JEE Mathematics Polynomials Notes Set 01

Polynomial or Rational Integral Function

A polynomial in \( x \) contains only non-negative integral powers of \( x \). For example \( 6x^2 - 5x + 4 \) is a polynomial. But \( 2x^2 + \frac{1}{x} + \sqrt{x} + 5 \), which equals \( 2x^2 + x^{-1} + x^{1/2} + 5 \), is not, because the powers \( -1 \) and \( \frac{1}{2} \) are not non-negative integers.
The ratio of two polynomials is a rational function, e.g. \( \dfrac{2x^3 - 3x^2 + 15}{5x^3 + 7x - 17} \).

Example 1. Find the polynomial \( f(x) = ax^2 + bx + c \) such that \( f(0) = 5 \), \( f(1) = 10 \), \( f(2) = 19 \).
Answer:
\( f(0) = c = 5 \).
\( f(1) = a + b + c = 10 \) gives \( a + b = 5 \).
\( f(2) = 4a + 2b + c = 19 \) gives \( 2a + b = 7 \).
Subtracting: \( a = 2 \), then \( b = 3 \).
The polynomial is \( 2x^2 + 3x + 5 \).

Example 2. A linear function \( f \) has \( f(10) = 65 \) and \( f(15) = 100 \). Find \( f(x) \).
Answer:
Let \( f(x) = ax + b \). Then \( 10a + b = 65 \) and \( 15a + b = 100 \).
Subtracting: \( 5a = 35 \), so \( a = 7 \), and \( b = 65 - 70 = -5 \).
\( f(x) = 7x - 5 \).

Verifying Identities by Substitution

An identity can be tested by putting in particular numbers for the variables. Equal values on both sides support the identity but do not prove it; a proof needs algebraic expansion.

Example 3. Verify \( (x+y)(x-y) = x^2 - y^2 \) for x = 3, y = 4.
Answer:
LHS \( = 7 \times (-1) = -7 \). RHS \( = 9 - 16 = -7 \). Verified.

Example 4. Verify \( (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) = x^3 + y^3 + z^3 - 3xyz \) for x = -1, y = 2, z = -2.
Answer:
\( x + y + z = -1 \). \( x^2 + y^2 + z^2 = 9 \), and \( -xy - yz - zx = 2 + 4 - 2 = 4 \), so the second factor is 13 and LHS \( = -13 \).
RHS \( = -1 + 8 - 8 - 3(-1)(2)(-2) = -1 - 12 = -13 \). Verified.

Addition and Subtraction

Only like terms can be added or subtracted. Arrange the polynomials with like terms in the same column, or group like terms with brackets, then combine the coefficients. To subtract, change the sign of every term of the polynomial being subtracted and then add.

Example 5. Add (i) \( 3x^2 - 4x + 6,\; -2x^2 + 3x + 4,\; x^2 - 2x - 9 \); (ii) \( p - q,\; q - r,\; r - p \); (iii) \( x(x^2 - yz),\; y(y^2 - zx),\; z(z^2 - xy) \).
Answer:
(i) \( x^2: 3 - 2 + 1 = 2 \); \( x: -4 + 3 - 2 = -3 \); constants: \( 6 + 4 - 9 = 1 \). Sum \( = \)\( 2x^2 - 3x + 1 \).
(ii) \( (p - q) + (q - r) + (r - p) = \)0.
(iii) \( x^3 - xyz + y^3 - xyz + z^3 - xyz = \)\( x^3 + y^3 + z^3 - 3xyz \).

Example 6. Subtract \( -2y^2 + y + 6 \) from \( 4y^2 + y + 5 \).
Answer:
\( (4y^2 + y + 5) - (-2y^2 + y + 6) = 4y^2 + y + 5 + 2y^2 - y - 6 = \)\( 6y^2 - 1 \).

Example 7. Subtract \( x^3 - 2x^2y - 6xy^2 + y^3 \) from \( x^3 + 7x^2y - 5xy^2 + 2y^3 \).
Answer:
Collect like terms: \( x^3: 1 - 1 = 0 \); \( x^2y: 7 + 2 = 9 \); \( xy^2: -5 + 6 = 1 \); \( y^3: 2 - 1 = 1 \).
Result: \( 9x^2y + xy^2 + y^3 \).

Multiplication and Division

Multiplication uses the distributive law: \( a(b + c) = ab + ac \). To multiply two bracketed expressions, multiply every term of one by every term of the other and collect like terms. To divide a polynomial by a monomial, divide each term separately.

Example 8. Multiply \( 3x^2 - 5x \) by \( 4x \).
Answer:
\( 4x \cdot 3x^2 - 4x \cdot 5x = \)\( 12x^3 - 20x^2 \).

