeroes as:
x2 - (α + β)x + αβ
Where α and β are the roots of the polynomial.
Here, α = 0 and β = √5
So, the polynomial will be:
x2 - (0 + √5)x + 0(√5)
= x2 - √5x

Q4: Find the value of "x" in the polynomial 2a2 + 2xa + 5a + 10 if (a + x) is one of its factors.
Ans: Let f(a) = 2a2 + 2xa + 5a + 10
Since, (a + x) is a factor of 2a2 + 2xa + 5a + 10, f(-x) = 0
So, f(-x) = 2x2 - 2x2 - 5x + 10 = 0
-5x + 10 = 0
5x = 10
x = 10/5
Therefore, x = 2

Q5: How many zeros does the polynomial (x - 3)2 - 4 have? Also, find its zeroes.
Ans: Given polynomial is (x - 3)2 - 4
Now, expand this expression.
=> x2 + 9 - 6x - 4
= x2 - 6x + 5
As the polynomial has a degree of 2, the number of zeroes will be 2.
Now, solve x2 - 6x + 5 = 0 to get the roots.
So, x2 - x - 5x + 5 = 0
=> x(x - 1) -5(x - 1) = 0
=> (x - 1)(x - 5) = 0
x = 1, x = 5
So, the roots are 1 and 5.

Q6: α and β are zeroes of the quadratic polynomial x2 - 6x + y. Find the value of 'y' if 3α + 2β = 20.
Ans: Let, f(x) = x² - 6x + y
From the given,
3α + 2β = 20-------(i)
From f(x),
α + β = 6-------(ii)
And,
αβ = y-------(iii)
Multiply equation (ii) by 2. Then, subtract the whole equation from equation (i),
=> α = 20 - 12 = 8
Now, substitute this value in equation (ii),
=> β = 6 - 8 = -2
Substitute the values of α and β in equation (iii) to get the value of y, such as;
y = αβ = (8)(-2) = -16

Q7: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.
(i) 1/4, -1
(ii) 1, 1
(iii) 4, 1
Ans:
(i) From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α + β
Product of zeroes = αβ
Given,
Sum of zeroes = 1/4
Product of zeroes = -1
Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-
x2 - (α + β)x + αβ
= x2 - (1/4)x + (-1)
= 4x2 - x - 4
Thus, 4x2 - x - 4 is the required quadratic polynomial.
(ii) Given,
Sum of zeroes = 1 = α + β
Product of zeroes = 1 = αβ
Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-
x2 - (α + β)x + αβ
= x2 - x + 1
Thus, x2 - x + 1 is the quadratic polynomial.
(iii) Given,
Sum of zeroes, α + β = 4
Product of zeroes, αβ = 1
Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:-
x2 - (α + β)x + αβ
= x2 - 4x + 1
Thus, x2 - 4x +1 is the quadratic polynomial.

Q8: Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x) = ax2 + bx + c, a ≠ 0, c ≠ 0.
Ans: Let α and β be the zeroes of the polynomial f(x) = ax2 + bx + c.
So, α + β = -b/a
αβ = c/a
According to the given, 1/α and 1/β are the zeroes of the required quadratic polynomial.
Now, the sum of zeroes = (1/α) + (1/β) 
= (α + β)/αβ
= (-b/a)/ (c/a)
= -b/c
Product of two zeroes = (1/α) (1/β)
= 1/αβ
= 1/(c/a)
= a/c
The required quadratic polynomial = k[x2 - (sum of zeroes)x + (product of zeroes)]
= k[x2 - (-b/c)x + (a/c)]
= k[x2 + (b/c)x + (a/c)]

Q9: If α and β are zereos of the polynomial 2x2 - 5x + 7, then find the value of α-1 + β-1.
Ans: Here p(x) = 2x2 - 5x + 7
α, β are zeroes of p(x)

Q10: If p and q are the roots of ax2 - bx + c = 0, a ≠ 0, then find the value of p + q.
Ans: Here, p and q are the roots of ax2 - bx + c = 0.
Sum of roots = -b/a
∴ p + q = -b/a

RS Aggarwal Solutions: Polynomials (Exercise 2A)

Unit Test (Solutions): Polynomials

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers  9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: Which of the following is NOT a polynomial?  (1 Mark) 
(a) 2x + 3
(b) 4x2 - 7x + 1
(c) 2√x + 1
(d) x3 - 5x2 + 4x - 7
Ans: (c)
A polynomial is an expression consisting of variables and coefficients, combined using addition, subtraction, and multiplication, where the exponents of variables are whole numbers (non-negative integers). The option 2√x + 1 contains a square root (√) in the expression, making it not a polynomial. All other options are polynomials.

Q2: If a polynomial has exactly three terms, what is it called?  (1 Mark) 
(a) Monomial
(b) Binomial
(c) Trinomial
(d) Polynomial
Ans: (c)
A polynomial with exactly three terms is called a trinomial. If it has one term, it's a monomial, and if it has two terms, it's a binomial. Any expression with more than three terms is still considered a polynomial.

Q3: What is the degree of the polynomial 2x3 - 5x2 + 7x - 9? (1 Mark) 
(a) 2
(b) 3
(c) 4
(d) 1
Ans: (b)
The degree of a polynomial is the highest power of the variable present in the expression. In this case, the highest power of x is 3 (in the term 2x^3), so the degree of the polynomial is 3.

