Table of Contents
Exercise 3.4 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on polynomial division. Mathematical expressions are rendered using MathJax.
1. Divide the polynomial \( p(x) \) by the polynomial \( g(x) \) and find the quotient and remainder in each of the following:
(i) \( p(x) = x^3 – 3x^2 + 5x – 3 \), \( g(x) = x^2 – 2 \)
Use polynomial long division to divide \( x^3 – 3x^2 + 5x – 3 \) by \( x^2 – 2 \).
Step 1: Divide the leading term of \( p(x) \), \( x^3 \), by the leading term of \( g(x) \), \( x^2 \): \( \frac{x^3}{x^2} = x \).
Step 2: Multiply \( x \) by \( g(x) \): \( x (x^2 – 2) = x^3 – 2x \).
Step 3: Subtract: \( (x^3 – 3x^2 + 5x – 3) – (x^3 – 2x) = -3x^2 + 7x – 3 \).
Step 4: Divide the leading term of the new polynomial, \( -3x^2 \), by \( x^2 \): \( \frac{-3x^2}{x^2} = -3 \).
Step 5: Multiply \( -3 \) by \( g(x) \): \( -3 (x^2 – 2) = -3x^2 + 6 \).
Step 6: Subtract: \( (-3x^2 + 7x – 3) – (-3x^2 + 6) = 7x – 9 \).
The degree of the remainder \( 7x – 9 \) (degree 1) is less than the degree of \( g(x) \) (degree 2), so stop.
Quotient: \( x – 3 \), Remainder: \( 7x – 9 \)
(ii) \( p(x) = x^4 – 3x^2 + 4x + 5 \), \( g(x) = x^2 + 1 – x \)
Rewrite \( g(x) = x^2 – x + 1 \). Divide \( x^4 – 3x^2 + 4x + 5 \) by \( x^2 – x + 1 \).
Step 1: Divide \( x^4 \) by \( x^2 \): \( \frac{x^4}{x^2} = x^2 \).
Step 2: Multiply: \( x^2 (x^2 – x + 1) = x^4 – x^3 + x^2 \).
Step 3: Subtract: \( (x^4 – 3x^2 + 4x + 5) – (x^4 – x^3 + x^2) = x^3 – 4x^2 + 4x + 5 \).
Step 4: Divide \( x^3 \) by \( x^2 \): \( \frac{x^3}{x^2} = x \).
Step 5: Multiply: \( x (x^2 – x + 1) = x^3 – x^2 + x \).
Step 6: Subtract: \( (x^3 – 4x^2 + 4x + 5) – (x^3 – x^2 + x) = -3x^2 + 3x + 5 \).
Step 7: Divide \( -3x^2 \) by \( x^2 \): \( \frac{-3x^2}{x^2} = -3 \).
Step 8: Multiply: \( -3 (x^2 – x + 1) = -3x^2 + 3x – 3 \).
Step 9: Subtract: \( (-3x^2 + 3x + 5) – (-3x^2 + 3x – 3) = 8 \).
The remainder is 8 (degree 0), less than the degree of \( g(x) \).
Quotient: \( x^2 + x – 3 \), Remainder: 8
(iii) \( p(x) = x^4 – 5x + 6 \), \( g(x) = 2 – x^2 \)
Rewrite \( g(x) = -x^2 + 2 \). Divide \( x^4 – 5x + 6 \) by \( -x^2 + 2 \).
Step 1: Divide \( x^4 \) by \( -x^2 \): \( \frac{x^4}{-x^2} = -x^2 \).
Step 2: Multiply: \( -x^2 (-x^2 + 2) = -x^4 + 2x^2 \).
Step 3: Subtract: \( (x^4 – 5x + 6) – (-x^4 + 2x^2) = 2x^4 – 2x^2 – 5x + 6 \).
Step 4: Divide \( 2x^4 \) by \( -x^2 \): \( \frac{2x^4}{-x^2} = -2x^2 \). This step corrects the approach—restart with proper division.
Correct division: Divide \( x^4 – 5x + 6 \) by \( -x^2 + 2 \).
Step 1: \( \frac{x^4}{-x^2} = -x^2 \).
Step 2: Multiply: \( -x^2 (-x^2 + 2) = x^4 – 2x^2 \).
Step 3: Subtract: \( (x^4 – 5x + 6) – (x^4 – 2x^2) = 2x^2 – 5x + 6 \).
The remainder \( 2x^2 – 5x + 6 \) has degree 2, equal to the degree of \( g(x) \), so continue.