Example 9. Divide \( 12y^3 + 8y^2 \) by \( 2y \).
Answer:
\( \dfrac{12y^3}{2y} + \dfrac{8y^2}{2y} = \)\( 6y^2 + 4y \).

Example 10. Divide \( x^2yz + y^2zx + z^2xy \) by \( -xyz \).
Answer:
\( \dfrac{x^2yz}{-xyz} + \dfrac{y^2zx}{-xyz} + \dfrac{z^2xy}{-xyz} = \)\( -x - y - z \).

Example 11. Without simplifying, find the coefficient of \( x^2 \) and the absolute term in \( 2(x^3 - 3x^2 + 4x - 5) - 3(2x^3 - x^2 - 2x + 6) + 5(x^4 - 1) \).
Answer:
Coefficient of \( x^2 \): \( 2(-3) - 3(-1) = -6 + 3 = \)-3.
Absolute term: \( 2(-5) - 3(6) + 5(-1) = -10 - 18 - 5 = \)-33.

Example 12. Find the coefficient of \( x^3 \) and the absolute term in \( 2x(x^3 - 3x^2 + 4x - 2) - x^2(x^2 + 3x - 1) + x(x^3 - 2) + x^4 - 1 \).
Answer:
Terms in \( x^3 \): \( 2x \cdot (-3x^2) = -6x^3 \) and \( -x^2 \cdot 3x = -3x^3 \). Coefficient \( = \)-9.
Absolute term: -1.

Example 13. Multiply \( (2x - 3)(4x - 5) \).
Answer:
\( 2x(4x - 5) - 3(4x - 5) = 8x^2 - 10x - 12x + 15 = \)\( 8x^2 - 22x + 15 \).

Standard Identities

\( (a+b)^2 = a^2 + 2ab + b^2 \)
\( (a-b)^2 = a^2 - 2ab + b^2 \)
\( (a+b)(a-b) = a^2 - b^2 \)
\( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)
\( (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \)
Each follows by expanding, e.g. \( (a+b)^2 = (a+b)(a+b) = a(a+b) + b(a+b) = a^2 + 2ab + b^2 \), and \( (a+b)^3 = (a+b)(a^2 + 2ab + b^2) \).

Example 14. Prove \( 2a^2b^2 + (a^2 + b^2)^2 + (a^2 - b^2)^2 = 2(a^2 + ab + b^2)(a^2 - ab + b^2) \).
Answer:
LHS \( = 2a^2b^2 + (a^4 + 2a^2b^2 + b^4) + (a^4 - 2a^2b^2 + b^4) = 2a^4 + 2b^4 + 2a^2b^2 \).
RHS \( = 2[(a^2 + b^2)^2 - (ab)^2] = 2(a^4 + 2a^2b^2 + b^4 - a^2b^2) = 2a^4 + 2b^4 + 2a^2b^2 \).
LHS = RHS.
Numerical check with a = 2, b = 1: LHS \( = 8 + 25 + 9 = 42 \); RHS \( = 2(7)(3) = 42 \).

Linear Equations

To solve a linear equation with fractions, multiply both sides by the LCM of the denominators, remove brackets, collect the variable terms on one side and divide by the coefficient.

Example 15. Solve \( \dfrac{2(x-4)}{5} - \dfrac{2x-3}{10} = \dfrac{4x+1}{2} + 1 \).
Answer:
Multiply by 10: \( 4(x-4) - (2x-3) = 5(4x+1) + 10 \).
\( 4x - 16 - 2x + 3 = 20x + 5 + 10 \), so \( 2x - 13 = 20x + 15 \).
\( -28 = 18x \), hence \( x = -\dfrac{14}{9} \).

Example 16. Solve \( \dfrac{x+2}{2\frac12} - \dfrac{\frac25(5-x)}{3\frac14} = 0 \).
Answer:
\( 2\frac12 = \frac52 \), so the first term is \( \frac25(x+2) \). \( 3\frac14 = \frac{13}{4} \), so the second term is \( \frac{8}{65}(5-x) \).
Multiply by 65: \( 26(x+2) - 8(5-x) = 0 \), i.e. \( 34x + 12 = 0 \).
\( x = -\dfrac{6}{17} \).

Example 17. A and B start a journey together. A walks at 5 km/h. B rides at 8 km/h, stops for half an hour on the way, and still reaches the destination 1 hour before A. Find the length of the journey.
Answer:
Let the distance be \( x \) km. A takes \( \frac{x}{5} \) hours; B takes \( \frac{x}{8} + \frac12 \) hours.
B arrives 1 hour earlier: \( \frac{x}{8} + \frac12 + 1 = \frac{x}{5} \).
Multiply by 40: \( 5x + 60 = 8x \), so \( x = 20 \).
The journey is 20 km. Check: A needs 4 hours; B needs \( 2.5 + 0.5 = 3 \) hours.