Q4: State whether the following statement is True or False: "A polynomial of degree 0 is called a constant polynomial."  (1 Mark) 
Ans: True
A polynomial of degree 0 is called a constant polynomial. It is a polynomial with no variable term, and the expression consists only of a constant value.

Q5: Identify the coefficients of the following polynomial: 3x2 - 8x + 5.  (1 Mark) 
Ans: The coefficients are 3, -8, and 5.
The coefficients of a polynomial are the numerical values multiplied by each variable term. In this case, the coefficients are 3 (multiplied by x^2), -8 (multiplied by x), and 5 (constant term).

Q6: Find the value of 'k' in the polynomial x2 - kx + 9, if one of its zeros is 3.  (2 Marks) 
Ans: k = 6
If one of the zeros of a polynomial is 3, it means that when x = 3, the polynomial becomes equal to 0. Substituting x = 3 in the given polynomial and solving for 'k':
32 - 3k + 9 = 0
9 - 3k + 9 = 0
-3k + 18 = 0
-3k = -18
k = -18 / -3
k = 6

Q7: Factorize the polynomial completely: 2x3 - 8x2 + 4x. (2 Marks) 
Ans: 2x(x2 - 4x + 2)
To factorize the polynomial completely, first, we take out the common factor '2x' from each term:
2x3 - 8x2 + 4x = 2x(x2 - 4x + 2)
The expression (x2 - 4x + 2) cannot be further factorized using integers. So, the polynomial is completely factored as 2x(x2 - 4x + 2).

Q8: If (x + 2) is a factor of the polynomial 2x3 + kx2 - 4x + 16, find the value of 'k'. (2 Marks) 
Ans: k = -2
If (x + 2) is a factor of the polynomial, it means that when x = -2, the polynomial becomes equal to 0. Substituting x = -2 in the given polynomial and solving for 'k':
2(-2)3 + k(-2)2 - 4(-2) + 16 = 0
-16 + 4k + 8 + 16 = 0
4k + 8 = 0
4k = -8
k = -8 / 4
k = -2

Q9: If the sum of the zeroes of the polynomial x2 - kx + 15 is 5, find the value of 'k'. (3 Marks) 
Ans: k = 5
The sum of the zeroes of the polynomial x2 - kx + 15 is given by the formula: Sum of zeroes = -Coefficient of x / Coefficient of x2.
In this case, the sum of zeroes is 5. So,
5 = -(-k) / 1
5 = k
Therefore, the value of k is 5.

Q10: The area of a rectangular garden is given by the polynomial 2x2 + 7x - 15. Find the dimensions of the garden if its length is 2 units more than its breadth.  (3 Marks) 
Ans:

Substitute $x=6$x=6 back into the dimensions:
Breadth = 2(6) - 3 = 9 units​Length = 6 + 5 = 11 unit

$=2\left(6\right)-3=9,=6+5=\mathrm{<br\; style="box-sizing:\; border-box;\; font-family:\; Lato,\; sans-serif;\; letter-spacing:\; inherit;\; line-height:\; 1.8;\; text-align:\; inherit;\; user-select:\; text;\; word-break:\; break-word;">}$

Q11: Find the value of 'p' for which (2p - 1) is a factor of the polynomial 4p3 - 6p2 + 3p - 1/2. (3 Marks) 
Ans:

Q12: Find the zeroes of the polynomial 2x2 - 7x + 3 and verify the relationship between zeroes and coefficients.  (5 Marks) 

Ans: Given polynomial: p(x) = 2x² - 7x + 3

Step 1: Find the zeroes

Factorize the polynomial:

2x² - 7x + 3
= 2x² - 6x - x + 3
= 2x(x - 3) - 1(x - 3)
= (2x - 1)(x - 3)

So, the zeroes are:
2x - 1 = 0 ⇒ x = 1/2
x - 3 = 0 ⇒ x = 3

Step 2: Verify relationship between zeroes and coefficients

For polynomial ax² + bx + c:
a = 2, b = -7, c = 3

Let zeroes be α = 1/2 and β = 3

Sum of zeroes:

α + β = 1/2 + 3 = 7/2

According to formula:
-b/a = -(-7)/2 = 7/2
Hence, verified.

Product of zeroes:

αβ = (1/2) × 3 = 3/2

According to formula:
c/a = 3/2
Hence, verified.

Q13: Find a quadratic polynomial whose zeroes are 3 and -2. Also verify the relationship between the zeroes and the coefficients.  (5 Marks)

Ans: Given zeroes: α = 3, β = -2

Step 1: Find the polynomial

Sum of zeroes = α + β = 3 + (-2) = 1
Product of zeroes = αβ = 3 × (-2) = -6

Quadratic polynomial =
x² - (sum)x + (product)

= x² - (1)x + (-6)
= x² - x - 6

Step 2: Verify relationship between zeroes and coefficients

For polynomial ax² + bx + c:
a = 1, b = -1, c = -6

Sum of zeroes:

α + β = 1

According to formula:
-b/a = -(-1)/1 = 1

Hence, verified.

Product of zeroes:

αβ = -6

According to formula:
c/a = -6/1 = -6

Hence, verified.