CBSE Class 10 Mathematics Polynomials Assignment Set 01

CBSE Class 10 Mathematics Polynomials Assignment Set A. Students are advised to refer to the attached assignments and practise them regularly. This will help them to identify their weak areas and will help them to score better in examination. Parents should download and give the assignments to their children for practice.

Question. The zeroes of the quadratic polynomial x² + 7x + 10 are
(a) -4, -3
(b) 2, 5
(c) -2, -5
(d) -2, 5
Answer : C

Question. A quadratic polynomial, whose zeroes are -2 and 4, is
(a) x² – 2x + 8
(b) x²+ 2x + 8
(c) x²– 2x- 8
(d) 2x²+ 2x-24
Answer : C

Question. The zeroes of the quadratic polynomial x² + kx + k where k ≠ 0,
(a) cannot both be positive
(b) cannot both be negative
(c) are always unequal
(d) are always equal
Answer : A

Question. The zeroes of the quadratic polynomial x² - 27 are
(a) +3√3 ,- 3√3
(b) 3 , 3
(c) 9 ,9
(d) +√3 , - √3
Answer : A

Question. The sum and the product of the zeroes of polynomial 6x² – 5 respectively are
(a) 0, (-6)/5
(b) 0, 6/5
(c) 0, 5/6
(d) 0, (-5)/6
Answer : D

Question. A quadratic polynomial can have at most ______ zeroes.
(a) 0
(b) 1
(c) 2
(d) infinite
Answer : C

Question. If the zeroes of the quadratic polynomial ax²+ bx+ c, where c≠0, are equal, then
(a) c and a have opposite signs
(b) c and b have opposite signs
(c) c and a have same signs
(d) c and b have the same signs
Answer : C

Level 1 (1 mark)

Question. The number of zeroes, the polynomial f(x) = (x – 3)²+ 1 can have is :
(a) 0
(b) 1
(c) 2
(d) 3
Answer : C

Question. The graph of the polynomial p(x) cuts the x-axis 5 times and touches it 3 times. The number of zeroes of p(x) is : (a) 5 (b) 3 (c) 8 (d) 2
Answer : C

3. If the zeroes of the quadratic polynomial x²+ (a + 1)x+ b are 2 and –3, then :
(a) a = –7, b = –1 (b) a = 5, b = –1
(c) a = 2, b = –6 (d) a = 0, b = –6
Answer : D

Question. The zeroes of the quadratic polynomial x² + 89x + 720 are :
(a) both are negative
(b) both are positive
(c) one is positive and one is negative
(d) both are equal
Answer : A

Very Short Answer type Questions

Question. Obtain the zeros of the quadratic polynomial √3x² − 8x + 4 √3 and verify the relation between its zeros and coefficients.
Answer : Zeros of are 2√3 and 2/√3

Question. Find a quadratic polynomial, the sum and product of whose zeroes are √2 and 1/3, respectively.
Answer : 3x² − 3√2 x + 1

Question. Find a quadratic polynomial, the sum and product of whose zeroes are 0 and √5, respectively.
Answer : x² + √5

Question. Find the zeros of the polynomial 6x² − 3 − 7x and verify the relationship between the zeros and the coefficients.
Answer : Zeros of are 3/2, 1/3

Question. If the product of the zeros of the polynomial (ax² − 6x − 6 )is 4, find the value of a.
Answer : -(3/2)

Question. If one zero of the polynomial (a₂ + 9)x² + 13x + 6a is reciprocal of the other, find the value of a.
Answer : a = 3.

Level 1 (1 mark)

Question 1. The number of zeroes, the polynomial \(f(x) = (x - 3)^2 + 1\) can have is:
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (a) 0
For any real value of \(x\), \((x - 3)^2 \geq 0\).
Adding 1 gives \(f(x) = (x - 3)^2 + 1 \geq 1\).
Thus, \(f(x)\) is always positive and never equals 0 for any real number \(x\). Hence, it has no real zeroes.
In simple words: The graph of this polynomial lies entirely above the x-axis and never intersects it, so there are no zeroes.

Exam Tip: Whenever a squared term is added to a positive constant, the expression cannot yield zero for real inputs.