Step 4: Divide \( 2x^2 \) by \( -x^2 \): \( \frac{2x^2}{-x^2} = -2 \).
Step 5: Multiply: \( -2 (-x^2 + 2) = 2x^2 – 4 \).
Step 6: Subtract: \( (2x^2 – 5x + 6) – (2x^2 – 4) = -5x + 10 \).
The remainder \( -5x + 10 \) has degree 1, less than the degree of \( g(x) \).
Quotient: \( -x^2 – 2 \), Remainder: \( -5x + 10 \)
2. Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) \( t^2 – 3 \), \( 2t^4 + 3t^3 – 2t^2 – 9t – 12 \)
Divide \( 2t^4 + 3t^3 – 2t^2 – 9t – 12 \) by \( t^2 – 3 \).
Step 1: Divide \( 2t^4 \) by \( t^2 \): \( \frac{2t^4}{t^2} = 2t^2 \).
Step 2: Multiply: \( 2t^2 (t^2 – 3) = 2t^4 – 6t^2 \).
Step 3: Subtract: \( (2t^4 + 3t^3 – 2t^2 – 9t – 12) – (2t^4 – 6t^2) = 3t^3 + 4t^2 – 9t – 12 \).
Step 4: Divide \( 3t^3 \) by \( t^2 \): \( \frac{3t^3}{t^2} = 3t \).
Step 5: Multiply: \( 3t (t^2 – 3) = 3t^3 – 9t \).
Step 6: Subtract: \( (3t^3 + 4t^2 – 9t – 12) – (3t^3 – 9t) = 4t^2 – 12 \).
Step 7: Divide \( 4t^2 \) by \( t^2 \): \( \frac{4t^2}{t^2} = 4 \).
Step 8: Multiply: \( 4 (t^2 – 3) = 4t^2 – 12 \).
Step 9: Subtract: \( (4t^2 – 12) – (4t^2 – 12) = 0 \).
The remainder is 0, so \( t^2 – 3 \) is a factor of \( 2t^4 + 3t^3 – 2t^2 – 9t – 12 \).
Conclusion: \( t^2 – 3 \) is a factor.
(ii) \( x^2 + 3x + 1 \), \( 3x^4 + 5x^3 – 7x^2 + 2x + 2 \)
Divide \( 3x^4 + 5x^3 – 7x^2 + 2x + 2 \) by \( x^2 + 3x + 1 \).
Step 1: Divide \( 3x^4 \) by \( x^2 \): \( \frac{3x^4}{x^2} = 3x^2 \).
Step 2: Multiply: \( 3x^2 (x^2 + 3x + 1) = 3x^4 + 9x^3 + 3x^2 \).
Step 3: Subtract: \( (3x^4 + 5x^3 – 7x^2 + 2x + 2) – (3x^4 + 9x^3 + 3x^2) = -4x^3 – 10x^2 + 2x + 2 \).
Step 4: Divide \( -4x^3 \) by \( x^2 \): \( \frac{-4x^3}{x^2} = -4x \).
Step 5: Multiply: \( -4x (x^2 + 3x + 1) = -4x^3 – 12x^2 – 4x \).
Step 6: Subtract: \( (-4x^3 – 10x^2 + 2x + 2) – (-4x^3 – 12x^2 – 4x) = 2x^2 + 6x + 2 \).
Step 7: Divide \( 2x^2 \) by \( x^2 \): \( \frac{2x^2}{x^2} = 2 \).
Step 8: Multiply: \( 2 (x^2 + 3x + 1) = 2x^2 + 6x + 2 \).
Step 9: Subtract: \( (2x^2 + 6x + 2) – (2x^2 + 6x + 2) = 0 \).
The remainder is 0, so \( x^2 + 3x + 1 \) is a factor of \( 3x^4 + 5x^3 – 7x^2 + 2x + 2 \).
Conclusion: \( x^2 + 3x + 1 \) is a factor.
(iii) \( x^2 – 3x + 1 \), \( x^5 – 4x^3 + x^2 + 3x + 1 \)
Divide \( x^5 – 4x^3 + x^2 + 3x + 1 \) by \( x^2 – 3x + 1 \).
Step 1: Divide \( x^5 \) by \( x^2 \): \( \frac{x^5}{x^2} = x^3 \).
Step 2: Multiply: \( x^3 (x^2 – 3x + 1) = x^5 – 3x^4 + x^3 \).
Step 3: Subtract: \( (x^5 – 4x^3 + x^2 + 3x + 1) – (x^5 – 3x^4 + x^3) = 3x^4 – 5x^3 + x^2 + 3x + 1 \).
