CBSE Class 10 Mathematics Polynomials Worksheet Set K

Points to Remember

• If \( \alpha \) and \( \beta \) are the zeros of the polynomial \( p(x) = ax^2 + bx + c \) and \( a \neq 0 \), then \( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \).
• A quadratic polynomial, whose zeros are \( \alpha \) and \( \beta \), is given by \( p(x) = x^2 – (\alpha + \beta)x + \alpha\beta \).
• If \( \alpha, \beta \) and \( \gamma \) are the zeros of polynomial \( p(x) = ax^3 + bx^2 + cx + d \) and \( a \neq 0 \), then \( \alpha + \beta + \gamma = -\frac{b}{a} \), \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} \) and \( \alpha\beta\gamma = -\frac{d}{a} \).
• A cubic polynomial whose zeros are \( \alpha, \beta \) and \( \gamma \) is given by \( p(x) = x^3 – (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x – \alpha\beta\gamma \).

Multiple Choice Questions

Question. The zeroes of the polynomial \( x^2 – 3x – m(m + 3) \) are 
(a) \( m, m + 3 \)
(b) \( -m, m + 3 \)
(c) \( m, -(m + 3) \)
(d) \( -m, -(m + 3) \)
Answer: (b)

Question. If \( x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times 2}}{2 \times 3} \) then the required polynomial is :
(a) \( 3x^2 – 8x + 2 = 0 \)
(b) \( 2x^2 – 8x – 2 = 0 \)
(c) \( 3x^2 + 8x – 2 = 0 \)
(d) \( 3x^2 + 8x + 2 = 0 \)
Answer: (a)

Question. Find the coefficient of \( x^0 \) in \( x^2 + 3x + 2 = 0 \).
(a) 3
(b) – 3
(c) 2
(d) – 2
Answer: (c)

Question. In which condition will the polynomial \( ax^2 + bx + c = 0 \) be a quadratic equation?
(a) \( a \neq 0 \)
(b) \( a = b \)
(c) \( a = c \)
(d) \( a = 0 \)
Answer: (a)

Question. The roots of the equation \( x^2 + x – p(p + 1) = 0 \) where \( p \) is a constant are :
(a) \( p, p + 1 \)
(b) \( -p, p + 1 \)
(c) \( p, -(p + 1) \)
(d) \( -p, -(p + 1) \)
Answer: (c)

Question. Polynomial \( x^4 – 2x^2 + x + 3 \) is which type of equation?
(a) Bi-quadratic
(b) Linear
(c) Quadratic
(d) Cubic
Answer: (a)

Fill in the Blanks

Question. Quadratic polynomial with sum and product of zeros as 3 and – 2 respectively is ......................
Answer: \( x^2 – 3x – 2 \)

Question. If \( p(x) = ax + b \), then the zero of \( p(x) \) will be ................
Answer: \( -\frac{b}{a} \)

Question. The graph of a quadratic polynomial is a .................
Answer: Parabola

Question. A polynomial of degree \( n \) has exactly ........... number of zeros.
Answer: \( n \)

Question. Number of zeros of a polynomial is equal to the number of points where the graph of polynomial intersects ............ axis.
Answer: X

True/False

Question. The degree of polynomial \( (x^2 + 4) (x^3 + x^4 – x – 1) \) is 6.
Answer: True

Question. A monomial contains one term.
Answer: True

Question. Any graph which touches X-axis at three distinct points has three zeros.
Answer: True

Very Short Answer Type Questions

Question. If one root of \( 5x^2 + 13x + k \) is reciprocal of the other root, then find the value of \( k \). 
Answer: Let \( \alpha \) and \( \beta \) be the roots of given polynomial \( 5x^2 + 13x + k \).
Then \( \alpha = \frac{1}{\beta} \) [Given]
or \( \alpha\beta = 1 \)
We know, Product of roots, \( \alpha\beta = \frac{\text{Constant term of quadratic polynomial}}{\text{Coefficient of } x^2 \text{ of quadratic polynomial}} \)
\( \Rightarrow 1 = \frac{k}{5} \)
\( \Rightarrow k = 5 \)