Question 2. The graph of the polynomial \(p(x)\) cuts the x-axis 5 times and touches it 3 times. The number of zeroes of \(p(x)\) is:
(a) 5
(b) 3
(c) 8
(d) 2
Answer: (c) 8
The total number of zeroes of a polynomial is equal to the total number of points where its graph intersects or touches the x-axis.
Number of intersections = 5
Number of touches = 3
Total zeroes = \(5 + 3 = 8\).
In simple words: Count every single time the graph crosses or touches the x-axis to find the total zeroes.

Exam Tip: Both crossing and touching the x-axis contribute to the count of zeroes in polynomial graphs.

Question 3. If the zeroes of the quadratic polynomial \(x^2 + (a + 1)x + b\) are 2 and -3, then:
(a) \(a = -7, b = -1\)
(b) \(a = 5, b = -1\)
(c) \(a = 2, b = -6\)
(d) \(a = 0, b = -6\)
Answer: (d) \(a = 0, b = -6\)
Let the polynomial be \(p(x) = x^2 + (a + 1)x + b\).
Sum of zeroes = \(2 + (-3) = -1\)
Using coefficients, sum of zeroes = \(-\frac{a + 1}{1} = -1\)
\(-(a + 1) = -1 \implies a + 1 = 1 \implies a = 0\)
Product of zeroes = \((2)(-3) = -6\)
Using coefficients, product of zeroes = \(\frac{b}{1} = b\)
\(b = -6\).
In simple words: Use the relationship between zeroes and coefficients to set up equations for sum and product.

Question 4. The zeroes of the quadratic polynomial \(x^2 + 89x + 720\) are:
(a) both are negative
(b) both are positive
(c) one is positive and one is negative
(d) both are equal
Answer: (a) both are negative
Let \(p(x) = x^2 + 89x + 720\).
Here, the coefficient of \(x\) is positive (+89) and the constant term is positive (+720).
If the roots are \(\alpha\) and \(\beta\), then sum \(\alpha + \beta = -89\) (negative) and product \(\alpha\beta = 720\) (positive).
For the product to be positive and the sum to be negative, both numbers must be negative.
In simple words: When all coefficients in a quadratic polynomial are positive, both its zeroes are negative.

Exam Tip: Quickly determine sign properties using sum and product rules without fully factoring large numbers.

Level 2 (2 marks)

Question 5. If \(\alpha\) and \(\beta\) are zeros of the polynomial \(3x^2 + 5x + 2\), find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).
Answer:
Given polynomial: \(3x^2 + 5x + 2\)
Here, \(a = 3, b = 5, c = 2\).
Sum of zeroes: \(\alpha + \beta = -\frac{b}{a} = -\frac{5}{3}\)
Product of zeroes: \(\alpha\beta = \frac{c}{a} = \frac{2}{3}\)
Now, find \(\frac{1}{\alpha} + \frac{1}{\beta}\):
\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}\)
Substitute the values:
\(= \frac{-\frac{5}{3}}{\frac{2}{3}} = -\frac{5}{2}\)
In simple words: Convert the target expression into sum and product form, then substitute the coefficient values.

Exam Tip: Always write down \(\alpha + \beta\) and \(\alpha\beta\) first before simplifying fractional algebraic expressions.

Question 6. Find the zeros of the quadratic polynomial \(6x^2 - 7x - 3\) and verify the relationship between the zeros and the coefficients.
Answer:
Let \(P(x) = 6x^2 - 7x - 3\)
Split the middle term:
\(= 6x^2 - 9x + 2x - 3\)
\(= 3x(2x - 3) + 1(2x - 3)\)
\(= (2x - 3)(3x + 1)\)
Setting \(P(x) = 0\):
\(2x - 3 = 0 \implies x = \frac{3}{2}\)
\(3x + 1 = 0 \implies x = -\frac{1}{3}\)
So, the zeroes are \(\frac{3}{2}\) and \(-\frac{1}{3}\).
Verification:
Sum of zeroes = \(\frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9 - 2}{6} = \frac{7}{6}\)
From coefficients, \(-\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6}\). (Hence verified)
Product of zeroes = \(\left(\frac{3}{2}\right) \times \left(-\frac{1}{3}\right) = -\frac{1}{2}\)
From coefficients, \(\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}\). (Hence verified)
In simple words: Factor the quadratic by splitting the middle term, find the roots, and check them against standard coefficient formulas.

