RBSE Solutions Class 10 Maths Chapter 3 Polynomials Exercise 3.2

Polynomials Ex 3.2

Question 1. Using division algorithm, find quotient and remainder dividing f(x) by g(x).
(i) \( f(x) = 3x^3 + x^2 + 2x + 5 \), \( g(x) = 1 + 2x + x^2 \)
(ii) \( f(x) = x^3 - 3x^2 + 5x + 3 \), \( g(x) = x^2 - 2 \)
(iii) \( f(x) = x^3 - 6x^2 + 11x - 6 \), \( g(x) = x + 2 \)
(iv) \( f(x) = 9x^4 - 4x^2 + 4 \), \( g(x) = 3x^2 + x - 1 \)
Answer:
(i) Given: \( f(x) = 3x^3 + x^2 + 2x + 5 \)
and \( g(x) = 1 + 2x + x^2 \). This can be written as \( g(x) = x^2 + 2x + 1 \).
Now, we divide \( f(x) \) by \( g(x) \):
\[ \require{enclose} \begin{array}{r} 3x-5 \\ x^2+2x+1 \enclose{longdiv}{3x^3 + x^2 + 2x + 5} \\ \underline{-(3x^3 + 6x^2 + 3x)} \\ -5x^2 - x + 5 \\ \underline{-(-5x^2 - 10x - 5)} \\ 9x + 10 \\ \end{array} \] The quotient \( q(x) = 3x - 5 \)
The remainder \( r(x) = 9x + 10 \)
We can verify this using the division algorithm: \( \text{Dividend} = \text{Quotient} \times \text{Divisor} + \text{Remainder} \).
So, \( (3x - 5)(1 + 2x + x^2) + (9x + 10) \)
\( = 3x + 6x^2 + 3x^3 - 5 - 10x - 5x^2 + 9x + 10 \)
\( = 3x^3 + x^2 - 7x - 5 + 9x + 10 \)
\( = 3x^3 + x^2 + 2x + 5 \)
\( = f(x) \)
The division algorithm is verified.

(ii) Given: \( f(x) = x^3 - 3x^2 + 5x + 3 \)
and \( g(x) = x^2 - 2 \)
Now, we divide \( f(x) \) by \( g(x) \):
\[ \require{enclose} \begin{array}{r} x-3 \\ x^2-2 \enclose{longdiv}{x^3 - 3x^2 + 5x + 3} \\ \underline{-(x^3 - 2x)} \\ -3x^2 + 7x + 3 \\ \underline{-(-3x^2 + 6)} \\ 7x - 3 \\ \end{array} \] The quotient \( q(x) = x - 3 \)
The remainder \( r(x) = 7x - 9 \)
We verify using the Euclid division algorithm: \( f(x) = g(x) \cdot q(x) + r(x) \)
\( = (x^2 - 2)(x - 3) + (7x - 9) \)
\( = x^3 - 3x^2 - 2x + 6 + 7x - 9 \)
\( = x^3 - 3x^2 + 5x - 3 \)
\( = f(x) \)
The division algorithm is verified.

(iii) Given: \( f(x) = x^3 - 6x^2 + 11x - 6 \)
and \( g(x) = x + 2 \)
Now, we divide \( f(x) \) by \( g(x) \):
\[ \require{enclose} \begin{array}{r} x^2 - 8x + 27 \\ x+2 \enclose{longdiv}{x^3 - 6x^2 + 11x - 6} \\ \underline{-(x^3 + 2x^2)} \\ -8x^2 + 11x - 6 \\ \underline{-(-8x^2 - 16x)} \\ 27x - 6 \\ \underline{-(27x + 54)} \\ -60 \\ \end{array} \] The quotient \( q(x) = x^2 - 8x + 27 \)
The remainder \( r(x) = -60 \)
We verify using the Euclid division algorithm: \( f(x) = g(x) \cdot q(x) + r(x) \)
\( = (x + 2)(x^2 - 8x + 27) + (-60) \)
\( = x(x^2 - 8x + 27) + 2(x^2 - 8x + 27) - 60 \)
\( = x^3 - 8x^2 + 27x + 2x^2 - 16x + 54 - 60 \)
\( = x^3 - 6x^2 + 11x - 6 \)
\( = f(x) \)
The division algorithm is verified.