Question. Write a quadratic polynomial, the sum and product of whose zeros are – 3 and 2, respectively.  
Answer: We know a quadratic polynomial, with sum and product of its zeros is given as,
\( x^2 – (\text{Sum of zeros})x + \text{Product of zeros} \)
i.e., \( x^2 – (- 3)x + 2 \)
or \( x^2 + 3x + 2 \)

Question. If both the zeros of the quadratic polynomial \( ax^2 + bx + c \) are equal and opposite in sign, then find the value of \( b \).
Answer: Let \( \alpha \) and \( (-\alpha) \) be the roots of given polynomial \( ax^2 + bx + c \).
Then, Sum of zeros \( = \alpha + (-\alpha) = -\frac{b}{a} \)
\( \Rightarrow -\frac{b}{a} = 0 \Rightarrow b = 0 \)

Question. What number should be added to the polynomial \( x^2 + 7x – 35 \), so that 3 is the zero of the polynomial?
Answer: Let \( p(x) = x^2 + 7x – 35 \)
So, \( p(3) = (3)^2 + 7(3) – 35 \)
\( = 9 + 21 – 35 = -5 \)
Thus, if we add 5 to the polynomial, 3 will be one of the zeros.

Question. If \( x = -\frac{1}{2} \) is a solution of the quadratic polynomial \( 3x^2 + 2kx – 3 \), then find the value of \( k \).
Answer: Let \( p(x) = 3x^2 + 2kx – 3 \)
Since, \( x = -\frac{1}{2} \) is a solution of given polynomial
\( \therefore p(-\frac{1}{2}) = 0 \)
\( \Rightarrow 3(-\frac{1}{2})^2 + 2k(-\frac{1}{2}) - 3 = 0 \)
\( \Rightarrow \frac{3}{4} - k - 3 = 0 \)
\( \Rightarrow -k - \frac{9}{4} = 0 \Rightarrow k = -\frac{9}{4} \)

Question. Find the value of \( a \) if the product of zeros of the polynomial \( f(x) = ax^3 – 6x^2 + 11x – 6 \) is 4.
Answer: Given : \( f(x) = ax^3 – 6x^2 + 11x – 6 \) and Product of zeros = 4
We know, Product of zeros \( = -\frac{\text{Constant term}}{\text{Coefficient of } x^3} \)
\( \Rightarrow 4 = -\frac{(-6)}{a} \)
\( \Rightarrow a = \frac{6}{4} = \frac{3}{2} \)

Question. If \( \alpha, \beta, \gamma \) are the zeros of the polynomial \( f(x) = ax^3 + bx^2 + cx + d \), then find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \).
Answer: Given : \( f(x) = ax^3 + bx^2 + cx + d \)
Now, \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma} \)
\( = \frac{\frac{c}{a}}{-\frac{d}{a}} = -\frac{c}{d} \)

Question. If the sum of roots of the quadratic polynomial \( 3x^2 – (3k – 2)x – (k – 6) \) is equal to the product of its roots, then find the value of \( k \).
Answer: Let \( p(x) = 3x^2 – (3k – 2)x – (k – 6) \)
It is given that, Sum of roots = Product of roots
\( \Rightarrow -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \)
\( \Rightarrow \frac{-( -(3k - 2) )}{3} = \frac{-(k - 6)}{3} \)
\( \Rightarrow 3k – 2 = – k + 6 \)
\( \Rightarrow 4k = 8 \Rightarrow k = 2 \)

Question. If \( x + k \) is the G.C.D. of \( x^2 – 2x – 15 \) and \( x^3 + 27 \), then find the value of \( k \).
Answer: We have \( x^3 + 27 = x^3 + 3^3 = (x + 3)(x^2 – 3x + 9) \)
\( x^2 – 2x – 15 = x^2 – 5x + 3x – 15 = (x + 3)(x – 5) \)
\( \therefore (x + 3) \) is the G.C.D. of both the equations.
So \( x + k = x + 3 \) or \( k = 3 \).

