CBSE Class 10 Mathematics Polynomials Notes Set 09

Polynomial and Its Parts

An expression of the form \( p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \) is a polynomial when the leading coefficient \( a_n \ne 0 \), every coefficient is a real number and every power of x is a whole number.
The part \( a_n x^n \) is the leading term and \( a_n \) is the leading coefficient. For example, in \( p(u) = \frac{1}{2}u^3 - 3u^2 + 2u - 4 \) the terms are \( \frac{1}{2}u^3, -3u^2, 2u, -4 \) and the coefficients are \( \frac{1}{2}, -3, 2, -4 \).

Types of Polynomials

A polynomial can be named in three ways: by the number of variables, by its degree and by its number of terms. The degree is the highest power of the variable that appears.

The zero polynomial 0 has no terms, so its degree is not defined. A single-variable polynomial is usually written in descending powers of x (or, less often, ascending powers).

Value of a Polynomial and Zero of a Polynomial

The value of \( p(x) \) at \( x = \alpha \) is the number \( p(\alpha) \) obtained by replacing x with \( \alpha \). A real number \( \alpha \) is a zero of \( p(x) \) when \( p(\alpha) = 0 \).
Example: for \( p(x) = x^3 - 6x^2 + 11x - 6 \) we get \( p(1) = 1 - 6 + 11 - 6 = 0 \), \( p(2) = 8 - 24 + 22 - 6 = 0 \) and \( p(3) = 27 - 54 + 33 - 6 = 0 \), so 1, 2 and 3 are zeroes of p(x).

Finding Zeroes of a Quadratic Polynomial by Factorisation

1. Split the middle term (or use any other suitable method) and factorise the quadratic.
2. Put each linear factor equal to zero.
3. Solve the two linear equations; their solutions are the zeroes.

Geometrical Meaning of Zeroes

The zeroes of \( f(x) \) are the x-coordinates of the points where the graph of \( y = f(x) \) meets the x-axis.
Linear polynomial: the graph of \( y = 2x - 5 \) is a straight line through A(1, -3) and B(4, 3). It cuts the x-axis once, at \( P(\frac{5}{2}, 0) \), so the only zero of \( 2x - 5 \) is \( \frac{5}{2} \). In general a linear polynomial \( ax + b \) has exactly one zero, \( -\frac{b}{a} \).

Quadratic polynomial: for \( y = x^2 - 2x - 8 \) the table of values is given below, and the smooth curve through these points is a parabola. It meets the x-axis at \( (-2, 0) \) and \( (4, 0) \), so the zeroes are -2 and 4.

The graph of \( y = ax^2 + bx + c \) (\( a \ne 0 \)) opens upward when \( a > 0 \) and downward when \( a < 0 \). There are three possible cases, shown below with the zeroes marked in red.
Case I: the parabola cuts the x-axis at two distinct points, so there are two distinct zeroes.
Case II: the parabola touches the x-axis at one point, so the two zeroes coincide and there is exactly one zero.
Case III: the parabola lies wholly above or wholly below the x-axis, so there is no real zero.
A quadratic polynomial therefore has at most two zeroes. More generally the graph of a polynomial of degree n meets the x-axis at most n times, so it has at most n zeroes.

Relationship Between Zeroes and Coefficients

Quadratic polynomial \( ax^2 + bx + c \) with zeroes \( \alpha, \beta \): since \( (x - \alpha) \) and \( (x - \beta) \) are factors, \( ax^2 + bx + c = a(x - \alpha)(x - \beta) \). Comparing coefficients gives
\[ \alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a} \]

Conversely, a quadratic polynomial whose zeroes have sum S and product Q is \( k\{x^2 - Sx + Q\} \), where k is any non-zero constant.
Cubic polynomial \( ax^3 + bx^2 + cx + d \) with zeroes \( \alpha, \beta, \gamma \):
\[ \alpha + \beta + \gamma = -\frac{b}{a}, \quad \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}, \quad \alpha\beta\gamma = -\frac{d}{a} \]
A cubic with these zeroes is \( k\{x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma\} \).