Exam Tip: Clearly show both the factored steps and the verification formulas to secure full marks.

Question 7. Write the zeroes of the polynomial \(x^2 - x - 6\).
Answer:
Let \(p(x) = x^2 - x - 6\)
Split the middle term:
\(x^2 - 3x + 2x - 6 = 0\)
\(x(x - 3) + 2(x - 3) = 0\)
\((x - 3)(x + 2) = 0\)
Equating each factor to zero:
\(x - 3 = 0 \implies x = 3\)
\(x + 2 = 0 \implies x = -2\)
Therefore, the zeroes of the polynomial are 3 and -2.
In simple words: Factor the quadratic expression into two linear binomials and solve for x.

Exam Tip: Double-check middle-term splitting signs carefully to avoid sign errors.

Question 8. Find a quadratic polynomial with sum of zeroes = \(\frac{1}{4}\) and product of zeroes = \(\frac{1}{4}\).
Answer:
Let sum of zeroes be \(S = \frac{1}{4}\) and product of zeroes be \(P = \frac{1}{4}\).
A quadratic polynomial is given by the formula:
\(P(x) = k(x^2 - Sx + P)\) where \(k\) is a non-zero real constant.
\(P(x) = k\left(x^2 - \frac{1}{4}x + \frac{1}{4}\right)\)
Taking \(k = 4\):
\(P(x) = 4x^2 - x + 1\)
Therefore, the required quadratic polynomial is \(4x^2 - x + 1\).
In simple words: Substitute the given sum and product directly into the standard quadratic template.

Exam Tip: Multiplying by the denominator clears fractions and presents the polynomial in standard integer coefficient form.

Level 3 (3 marks)

Question 9. Find the zeroes of quadratic \(x^2 - 2x - 8\) and verify the relationship between the zeroes and their coefficient.
Answer:
We have \(f(x) = x^2 - 2x - 8\)
\(f(x) = x^2 - 4x + 2x - 8\)
\(f(x) = x(x - 4) + 2(x - 4)\)
\(f(x) = (x - 4)(x + 2)\)
For zeroes of \(f(x)\), set \(f(x) = 0\):
\(x + 2 = 0 \implies x = -2\)
\(x - 4 = 0 \implies x = 4\)
Let \(\alpha = -2\) and \(\beta = 4\).
Sum of zeroes = \(\alpha + \beta = -2 + 4 = 2\)
From coefficients: \(-\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{-2}{1} = 2\) (Verified)
Product of zeroes = \(\alpha\beta = (-2)(4) = -8\)
From coefficients: \(\frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{-8}{1} = -8\) (Verified)
In simple words: Factor the polynomial, find the roots, and verify them using standard coefficient formulas.

Exam Tip: State the verification formulas clearly before plugging in the values.

Question 10. Obtain all other zeroes of \(3x^4 + 6x^3 - 2x^2 - 10x - 5\), if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\).
Answer:
Since \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\) are two zeroes of the polynomial, the factors are:
\(\left(x - \sqrt{\frac{5}{3}}\right)\left(x + \sqrt{\frac{5}{3}}\right) = x^2 - \frac{5}{3}\)
Thus, \(3x^2 - 5\) is a factor of the given polynomial \(p(x)\).
Now, divide \(3x^4 + 6x^3 - 2x^2 - 10x - 5\) by \(3x^2 - 5\):
Quotient = \(x^2 + 2x + 1\)
Factorize the quotient:
\(x^2 + 2x + 1 = (x + 1)(x + 1)\)
Setting the factors to zero gives \(x = -1\) and \(x = -1\).
Therefore, all the zeroes of the polynomial are \(\sqrt{\frac{5}{3}}, -\sqrt{\frac{5}{3}}, -1,\) and \(-1\).
In simple words: Use the given zeroes to form a quadratic factor, divide the polynomial by it, and factorize the resulting quotient.