(iv) Given: \( f(x) = 9x^4 - 4x^2 + 4 \)
and \( g(x) = 3x^2 + x - 1 \)
Now, we divide \( f(x) \) by \( g(x) \):
\[ \require{enclose} \begin{array}{r} 3x^2 - x \\ 3x^2+x-1 \enclose{longdiv}{9x^4 + 0x^3 - 4x^2 + 0x + 4} \\ \underline{-(9x^4 + 3x^3 - 3x^2)} \\ -3x^3 - x^2 + 0x + 4 \\ \underline{-(-3x^3 - x^2 + x)} \\ -x + 4 \\ \end{array} \] The quotient \( q(x) = 3x^2 - x \)
The remainder \( r(x) = -x + 4 \)
We verify using the Euclid division algorithm: \( f(x) = g(x) \cdot q(x) + r(x) \)
\( = (3x^2 + x - 1)(3x^2 - x) + (-x + 4) \)
\( = 3x^2(3x^2 - x) + x(3x^2 - x) - 1(3x^2 - x) - x + 4 \)
\( = 9x^4 - 3x^3 + 3x^3 - x^2 - 3x^2 + x - x + 4 \)
\( = 9x^4 - 4x^2 + 4 \)
\( = f(x) \)
The division algorithm is verified.
In simple words: For each part, we perform polynomial long division of \( f(x) \) by \( g(x) \) to find the quotient and remainder. Then, we use the formula: Dividend = Divisor × Quotient + Remainder to check if our answer is correct. This helps ensure the division was done right.

🎯 Exam Tip: Always remember to write the polynomials in descending order of powers and include terms with zero coefficients (like \( 0x^3 \)) if a power is missing, to make long division easier and prevent errors.

Question 2. Dividing second polynomial by first polynomial and test whether first polynomial is a factor of second polynomial.
(i) First polynomial \( g(x) = x^2 + 3x + 1 \), second polynomial \( f(x) = 3x^4 + 5x^3 – 7x^2 + 2x + 2 \)
(ii) First polynomial \( g(t) = t^2 - 3 \), second polynomial \( f(t) = 2t^4 + 3t^3 – 2t^2 – 9t - 12 \)
(iii) First polynomial \( g(x) = x^3 – 3x + 1 \), second polynomial \( f(x) = x^5 – 4x^3 + x^2 + 3x + 1 \)
Answer:
(i) Given: First polynomial \( g(x) = x^2 + 3x + 1 \)
Second polynomial \( f(x) = 3x^4 + 5x^3 – 7x^2 + 2x + 2 \)
For \( g(x) \) to be a factor of \( f(x) \), the remainder after division must be zero.
Dividing \( f(x) \) by \( g(x) \):
\[ \require{enclose} \begin{array}{r} 3x^2 - 4x + 2 \\ x^2+3x+1 \enclose{longdiv}{3x^4 + 5x^3 - 7x^2 + 2x + 2} \\ \underline{-(3x^4 + 9x^3 + 3x^2)} \\ -4x^3 - 10x^2 + 2x + 2 \\ \underline{-(-4x^3 - 12x^2 - 4x)} \\ 2x^2 + 6x + 2 \\ \underline{-(2x^2 + 6x + 2)} \\ 0 \\ \end{array} \] The remainder is zero.
By the division algorithm theorem, \( 3x^4 + 5x^3 – 7x^2 + 2x + 2 = (x^2 + 3x + 1)(3x^2 – 4x + 2) + 0 \).
Thus, \( x^2 + 3x + 1 \) is a factor of \( 3x^4 + 5x^3 – 7x^2 + 2x + 2 \). This means the first polynomial divides the second one completely.

(ii) Given: First polynomial \( g(t) = t^2 - 3 \)
Second polynomial \( f(t) = 2t^4 + 3t^3 - 2t^2 – 9t - 12 \)
For \( g(t) \) to be a factor of \( f(t) \), the remainder after division must be zero.
Dividing \( f(t) \) by \( g(t) \):
\[ \require{enclose} \begin{array}{r} 2t^2 + 3t + 4 \\ t^2-3 \enclose{longdiv}{2t^4 + 3t^3 - 2t^2 - 9t - 12} \\ \underline{-(2t^4 - 6t^2)} \\ 3t^3 + 4t^2 - 9t - 12 \\ \underline{-(3t^3 - 9t)} \\ 4t^2 - 12 \\ \underline{-(4t^2 - 12)} \\ 0 \\ \end{array} \] The remainder is zero.
By the division algorithm theorem, \( 2t^4 + 3t^3 - 2t^2 - 9t – 12 = (t^2 – 3)(2t^2 + 3t + 4) + 0 \).
Therefore, \( t^2 - 3 \) is a factor of \( 2t^4 + 3t^3 – 2t^2 - 9t - 12 \). This shows that the first polynomial completely divides the second polynomial.