Question. Sum of zeros of the polynomial \( 2x^2 – 4x + 5 \) is 4. Navdeep at once said ‘‘it is false’’. Do you agree with Navdeep? Justify.  
Answer: Yes. Here \( a = 2, b = - 4, c = 5 \)
Sum of zeros \( = -\frac{b}{a} = \frac{-(-4)}{2} = \frac{4}{2} = 2 \)
\( \therefore \) Sum of zeros is 2, not 4. Navdeep is right.

Short Answer Type Questions-I

Question. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper. The following were the answers given by the students : \( 2x + 3, 3x^2 + 7x + 2, 4x^3 + 3x^2 + 2, x + \sqrt{3x} + 7, 7x + \sqrt{7}, 5x^3 – 7x + 2, 2x^2 + 3 - \frac{5}{x}, 5x - \frac{1}{2}, ax^3 + bx^2 + cx + d, x + \frac{1}{x} \) Answer the following questions : (i) How many of the above ten, are not polynomials? (ii) How many of the above ten, are quadratic polynomials? 
Answer: (i) 3 are not polynomials: \( x + \sqrt{3x} + 7, 2x^2 + 3 - \frac{5}{x}, \) and \( x + \frac{1}{x} \). (ii) There is only one quadratic polynomial which is \( 3x^2 + 7x + 2 \).

Question. If \( x = \frac{2}{3} \) and \( x = – 3 \) are the roots of quadratic polynomial \( ax^2 + 7x + b \), find the values of \( a \) and \( b \).
Answer: Since \( \frac{2}{3} \) and \( – 3 \) are the roots of given polynomial.
\( \therefore \) Sum of roots \( = \frac{2}{3} + (-3) = -\frac{7}{a} \)
\( \Rightarrow -\frac{7}{3} = -\frac{7}{a} \Rightarrow a = 3 \)
Also, Product of roots \( = \frac{2}{3} \times (-3) = \frac{b}{a} \)
\( \Rightarrow -2 = \frac{b}{3} \Rightarrow b = -6 \)
Thus, \( a = 3 \) and \( b = -6 \).

Question. Find the values of \( p \) for which the quadratic polynomial \( 4x^2 + px + 3 = 0 \) has equal zeros.
Answer: Let \( \alpha, \beta \) be the zeros of the given polynomial. \( \therefore \alpha + \beta = -\frac{p}{4} \) and \( \alpha\beta = \frac{3}{4} \).
Also as per the question, \( \alpha = \beta \). Substituting \( \alpha \) for \( \beta \), we get:
\( 2\alpha = -\frac{p}{4} \Rightarrow \alpha = -\frac{p}{8} \) and \( \alpha^2 = \frac{3}{4} \).
Now \( (-\frac{p}{8})^2 = \frac{3}{4} \Rightarrow \frac{p^2}{64} = \frac{3}{4} \Rightarrow p^2 = 48 \Rightarrow p = \pm 4\sqrt{3} \).

Question. Solve the quadratic polynomial \( 2x^2 + ax – a^2 \) for \( x \).
Answer: Let \( f(x) = 2x^2 + ax – a^2 = 0 \)
\( \Rightarrow 2x^2 + 2ax – ax – a^2 = 0 \)
\( \Rightarrow 2x(x + a) – a(x + a) = 0 \)
\( \Rightarrow (x + a)(2x – a) = 0 \)
\( \therefore x = -a \) and \( \frac{a}{2} \).
Thus, the values of \( x \) for the polynomial \( 2x^2 + ax – a^2 \) will be \( -a \) and \( \frac{a}{2} \).