Exam Tip: Polynomial long division is essential here; align powers of x carefully.

Question 11. Find the zeros of the polynomial \(4\sqrt{3}x^2 + 5x - 2\sqrt{3}\).
Answer:
Let \(F(x) = 4\sqrt{3}x^2 + 5x - 2\sqrt{3}\)
Product of coefficients = \(4\sqrt{3} \times (-2\sqrt{3}) = -24\)
Sum = \(5\)
Split the middle term using 8 and -3:
\(F(x) = 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3}\)
\(F(x) = 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2)\)
\(F(x) = (\sqrt{3}x + 2)(4x - \sqrt{3})\)
To find the zeroes, set \(F(x) = 0\):
\(\sqrt{3}x + 2 = 0 \implies x = -\frac{2}{\sqrt{3}}\)
\(4x - \sqrt{3} = 0 \implies x = \frac{\sqrt{3}}{4}\)
Therefore, the zeroes are \(-\frac{2}{\sqrt{3}}\) and \(\frac{\sqrt{3}}{4}\).
In simple words: Factor the quadratic by splitting the middle term with radical coefficients, then solve each linear factor.

Exam Tip: Be careful with square roots during grouping and factoring steps.

Question 12. If \(m\) and \(n\) are the zeros of the polynomial \(3x^2 + 11x - 4\), find the value of \(\frac{m}{n} + \frac{n}{m}\).
Answer:
Given polynomial: \(3x^2 + 11x - 4\)
Here, \(a = 3, b = 11, c = -4\).
Sum of zeroes: \(m + n = -\frac{11}{3}\)
Product of zeroes: \(mn = -\frac{4}{3}\)
Now, simplify the expression \(\frac{m}{n} + \frac{n}{m}\):
\(\frac{m}{n} + \frac{n}{m} = \frac{m^2 + n^2}{mn} = \frac{(m + n)^2 - 2mn}{mn}\)
Substitute the values:
\(= \frac{\left(-\frac{11}{3}\right)^2 - 2\left(-\frac{4}{3}\right)}{-\frac{4}{3}}\)
\(= \frac{\frac{121}{9} + \frac{8}{3}}{-\frac{4}{3}} = \frac{\frac{121 + 24}{9}}{-\frac{4}{3}} = \frac{\frac{145}{9}}{-\frac{4}{3}}\)
\(= \frac{145}{9} \times \left(-\frac{3}{4}\right) = -\frac{145}{12}\)
In simple words: Express \(\frac{m}{n} + \frac{n}{m}\) in terms of \((m+n)\) and \(mn\), then substitute their values.

Exam Tip: Remember the algebraic identity \(m^2 + n^2 = (m + n)^2 - 2mn\).

Level 4 (4 marks)

Question 13. If \(p\) and \(q\) are the zeroes of polynomial \(f(x) = ax^2 - 5x + c\), find the values of \(a\) and \(c\), if \(p + q = pq = 10\).
Answer:
Given polynomial: \(f(x) = ax^2 - 5x + c\)
Sum of zeroes: \(p + q = -\frac{-5}{a} = \frac{5}{a}\)
Product of zeroes: \(pq = \frac{c}{a}\)
Given that \(p + q = pq = 10\).
From \(p + q = 10\):
\(\frac{5}{a} = 10 \implies a = \frac{5}{10} = \frac{1}{2}\) ---- (i)
From \(pq = 10\):
\(\frac{c}{a} = 10 \implies c = 10a\)
Substitute \(a = \frac{1}{2}\) into the equation for \(c\):
\(c = 10 \left(\frac{1}{2}\right) = 5\)
Therefore, the values are \(a = \frac{1}{2}\) and \(c = 5\).
In simple words: Set the sum and product formulas equal to 10 to solve directly for a and c.

Exam Tip: Equate coefficient formulas directly with the given numerical values.

