CBSE Class 10 Mathematics Polynomials Notes Set 07

Value and Zeros of a Polynomial

The value of a polynomial f(x) at \( x = \alpha \) is obtained by substituting \( x = \alpha \) in the polynomial and is denoted by \( f(\alpha) \). For example, if \( f(x) = 2x^3 - 13x^2 + 17x + 12 \), its value at x = 1 is \( f(1) = 2 - 13 + 17 + 12 = 18 \).

A real number \( \alpha \) is a zero of a polynomial f(x) if \( f(\alpha) = 0 \).

Solved Examples on Zeros of a Polynomial

Example 1. Show that x = 2 is a zero of \( 2x^3 + x^2 - 7x - 6 \).
Answer:
Let \( p(x) = 2x^3 + x^2 - 7x - 6 \). Then \( p(2) = 2(2)^3 + (2)^2 - 7(2) - 6 = 16 + 4 - 14 - 6 = 0 \).
Hence x = 2 is a zero of p(x).

Example 2. If \( x = \dfrac{4}{3} \) is a zero of the polynomial \( f(x) = 6x^3 - 11x^2 + kx - 20 \), find the value of k.
Answer:
\( f\left(\dfrac{4}{3}\right) = 6\left(\dfrac{64}{27}\right) - 11\left(\dfrac{16}{9}\right) + \dfrac{4k}{3} - 20 = 0 \).
\( \dfrac{128}{9} - \dfrac{176}{9} + \dfrac{4k}{3} - 20 = 0 \), so \( 128 - 176 + 12k - 180 = 0 \) (multiplying by 9).
\( 12k = 228 \), so k = 19.

Example 3. If x = 2 and x = 0 are zeros of the polynomial \( f(x) = 2x^3 - 5x^2 + ax + b \), find the values of a and b.
Answer:
\( f(2) = 2(8) - 5(4) + 2a + b = 0 \), so \( 2a + b = 4 \) ... (i)
\( f(0) = b = 0 \).
Putting b = 0 in (i): \( 2a = 4 \), so a = 2 and b = 0.

Check Your Progress: Zeros of a Polynomial

Check Your Progress 1. Is x = −4 a zero of the polynomial \( 2x^2 + 5x - 12 \)?
Answer:
\( 2(-4)^2 + 5(-4) - 12 = 32 - 20 - 12 = 0 \). Yes, x = −4 is a zero.

Check Your Progress 2. If x = −5 is a zero of \( x^3 + 2x^2 - 14x + K + 2 \), find K.
Answer:
\( (-5)^3 + 2(-5)^2 - 14(-5) + K + 2 = -125 + 50 + 70 + K + 2 = K - 3 = 0 \), so K = 3.

Geometrical Meaning of the Zeros of a Polynomial

Graph of a linear polynomial. The graph of \( y = ax + b \), \( a \ne 0 \), is a straight line, which is why \( f(x) = ax + b \) is called a linear polynomial. Two points are enough to draw it. The line crosses the x-axis at exactly one point, \( \left(-\dfrac{b}{a}, 0\right) \), so a linear polynomial has exactly one zero.

Graph of a quadratic polynomial. The graph of \( y = ax^2 + bx + c \), \( a \ne 0 \), is always a parabola. It opens upwards if \( a > 0 \) and downwards if \( a < 0 \).

Number of zeros. If \( \alpha \) is a real number with \( a\alpha^2 + b\alpha + c = 0 \), then the point \( (\alpha, 0) \) lies on the parabola; conversely, if the parabola meets the x-axis at \( (\alpha, 0) \), then \( a\alpha^2 + b\alpha + c = 0 \). So the x-coordinates of the points where the parabola meets the x-axis are exactly the real zeros of the polynomial.

1. If the parabola cuts the x-axis at two distinct points, the x-coordinates of these points are the two zeros of the polynomial.
2. If the parabola just touches the x-axis at one point (two coincident points), the x-coordinate of that point is the only zero.
3. If the parabola does not meet the x-axis at all, the polynomial has no real zero.

Note: A quadratic polynomial has either two zeros, one zero or no zero. In general, a polynomial of degree n has at most n zeros.

Solved Examples on Graphs of Polynomials

Check Your Progress: Geometrical Meaning of Zeros

Check Your Progress 1. Check whether the graph of \( 3(x - 2)(2 - x) = x^2 + 2x(1 - x) \) opens upwards or downwards.
Answer:
Left side: \( 3(x-2)(2-x) = -3(x-2)^2 = -3x^2 + 12x - 12 \). Right side: \( x^2 + 2x - 2x^2 = -x^2 + 2x \).
Left side minus right side: \( -2x^2 + 10x - 12 \). The coefficient of \( x^2 \) is \( -2 < 0 \), so the graph opens downwards (its zeros are 2 and 3).