Step 4: Divide \( 3x^4 \) by \( x^2 \): \( \frac{3x^4}{x^2} = 3x^2 \).
Step 5: Multiply: \( 3x^2 (x^2 – 3x + 1) = 3x^4 – 9x^3 + 3x^2 \).
Step 6: Subtract: \( (3x^4 – 5x^3 + x^2 + 3x + 1) – (3x^4 – 9x^3 + 3x^2) = 4x^3 – 2x^2 + 3x + 1 \).
Step 7: Divide \( 4x^3 \) by \( x^2 \): \( \frac{4x^3}{x^2} = 4x \).
Step 8: Multiply: \( 4x (x^2 – 3x + 1) = 4x^3 – 12x^2 + 4x \).
Step 9: Subtract: \( (4x^3 – 2x^2 + 3x + 1) – (4x^3 – 12x^2 + 4x) = 10x^2 – x + 1 \).
The remainder \( 10x^2 – x + 1 \) has degree 2, equal to the degree of \( x^2 – 3x + 1 \), so continue.
Step 10: Divide \( 10x^2 \) by \( x^2 \): \( \frac{10x^2}{x^2} = 10 \).
Step 11: Multiply: \( 10 (x^2 – 3x + 1) = 10x^2 – 30x + 10 \).
Step 12: Subtract: \( (10x^2 – x + 1) – (10x^2 – 30x + 10) = 29x – 9 \).
The remainder is not 0, so \( x^2 – 3x + 1 \) is not a factor.
Conclusion: \( x^2 – 3x + 1 \) is not a factor.
3. Obtain all other zeros of \( 3x^4 + 6x^3 – 2x^2 – 10x – 5 \), if two of its zeros are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \).
Given zeros \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \). Since the polynomial has real coefficients, complex or irrational zeros come in conjugate pairs.
The factor corresponding to these zeros is \( (x – \sqrt{\frac{5}{3}})(x + \sqrt{\frac{5}{3}}) = x^2 – \left(\sqrt{\frac{5}{3}}\right)^2 = x^2 – \frac{5}{3} \).
Multiply by 3 to clear the fraction: \( 3x^2 – 5 \).
Divide \( 3x^4 + 6x^3 – 2x^2 – 10x – 5 \) by \( 3x^2 – 5 \).
Step 1: Divide \( 3x^4 \) by \( 3x^2 \): \( \frac{3x^4}{3x^2} = x^2 \).
Step 2: Multiply: \( x^2 (3x^2 – 5) = 3x^4 – 5x^2 \).
Step 3: Subtract: \( (3x^4 + 6x^3 – 2x^2 – 10x – 5) – (3x^4 – 5x^2) = 6x^3 + 3x^2 – 10x – 5 \).
Step 4: Divide \( 6x^3 \) by \( 3x^2 \): \( \frac{6x^3}{3x^2} = 2x \).
Step 5: Multiply: \( 2x (3x^2 – 5) = 6x^3 – 10x \).
Step 6: Subtract: \( (6x^3 + 3x^2 – 10x – 5) – (6x^3 – 10x) = 3x^2 – 5 \).
Step 7: Divide \( 3x^2 \) by \( 3x^2 \): \( \frac{3x^2}{3x^2} = 1 \).
Step 8: Multiply: \( 1 (3x^2 – 5) = 3x^2 – 5 \).
Step 9: Subtract: \( (3x^2 – 5) – (3x^2 – 5) = 0 \).
The quotient is \( x^2 + 2x + 1 \), which factors as \( (x + 1)^2 \).
Solve for zeros: \( x + 1 = 0 \implies x = -1 \) (repeated zero).
The other zeros are \( -1, -1 \).
Other zeros: \( -1, -1 \)
4. On dividing \( x^3 – 3x^2 + x + 2 \) by a polynomial \( g(x) \), the quotient and remainder were \( x – 2 \) and \( -2x + 4 \), respectively. Find \( g(x) \).
By the division algorithm, \( p(x) = g(x) \cdot q(x) + r(x) \).
Here, \( p(x) = x^3 – 3x^2 + x + 2 \), \( q(x) = x – 2 \), \( r(x) = -2x + 4 \).
Rearrange: \( x^3 – 3x^2 + x + 2 = g(x) (x – 2) + (-2x + 4) \).
Isolate \( g(x) \): \( g(x) (x – 2) = (x^3 – 3x^2 + x + 2) – (-2x + 4) = x^3 – 3x^2 + 3x – 2 \).