Question 14. If the sum of the squares of zeroes of the polynomial \(6x^2 + x + k\) is \(\frac{25}{36}\), find the value of \(k\).
Answer:
Let the polynomial be \(6x^2 + x + k\). Here, \(a = 6, b = 1, c = k\).
Let the zeroes be \(\alpha\) and \(\beta\).
Sum of zeroes: \(\alpha + \beta = -\frac{1}{6}\)
Product of zeroes: \(\alpha\beta = \frac{k}{6}\)
Given that \(\alpha^2 + \beta^2 = \frac{25}{36}\).
Using the identity \((\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta\):
\(\left(-\frac{1}{6}\right)^2 = \frac{25}{36} + 2\left(\frac{k}{6}\right)\)
\(\frac{1}{36} = \frac{25}{36} + \frac{2k}{6}\)
\(\frac{2k}{6} = \frac{1}{36} - \frac{25}{36} = -\frac{24}{36}\)
\(\frac{k}{3} = -\frac{24}{36}\)
\(k = -\frac{24}{36} \times 3 = -\frac{72}{36} = -2\)
In simple words: Expand \((\alpha + \beta)^2\) to incorporate \(\alpha^2 + \beta^2\), then solve for \(k\).

Exam Tip: Substitute \(\alpha^2 + \beta^2\) and expand carefully while handling fractions.

Question 15. If \(\alpha\) and \(\beta\) are two zeroes of the quadratic polynomial \(p(x) = 2x^2 - 3x + 7\), find: (a) \(\frac{1}{\alpha} + \frac{1}{\beta}\) (b) \(\alpha^2 + \beta^2\).
Answer:
Given polynomial: \(p(x) = 2x^2 - 3x + 7\)
Here, \(a = 2, b = -3, c = 7\).
Sum of zeroes: \(\alpha + \beta = -\frac{-3}{2} = \frac{3}{2}\)
Product of zeroes: \(\alpha\beta = \frac{7}{2}\)

(a) Find \(\frac{1}{\alpha} + \frac{1}{\beta}\):
\(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{\frac{3}{2}}{\frac{7}{2}} = \frac{3}{2} \times \frac{2}{7} = \frac{3}{7}\)

(b) Find \(\alpha^2 + \beta^2\):
Using the identity \((\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta\):
\(\left(\frac{3}{2}\right)^2 = \alpha^2 + \beta^2 + 2\left(\frac{7}{2}\right)\)
\(\frac{9}{4} = \alpha^2 + \beta^2 + 7\)
\(\alpha^2 + \beta^2 = \frac{9}{4} - 7 = \frac{9 - 28}{4} = -\frac{19}{4}\)
In simple words: First find the sum and product of zeroes, then apply standard algebraic identities to find the required values.

Exam Tip: Clearly label parts (a) and (b) and write down the formulas used for each part.

Question 16. Find the value of \(a\) for which polynomial \(x^4 + 10x^3 + 25x^2 + 15x + a\) is exactly divisible by \(x + 7\).
Answer:
Let \(P(x) = x^4 + 10x^3 + 25x^2 + 15x + a\) and \(g(x) = x + 7\).
Since \(P(x)\) is exactly divisible by \(g(x)\), the remainder must be 0.
Using polynomial long division of \(P(x)\) by \(x + 7\):
1. Divide \(x^4\) by \(x\) to get \(x^3\). Multiply \(x^3(x + 7) = x^4 + 7x^3\). Subtract to get \(3x^3 + 25x^2\).
2. Divide \(3x^3\) by \(x\) to get \(3x^2\). Multiply \(3x^2(x + 7) = 3x^3 + 21x^2\). Subtract to get \(4x^2 + 15x\).
3. Divide \(4x^2\) by \(x\) to get \(4x\). Multiply \(4x(x + 7) = 4x^2 + 28x\). Subtract to get \(-13x + a\).
4. Divide \(-13x\) by \(x\) to get \(-13\). Multiply \(-13(x + 7) = -13x - 91\). Subtract to get the remainder \(a + 91\).
Since the polynomial is exactly divisible, remainder = 0:
\(a + 91 = 0 \implies a = -91\)
In simple words: Divide the polynomial by \(x + 7\), find the remainder expression, and equate it to zero to solve for \(a\).

Exam Tip: Be extremely careful with signs during subtraction steps in polynomial long division.