Question. If one zero of a polynomial \( (a^2 + 9)x^2 + 13x + 6a \) is the reciprocal of the other, then find the value of \( a \).
Answer: Let the zeros of the polynomial be \( \alpha \) and \( \beta \).
According to the question, \( \alpha = \frac{1}{\beta} \) or \( \alpha\beta = 1 \).
Now \( \alpha\beta = \frac{6a}{a^2 + 9} \)
\( \therefore \frac{6a}{a^2 + 9} = 1 \Rightarrow a^2 + 9 = 6a \Rightarrow a^2 – 6a + 9 = 0 \)
\( \Rightarrow (a – 3)^2 = 0 \Rightarrow a = 3 \).

Question. Find the roots of the polynomial : \( 2\sqrt{3}x^2 – 5x + \sqrt{3} = 0 \).
Answer: We have, \( 2\sqrt{3}x^2 – 5x + \sqrt{3} = 0 \)
\( \Rightarrow 2\sqrt{3}x^2 – 2x – 3x + \sqrt{3} = 0 \)
\( \Rightarrow 2x(\sqrt{3}x – 1) – \sqrt{3}(\sqrt{3}x – 1) = 0 \)
\( \Rightarrow (\sqrt{3}x – 1)(2x – \sqrt{3}) = 0 \)
If \( (\sqrt{3}x – 1) = 0 \Rightarrow x = \frac{1}{\sqrt{3}} \)
If \( 2x – \sqrt{3} = 0 \Rightarrow x = \frac{\sqrt{3}}{2} \)
\( \therefore \) Roots of given polynomial are \( \frac{1}{\sqrt{3}} \) and \( \frac{\sqrt{3}}{2} \).

Question. If two zeros of a polynomial \( x^3 – 4x^2 – 3x + 12 \) are \( \sqrt{3} \) and \( -\sqrt{3} \), then find its third zero. 
Answer: Since \( x = \sqrt{3} \) and \( x = -\sqrt{3} \) are zeros, \( (x - \sqrt{3})(x + \sqrt{3}) = x^2 – 3 \) is a factor.
Now dividing \( x^3 – 4x^2 – 3x + 12 \) by \( x^2 – 3 \):
\( x^3 – 4x^2 – 3x + 12 = (x^2 – 3)(x – 4) \)
Thus, third zero \( = x – 4 = 0 \Rightarrow x = 4 \).

Question. If three zeros of a polynomial \( x^4 – x^3 – 3x^2 + 3x \) are \( 0, \sqrt{3} \) and \( -\sqrt{3} \), then find the fourth zero.
Answer: Let \( P(x) = x^4 – x^3 – 3x^2 + 3x \). Given zeros are \( 0, \sqrt{3} \) and \( -\sqrt{3} \).
So \( x(x – \sqrt{3})(x + \sqrt{3}) = x(x^2 – 3) = x^3 – 3x \) will be the factor of \( P(x) \).
Dividing \( x^4 – x^3 – 3x^2 + 3x \) by \( x^3 – 3x \):
The quotient is \( (x - 1) \).
So, fourth zero \( = (x – 1) = 0 \Rightarrow x = 1 \).

Question. For what value of \( k \) are the roots of the quadratic polynomial \( kx(x – 2) + 6 = 0 \) are equal ? 
Answer: Given, \( kx(x – 2) + 6 = 0 \Rightarrow kx^2 – 2kx + 6 = 0 \)
Let \( \alpha \) and \( \beta \) be the roots. If roots are equal, \( \alpha = \beta \).
Sum of roots \( \alpha + \beta = 2\alpha = \frac{2k}{k} = 2 \Rightarrow \alpha = 1 \).
Product of roots \( \alpha\beta = \alpha^2 = \frac{6}{k} \Rightarrow 1^2 = \frac{6}{k} \Rightarrow k = 6 \).