Useful Identities Involving Two Zeroes

1. \( \alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta) \)
2. \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \)
3. \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \)
4. \( \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \)
5. \( \alpha^3 - \beta^3 = (\alpha - \beta)^3 + 3\alpha\beta(\alpha - \beta) \)

Division Algorithm for Polynomials

If f(x) is a polynomial and g(x) is a non-zero polynomial, there are polynomials q(x) and r(x) with
\[ f(x) = g(x) \cdot q(x) + r(x) \]
where either \( r(x) = 0 \) or the degree of r(x) is less than the degree of g(x). In words: dividend = divisor × quotient + remainder. If the remainder is 0, then g(x) is a factor of f(x).

Solved Examples

Example 1. Verify that -3, -2, 1, 2 are zeroes of \( x^4 + 2x^3 - 7x^2 - 8x + 12 \).
Answer:
Let \( p(x) = x^4 + 2x^3 - 7x^2 - 8x + 12 \).
\( p(-3) = 81 - 54 - 63 + 24 + 12 = 0 \)
\( p(-2) = 16 - 16 - 28 + 16 + 12 = 0 \)
\( p(1) = 1 + 2 - 7 - 8 + 12 = 0 \)
\( p(2) = 16 + 16 - 28 - 16 + 12 = 0 \)
Since the value is 0 in each case, -3, -2, 1 and 2 are zeroes of p(x).

Example 2. Find the zeroes of (i) \( 5x^2 + 4x \) (ii) \( 9x^2 + 6x + 1 \) (iii) \( 6x^2 + x - 12 \).
Answer:
(i) \( 5x^2 + 4x = x(5x + 4) \), so the zeroes are \( x = 0 \) and \( x = -\frac{4}{5} \).
(ii) \( 9x^2 + 6x + 1 = (3x + 1)^2 \), so the only zero is \( x = -\frac{1}{3} \).
(iii) \( 6x^2 + x - 12 = 6x^2 + 9x - 8x - 12 = (2x + 3)(3x - 4) \), so the zeroes are \( x = -\frac{3}{2} \) and \( x = \frac{4}{3} \).

Example 3. Find the zeroes of \( 5x^2 + 13x + 8 \).
Answer:
\( 5x^2 + 13x + 8 = 5x^2 + 5x + 8x + 8 = (x + 1)(5x + 8) \).
The zeroes are \( x = -1 \) and \( x = -\frac{8}{5} \).

Example 4. Find the zeroes of \( 9x^2 - 5 \) and verify the relation between zeroes and coefficients.
Answer:
\( 9x^2 - 5 = (3x)^2 - (\sqrt{5})^2 = (3x - \sqrt{5})(3x + \sqrt{5}) \), so the zeroes are \( \frac{\sqrt{5}}{3} \) and \( -\frac{\sqrt{5}}{3} \).
Sum of zeroes = 0 = \( -\frac{0}{9} \) = \( -\frac{b}{a} \).
Product of zeroes = \( -\frac{5}{9} \) = \( \frac{c}{a} \). Both relations hold.

Example 5. Find the zeroes of \( 6x^2 + 7x - 3 \) and verify the relation between zeroes and coefficients.
Answer:
\( 6x^2 + 7x - 3 = 6x^2 + 9x - 2x - 3 = (2x + 3)(3x - 1) \), so the zeroes are \( -\frac{3}{2} \) and \( \frac{1}{3} \).
Sum = \( -\frac{3}{2} + \frac{1}{3} = -\frac{7}{6} \) and \( -\frac{b}{a} = -\frac{7}{6} \).
Product = \( -\frac{3}{2} \times \frac{1}{3} = -\frac{1}{2} \) and \( \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2} \).