(iii) Given: First polynomial \( g(x) = x^3 – 3x + 1 \)
Second polynomial \( f(x) = x^5 – 4x^3 + x^2 + 3x + 1 \)
For \( g(x) \) to be a factor of \( f(x) \), the remainder after division must be zero.
Dividing \( f(x) \) by \( g(x) \):
\[ \require{enclose} \begin{array}{r} x^2 - 1 \\ x^3-3x+1 \enclose{longdiv}{x^5 + 0x^4 - 4x^3 + x^2 + 3x + 1} \\ \underline{-(x^5 - 3x^3 + x^2)} \\ -x^3 + 3x + 1 \\ \underline{-(-x^3 + 3x - 1)} \\ 2 \\ \end{array} \] The remainder \( r(x) = 2 \).
Since the remainder is not zero, \( g(x) \) is not a factor of \( f(x) \). This means the first polynomial does not divide the second one evenly.
In simple words: To check if the first polynomial is a factor of the second, we divide the second polynomial by the first. If the remainder is zero, it's a factor; if not, it's not a factor. This is like checking if a number divides another number perfectly.

🎯 Exam Tip: When performing polynomial division, ensure all terms are present (even with zero coefficients) and always double-check your subtraction steps, as sign errors are common.

Question 3. Following are the polynomials with their zeros, find all the other zeros.
(i) \( f(x) = 2x^4 - 3x^3 – 3x^2 + 6x – 2 \); given zeros are \( \sqrt{2} \) and \( -\sqrt{2} \)
(ii) \( f(x) = x^4 – 6x^3 – 26x^2 + 138x – 35 \); given zeros are \( 2 \pm \sqrt{3} \)
(iii) \( f(x) = x^3 + 13x^2 + 32x + 20 \); given zero is \( -2 \)
Answer:
(i) Given: \( f(x) = 2x^4 - 3x^3 – 3x^2 + 6x – 2 \)
The two given zeros are \( \sqrt{2} \) and \( -\sqrt{2} \).
If \( x = \sqrt{2} \) and \( x = -\sqrt{2} \) are zeros, then \( (x - \sqrt{2}) \) and \( (x + \sqrt{2}) \) are factors.
Multiplying these factors: \( (x - \sqrt{2})(x + \sqrt{2}) = x^2 - (\sqrt{2})^2 = x^2 - 2 \).
So, \( (x^2 - 2) \) is a factor of \( f(x) \).
Now, we divide \( f(x) \) by \( (x^2 - 2) \):
\[ \require{enclose} \begin{array}{r} 2x^2 - 3x + 1 \\ x^2-2 \enclose{longdiv}{2x^4 - 3x^3 - 3x^2 + 6x - 2} \\ \underline{-(2x^4 - 4x^2)} \\ -3x^3 + x^2 + 6x - 2 \\ \underline{-(-3x^3 + 6x)} \\ x^2 - 2 \\ \underline{-(x^2 - 2)} \\ 0 \\ \end{array} \] So, \( f(x) = (x^2 - 2)(2x^2 - 3x + 1) \).
To find the other zeros, we set the quadratic factor to zero: \( 2x^2 - 3x + 1 = 0 \)
We can factor this: \( 2x^2 - 2x - x + 1 = 0 \)
\( 2x(x - 1) - 1(x - 1) = 0 \)
\( (2x - 1)(x - 1) = 0 \)
This gives us two more zeros:
\( x - 1 = 0 \implies x = 1 \)
\( 2x - 1 = 0 \implies x = \frac{1}{2} \)
Thus, all the zeros of the polynomial \( f(x) \) are \( \sqrt{2}, -\sqrt{2}, 1 \), and \( \frac{1}{2} \). This means we found all the points where the polynomial crosses the x-axis.

(ii) Given: \( f(x) = x^4 – 6x^3 – 26x^2 + 138x – 35 \)
The two given zeros are \( 2 + \sqrt{3} \) and \( 2 - \sqrt{3} \).
If \( x = 2 + \sqrt{3} \) and \( x = 2 - \sqrt{3} \) are zeros, then \( [x - (2 + \sqrt{3})] \) and \( [x - (2 - \sqrt{3})] \) are factors.
Multiplying these factors: \( [x - 2 - \sqrt{3}][x - 2 + \sqrt{3}] \)
\( = [(x - 2) - \sqrt{3}][(x - 2) + \sqrt{3}] \)
\( = (x - 2)^2 - (\sqrt{3})^2 \)
\( = (x^2 - 4x + 4) - 3 \)
\( = x^2 - 4x + 1 \)
So, \( (x^2 - 4x + 1) \) is a factor of \( f(x) \).
Now, we divide \( f(x) \) by \( (x^2 - 4x + 1) \):
\[ \require{enclose} \begin{array}{r} x^2 - 2x - 35 \\ x^2-4x+1 \enclose{longdiv}{x^4 - 6x^3 - 26x^2 + 138x - 35} \\ \underline{-(x^4 - 4x^3 + x^2)} \\ -2x^3 - 27x^2 + 138x - 35 \\ \underline{-(-2x^3 + 8x^2 - 2x)} \\ -35x^2 + 140x - 35 \\ \underline{-(-35x^2 + 140x - 35)} \\ 0 \\ \end{array} \] So, \( f(x) = (x^2 - 4x + 1)(x^2 - 2x - 35) \).
To find the other zeros, we set the quadratic factor to zero: \( x^2 - 2x - 35 = 0 \)
We can factor this: \( x^2 - 7x + 5x - 35 = 0 \)
\( x(x - 7) + 5(x - 7) = 0 \)
\( (x + 5)(x - 7) = 0 \)
This gives us two more zeros:
\( x + 5 = 0 \implies x = -5 \)
\( x - 7 = 0 \implies x = 7 \)
Thus, the other zeros of the polynomial \( f(x) \) are \( -5 \) and \( 7 \). These are the remaining values of x where the polynomial equals zero.