Question. If the polynomial \( f(x) = (6x^4 + 8x^3 + 17x^2 + 21x + 7) \) is divided by another polynomial \( g(x) = 3x^2 + 4x + 1 \), the remainder is \( (ax + b) \). Find \( a \) and \( b \).
Answer: Dividing \( f(x) \) by \( g(x) \), we get the quotient \( 2x^2 + 5 \) and remainder \( x + 2 \).
Comparing \( x + 2 \) with \( ax + b \), we get \( a = 1, b = 2 \).

Question. What must be subtracted or added to \( p(x) = 8x^4 + 14x^3 – 2x^2 + 8x – 12 \) so that \( 4x^2 + 3x – 2 \) is a factor of \( p(x) \)?
Answer: Dividing \( p(x) \) by \( 4x^2 + 3x – 2 \), we get a remainder of \( 15x – 14 \).
Therefore, on subtracting \( 15x – 14 \) from \( p(x) \), \( 4x^2 + 3x – 2 \) is a factor of \( p(x) \).

Question. What real number should be subtracted from the polynomial \( (3x^3 + 10x^2 – 14x + 9) \) so that \( (3x – 2) \) divides it exactly?
Answer: Dividing \( 3x^3 + 10x^2 – 14x + 9 \) by \( 3x – 2 \), we get a remainder of 5.
Thus, 5 should be subtracted from the polynomial so that it is divisible by \( 3x – 2 \).

Question. Using division algorithm, find the quotient and remainder on dividing \( f(x) \) by \( g(x) \) where \( f(x) = 6x^3 + 13x^2 + x – 2 \) and \( g(x) = 2x + 1 \).
Answer: Dividing \( 6x^3 + 13x^2 + x – 2 \) by \( 2x + 1 \), we get:
Quotient \( = 3x^2 + 5x – 2 \)
Remainder \( = 0 \).

Short Answer Type Questions-II

Question. Find the zeros of the following polynomial: \( 5\sqrt{5}x^2 + 30x + 8\sqrt{5} \)  
Answer: \( f(x) = 5\sqrt{5}x^2 + 30x + 8\sqrt{5} = 5\sqrt{5}x^2 + 20x + 10x + 8\sqrt{5} \)
\( = 5x(\sqrt{5}x + 4) + 2\sqrt{5}(\sqrt{5}x + 4) = (5x + 2\sqrt{5})(\sqrt{5}x + 4) \).
Zeros are \( -\frac{2\sqrt{5}}{5} = -\frac{2}{\sqrt{5}} \) and \( -\frac{4}{\sqrt{5}} \).

Question. Find the value of \( k \) so that the polynomial equation \( x^2 – 4kx + k = 0 \) has equal roots.
Answer: For equal roots, Sum of zeros \( \alpha + \beta = 2\alpha = 4k \Rightarrow \alpha = 2k \).
Product of zeros \( \alpha\beta = \alpha^2 = k \).
Substituting \( \alpha \): \( (2k)^2 = k \Rightarrow 4k^2 = k \Rightarrow 4k^2 - k = 0 \Rightarrow k(4k - 1) = 0 \).
As \( k \neq 0 \) (for it to be quadratic), \( k = \frac{1}{4} \).

Question. Obtain all other zeros of the polynomial \( x^4 + 6x^3 + x^2 – 24x – 20 \), if two of its zeros are + 2 and – 5.
Answer: Since \( 2 \) and \( -5 \) are zeros, \( (x - 2)(x + 5) = x^2 + 3x - 10 \) is a factor.
Dividing \( x^4 + 6x^3 + x^2 – 24x – 20 \) by \( x^2 + 3x - 10 \), we get quotient \( x^2 + 3x + 2 \).
Factoring \( x^2 + 3x + 2 = (x + 2)(x + 1) \).
So, all zeros are \( 2, -5, -2 \) and \( -1 \).