Example 6. Find a quadratic polynomial whose zeroes are 5 and -2.
Answer:
Sum = 5 + (-2) = 3 and product = 5 × (-2) = -10.
The polynomial is \( k(x^2 - 3x - 10) \), for example \( x^2 - 3x - 10 \).

Example 7. If \( \alpha, \beta \) are the zeroes of \( x^2 + 4x + 3 \), form the polynomial whose zeroes are \( \alpha^2 + \beta^2 \) and \( \alpha\beta \).
Answer:
Here \( \alpha + \beta = -4 \) and \( \alpha\beta = 3 \), so \( \alpha^2 + \beta^2 = 16 - 6 = 10 \).
New sum = \( 10 + 3 = 13 \) and new product = \( 10 \times 3 = 30 \).
The polynomial is \( k(x^2 - 13x + 30) \).

Example 8. Show that 3, -1 and \( -\frac{1}{3} \) are the zeroes of \( p(x) = 3x^3 - 5x^2 - 11x - 3 \) and verify the relations between zeroes and coefficients.
Answer:
\( p(3) = 81 - 45 - 33 - 3 = 0 \), \( p(-1) = -3 - 5 + 11 - 3 = 0 \) and \( p(-\frac{1}{3}) = -\frac{1}{9} - \frac{5}{9} + \frac{11}{3} - 3 = 0 \).
Sum of zeroes = \( 3 - 1 - \frac{1}{3} = \frac{5}{3} = -\frac{b}{a} \).
Sum of products in pairs = \( -3 + \frac{1}{3} - 1 = -\frac{11}{3} = \frac{c}{a} \).
Product of zeroes = \( 3 \times (-1) \times (-\frac{1}{3}) = 1 = -\frac{d}{a} \).

Example 9. If \( \alpha, \beta \) are the zeroes of \( 2x^2 - 5x + 3 \), find \( \alpha^2 + \beta^2 \) and \( \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} \).
Answer:
\( \alpha + \beta = \frac{5}{2} \) and \( \alpha\beta = \frac{3}{2} \).
\( \alpha^2 + \beta^2 = \frac{25}{4} - 3 = \frac{13}{4} \).
\( \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta} = \frac{(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)}{\alpha\beta} = \frac{\frac{125}{8} - \frac{45}{4}}{\frac{3}{2}} = \frac{35}{12} \).

Example 10. If \( \alpha, \beta \) are the zeroes of \( x^2 - 3x + k \) and \( \alpha - \beta = 1 \), find k.
Answer:
\( \alpha + \beta = 3 \) and \( \alpha\beta = k \).
\( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \) gives \( 1 = 9 - 4k \), so \( k = 2 \).

Example 11. Divide \( 2x^2 + 3x + 1 \) by \( x + 2 \) and verify the division algorithm.
Answer:
Dividing: \( 2x^2 + 3x + 1 = (x + 2)(2x - 1) + 3 \), so quotient = \( 2x - 1 \) and remainder = 3.
Check: \( (x + 2)(2x - 1) + 3 = 2x^2 + 3x - 2 + 3 = 2x^2 + 3x + 1 \), which is the dividend.

Example 12. Is \( x^2 - 2x + 1 \) a factor of \( 2x^4 - 3x^3 - 3x^2 + 7x - 3 \)?
Answer:
Long division gives \( 2x^4 - 3x^3 - 3x^2 + 7x - 3 = (x^2 - 2x + 1)(2x^2 + x - 3) \) with remainder 0.
The remainder is zero, so \( x^2 - 2x + 1 \) is a factor.

Example 13. What must be subtracted from \( 8x^4 + 14x^3 - 2x^2 + 8x - 12 \) so that it is exactly divisible by \( 4x^2 + 3x - 2 \)?
Answer:
Dividing gives quotient \( 2x^2 + 2x - 1 \) and remainder \( 15x - 14 \).
So \( 15x - 14 \) must be subtracted (equivalently \( -15x + 14 \) must be added).