(iii) Given: \( f(x) = x^3 + 13x^2 + 32x + 20 \)
One given zero is \( -2 \).
If \( x = -2 \) is a zero, then \( (x - (-2)) = (x + 2) \) is a factor of \( f(x) \).
Now, we divide \( f(x) \) by \( (x + 2) \):
\[ \require{enclose} \begin{array}{r} x^2 + 11x + 10 \\ x+2 \enclose{longdiv}{x^3 + 13x^2 + 32x + 20} \\ \underline{-(x^3 + 2x^2)} \\ 11x^2 + 32x + 20 \\ \underline{-(11x^2 + 22x)} \\ 10x + 20 \\ \underline{-(10x + 20)} \\ 0 \\ \end{array} \] So, \( f(x) = (x + 2)(x^2 + 11x + 10) \).
To find the other zeros, we set the quadratic factor to zero: \( x^2 + 11x + 10 = 0 \)
We can factor this: \( x^2 + 10x + x + 10 = 0 \)
\( x(x + 10) + 1(x + 10) = 0 \)
\( (x + 1)(x + 10) = 0 \)
This gives us two more zeros:
\( x + 1 = 0 \implies x = -1 \)
\( x + 10 = 0 \implies x = -10 \)
Thus, the other zeros of the polynomial \( f(x) \) are \( -10 \) and \( -1 \). Finding these zeros helps us fully understand the behavior of the polynomial function.
In simple words: When some zeros of a polynomial are given, we use them to form a factor. Then we divide the polynomial by this factor. The result is a simpler polynomial. We then find the zeros of this simpler polynomial to get all the remaining zeros of the original polynomial.

🎯 Exam Tip: When given irrational or complex conjugate roots, always multiply them to form a quadratic factor, which will have rational coefficients, making the division much simpler.

Question 4. Dividing polynomial \( f(x) = x^3 - 3x^2 + x + 2 \) by polynomial \( g(x) \). Quotient \( q(x) \) and remainder \( r(x) \) are obtained as \( x - 2 \) and \( -2x + 4 \) respectively. Find polynomial \( g(x) \).
Answer:
Given:
Dividend \( f(x) = x^3 - 3x^2 + x + 2 \)
Quotient \( q(x) = x - 2 \)
Remainder \( r(x) = -2x + 4 \)
We need to find the divisor \( g(x) \).
We use the division algorithm formula: \( f(x) = g(x) \cdot q(x) + r(x) \)
Rearranging to find \( g(x) \): \( g(x) = \frac{f(x) - r(x)}{q(x)} \)
First, calculate \( f(x) - r(x) \):
\( f(x) - r(x) = (x^3 - 3x^2 + x + 2) - (-2x + 4) \)
\( = x^3 - 3x^2 + x + 2 + 2x - 4 \)
\( = x^3 - 3x^2 + 3x - 2 \)
Now, we divide this result by \( q(x) = x - 2 \):
\[ \require{enclose} \begin{array}{r} x^2 - x + 1 \\ x-2 \enclose{longdiv}{x^3 - 3x^2 + 3x - 2} \\ \underline{-(x^3 - 2x^2)} \\ -x^2 + 3x - 2 \\ \underline{-(-x^2 + 2x)} \\ x - 2 \\ \underline{-(x - 2)} \\ 0 \\ \end{array} \] So, \( g(x) = x^2 - x + 1 \). This polynomial is the divisor that was missing.
In simple words: We are given a polynomial, its quotient, and its remainder after division. To find the missing divisor, we subtract the remainder from the original polynomial, then divide that result by the given quotient. This helps us work backward to find the divisor.

🎯 Exam Tip: Always remember the division algorithm formula: Dividend = Divisor × Quotient + Remainder. This formula is key to solving problems where any one component of the division is unknown.