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Polynomials - Find zeros of polynomials graphically and verify them algebraically

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The zero of a polynomial p(x)p(x) is the xx-coordinate of the point where the graph of y=p(x)y = p(x) intersects or touches the xx-axis. For a linear polynomial ax+bax + b, the graph is a straight line and has exactly one zero.

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For a quadratic polynomial ax2+bx+cax^2 + bx + c, the graph is a parabola. It can intersect the xx-axis at two distinct points, touch it at one point, or not intersect it at all, corresponding to having 2, 1, or 0 real zeroes respectively.

A parabola intersecting the x-axis at two points: -1 and 1.
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To verify a zero kk algebraically, substitute x=kx = k into the polynomial. If p(k)=0p(k) = 0, then kk is confirmed as a zero of the polynomial.

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The number of zeroes of a polynomial p(x)p(x) of degree nn is at most nn. This means the graph of y=p(x)y = p(x) can intersect the xx-axis at most at nn points.

📐Formulae

p(k)=0p(k) = 0 (Condition for kk to be a zero)

x=−bax = -\frac{b}{a} (Zero of a linear polynomial ax+bax + b)

y=ax2+bx+cy = ax^2 + bx + c (General form of a quadratic polynomial)

Number of zeroes≤nNumber\ of\ zeroes \leq n (For a polynomial of degree nn)

💡Examples

Problem 1:

Look at a graph of y=p(x)y = p(x) that crosses the x-axis at (−2,0)(-2, 0) and (3,0)(3, 0), and crosses the y-axis at (0,−6)(0, -6). Determine the number of zeroes and specify what they are.

Solution:

  1. Identify the points where the graph intersects the x-axis: (−2,0)(-2, 0) and (3,0)(3, 0).
  2. The zeroes are the x-coordinates of these intersection points: x=−2x = -2 and x=3x = 3.
  3. Ignore the y-intercept (0,−6)(0, -6) as it does not indicate a zero.
  4. Total number of zeroes = 22.

Explanation:

The geometric meaning of a zero is the x-coordinate of the point where the graph meets the x-axis. Since there are two such points, the polynomial has two zeroes.

Problem 2:

Given the quadratic polynomial p(x)=x2−4x+4p(x) = x^2 - 4x + 4, describe the nature of its graph and identify the number of zeroes based on its algebraic form (x−2)2(x-2)^2.

Solution:

  1. The polynomial can be rewritten as p(x)=(x−2)2p(x) = (x-2)^2.
  2. To find the zeroes, set p(x)=0  ⟹  (x−2)2=0p(x) = 0 \implies (x-2)^2 = 0, which gives x=2,2x = 2, 2.
  3. Since there is only one unique value, the graph (a parabola opening upwards because a=1>0a=1 > 0) touches the x-axis at exactly one point: (2,0)(2, 0).
  4. Number of zeroes = 11 (or two coincident zeroes).

Explanation:

When a quadratic is a perfect square, its parabola just touches the x-axis at a single point, representing one unique real zero.

Problem 3:

Identify the zeroes of the polynomial p(x)=x2−2x−3p(x) = x^2 - 2x - 3 from its graph and verify them algebraically.

Graph of y = x^2 - 2x - 3 crossing the x-axis at -1 and 3.

Solution:

  1. Graphical Observation: From the graph, the parabola intersects the xx-axis at x=−1x = -1 and x=3x = 3. Therefore, the zeroes are −1-1 and 33.
  2. Algebraic Verification: Substitute x=−1x = -1: p(−1)=(−1)2−2(−1)−3=1+2−3=0p(-1) = (-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0. Substitute x=3x = 3: p(3)=(3)2−2(3)−3=9−6−3=0p(3) = (3)^2 - 2(3) - 3 = 9 - 6 - 3 = 0. Both values are verified as zeroes.

Explanation:

The points (−1,0)(-1, 0) and (3,0)(3, 0) are the x-intercepts. Since p(x)=0p(x) = 0 at these values, they are the zeroes.

Problem 4:

Identify the zeroes of the cubic polynomial p(x)=x3−4xp(x) = x^3 - 4x from its graph and verify your findings algebraically.

Graph of y = x^3 - 4x showing intersections with the x-axis at -2, 0, and 2.

Solution:

  1. Graphical Observation: From the graph, we observe that the curve y=x3−4xy = x^3 - 4x intersects the xx-axis at three distinct points: (−2,0)(-2, 0), (0,0)(0, 0), and (2,0)(2, 0). Therefore, the zeroes of the polynomial are −2-2, 00, and 22.

  2. Algebraic Verification: To find the zeroes of p(x)=x3−4xp(x) = x^3 - 4x, set p(x)=0p(x) = 0: x3−4x=0x^3 - 4x = 0 Factor out the common term xx: x(x2−4)=0x(x^2 - 4) = 0 Factor the difference of squares (x2−4)(x^2 - 4) as (x−2)(x+2)(x - 2)(x + 2): x(x−2)(x+2)=0x(x - 2)(x + 2) = 0 Setting each factor to zero gives: x=0,x−2=0  ⟹  x=2,x+2=0  ⟹  x=−2x = 0, \quad x - 2 = 0 \implies x = 2, \quad x + 2 = 0 \implies x = -2 The algebraic zeroes are {−2,0,2}\{-2, 0, 2\}, which matches the graphical observation.

Explanation:

The zeroes of a polynomial p(x)p(x) are the xx-coordinates of the points where the graph of y=p(x)y = p(x) intersects the xx-axis. For the cubic polynomial x3−4xx^3 - 4x, the graph crosses the axis at three points, indicating three real zeroes. Factorization confirms these specific values.