Question. When divided by \( (x – 3) \), the polynomials \( (x^3 – px^2 + x + 6) \) and \( [2x^3 – x^2 – (p + 3)x – 6] \) leave the same remainder. Find the value of \( p \).
Answer: Using Remainder Theorem, \( f(3) = g(3) \).
\( 3^3 – p(3^2) + 3 + 6 = 2(3^3) – (3^2) – (p + 3)(3) – 6 \)
\( 36 – 9p = 54 – 9 – 3p – 9 – 6 \)
\( 36 – 9p = 30 – 3p \Rightarrow 6p = 6 \Rightarrow p = 1 \).

Question. Find the values of \( a \) and \( b \) so that \( x^4 + x^3 + 8x^2 + ax – b \) is divisible by \( x^2 + 1 \).
Answer: Dividing the polynomial by \( x^2 + 1 \), we get a remainder of \( (a - 1)x - b - 7 \).
For exact divisibility, remainder must be zero.
\( a - 1 = 0 \Rightarrow a = 1 \) and \( -b - 7 = 0 \Rightarrow b = -7 \).

Case Study

The distance covered by the bike is given by the expression \( (2x^2 + 6x – 20) \). The time taken by the bike to cover this distance is given by the expression \( (x – 2) \).

Question. Calculate the zeroes of expression of distance covered by the bike.
(a) 4, 2
(b) 2, – 5
(c) 2, – 4
(d) 3, 2
Answer: (b)

Question. Calculate the speed of the bike in terms of \( x \).
(a) \( x + 5 \)
(b) \( 2x + 10 \)
(c) \( x – 10 \)
(d) \( x^2 + 5 \)
Answer: (b)

Archives

Question. If three zeroes of a polynomial \( x^4 – x^3 – 3x^2 + 3x \) are \( 0, \sqrt{3} \) and \( -\sqrt{3} \), then find the fourth zero.
Answer: 1

Question. Using division algorithm, find the quotient and remainder on dividing \( f(x) \) by \( g(x) \) where \( f(x) = 6x^3 + 13x^2 + x – 2 \) and \( g(x) = 2x + 1 \).
Answer: Quotient \( = 3x^2 + 5x – 2 \), Remainder \( = 0 \)

Question. Find a quadratic polynomial, the sum and product of whose zeroes are \( 0 \) and \( -\frac{3}{5} \) respectively. Hence find the zeroes.
Answer: \( x^2 - \frac{3}{5} \); Zeroes are \( \pm \frac{\sqrt{15}}{5} \)

Question. Obtain all other zeroes of the polynomial \( x^4 + 6x^3 + x^2 – 24x – 20 \), if two of its zeroes are \( + 2 \) and \( – 5 \).
Answer: – 2, – 1

Question. What must be subtracted from \( p(x) = 8x^4 + 14x^3 – 2x^2 + 8x – 12 \) so that \( 4x^2 + 3x – 2 \) is factor of \( p(x) \)?
Answer: \( 15x – 14 \)

Assertion and Reasoning Based Questions

Question. Statement 1 : \( \frac{5}{4} \) is the zero of the polynomial \( (4x – 5) \).
Statement 2 : Every real number is the zero of the zero polynomial.
Answer: (B)

Question. Statement 1 : A degree of the non-zero constant polynomial is zero.
Statement 2 : The degree of a zero polynomial is zero.
Answer: (C)

Question. Statement 1 : If \( f(x) = px + q, p \neq 0 \), is a linear polynomial, then \( x = -\frac{q}{p} \) is the only zero of \( f(x) \).
Statement 2 : A linear polynomial has one and only one zero.
Answer: (A)

Question. Statement 1 : The expression \( x^4 - 4x^{3/2} + x^2 = 2 \) is not a polynomial.
Statement 2 : The highest exponent in various terms of an algebraic expression in one variable is called its degree.
Answer: (B)