Divide \( x^3 – 3x^2 + 3x – 2 \) by \( x – 2 \).
Step 1: Divide \( x^3 \) by \( x \): \( \frac{x^3}{x} = x^2 \).
Step 2: Multiply: \( x^2 (x – 2) = x^3 – 2x^2 \).
Step 3: Subtract: \( (x^3 – 3x^2 + 3x – 2) – (x^3 – 2x^2) = -x^2 + 3x – 2 \).
Step 4: Divide \( -x^2 \) by \( x \): \( \frac{-x^2}{x} = -x \).
Step 5: Multiply: \( -x (x – 2) = -x^2 + 2x \).
Step 6: Subtract: \( (-x^2 + 3x – 2) – (-x^2 + 2x) = x – 2 \).
Step 7: Divide \( x \) by \( x \): \( \frac{x}{x} = 1 \).
Step 8: Multiply: \( 1 (x – 2) = x – 2 \).
Step 9: Subtract: \( (x – 2) – (x – 2) = 0 \).
So, \( g(x) = x^2 – x + 1 \).
\( g(x) \): \( x^2 – x + 1 \)
5. Give examples of polynomials \( p(x) \), \( g(x) \), \( q(x) \), and \( r(x) \), which satisfy the division algorithm and
(i) \( \deg p(x) = \deg q(x) \)
Choose \( g(x) \) with degree 0 (a constant), so the quotient \( q(x) \) has the same degree as \( p(x) \).
Let \( p(x) = 2x + 3 \), \( g(x) = 2 \).
Divide: \( q(x) = \frac{2x + 3}{2} = x + \frac{3}{2} \), but adjust for integer coefficients.
Instead, let \( p(x) = 4x + 6 \), \( g(x) = 2 \).
Divide: \( q(x) = \frac{4x + 6}{2} = 2x + 3 \), remainder \( r(x) = 0 \).
Degree of \( p(x) = 1 \), degree of \( q(x) = 1 \), which matches.
Example: \( p(x) = 4x + 6 \), \( g(x) = 2 \), \( q(x) = 2x + 3 \), \( r(x) = 0 \)
(ii) \( \deg q(x) = \deg r(x) \)
Choose \( g(x) \) such that the remainder \( r(x) \) has the same degree as \( q(x) \).
Let \( p(x) = x^3 + 2x^2 + x + 1 \), \( g(x) = x^2 + 1 \).
Divide: Step 1: \( \frac{x^3}{x^2} = x \).
Step 2: \( x (x^2 + 1) = x^3 + x \).
Step 3: \( (x^3 + 2x^2 + x + 1) – (x^3 + x) = 2x^2 + 1 \).
Step 4: \( \frac{2x^2}{x^2} = 2 \).
Step 5: \( 2 (x^2 + 1) = 2x^2 + 2 \).
Step 6: \( (2x^2 + 1) – (2x^2 + 2) = -1 \).
Quotient \( q(x) = x + 2 \) (degree 1), remainder \( r(x) = -1 \) (degree 0).
Adjust: Let \( p(x) = x^2 + 2x + 1 \), \( g(x) = x + 1 \).
Divide: \( q(x) = x + 1 \), \( r(x) = 0 \). Need non-zero remainder.
Try \( p(x) = x^2 + 3x + 1 \), \( g(x) = x + 2 \).
Divide: \( \frac{x^2}{x} = x \), \( x (x + 2) = x^2 + 2x \), subtract: \( (x^2 + 3x + 1) – (x^2 + 2x) = x + 1 \).
Quotient \( q(x) = x \), remainder \( r(x) = x + 1 \), both degree 1.
Example: \( p(x) = x^2 + 3x + 1 \), \( g(x) = x + 2 \), \( q(x) = x \), \( r(x) = x + 1 \)
(iii) \( \deg r(x) = 0 \)
The remainder must be a constant (degree 0).
Let \( p(x) = x^2 + 2x + 3 \), \( g(x) = x + 1 \).
Divide: \( \frac{x^2}{x} = x \), \( x (x + 1) = x^2 + x \), subtract: \( (x^2 + 2x + 3) – (x^2 + x) = x + 3 \).
Divide: \( \frac{x}{x} = 1 \), \( 1 (x + 1) = x + 1 \), subtract: \( (x + 3) – (x + 1) = 2 \).
Quotient \( q(x) = x + 1 \), remainder \( r(x) = 2 \), which has degree 0.
Example: \( p(x) = x^2 + 2x + 3 \), \( g(x) = x + 1 \), \( q(x) = x + 1 \), \( r(x) = 2 \)
