2.3-Relationship between Zeroes and Coefficients of a Polynomial

2.3-Relationship between Zeroes and Coefficients of a Polynomial Important Formulae

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Grade 10 → Math → Polynomials → 2.3-Relationship between Zeroes and Coefficients of a Polynomial

After successful completion of this topic, you should be able to:

  • Compute zeroes of the polynomials in order to verify the relationship between zeroes and the coefficients.
  • Compute the sum and product of zeroes of the polynomial in order to find the quadratic polynomial.

The relationship between the zeroes of a polynomial and its coefficients is a fundamental concept in algebra. This relationship allows us to understand the properties of polynomials and to find zeroes without explicitly solving them. In this section, we will explore this relationship in detail.

1. Definition of a Polynomial

A polynomial in one variable $x$ can be expressed as:

$$P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$$

where $a_n, a_{n-1}, \ldots, a_1, a_0$ are constants (coefficients) and $n$ is a non-negative integer (degree of the polynomial).

2. Zeroes of a Polynomial

The zeroes (or roots) of the polynomial $P(x)$ are the values of $x$ for which $P(x) = 0$. If a polynomial has $n$ zeroes, we can denote them as $r_1, r_2, \ldots, r_n$. The relationship between the zeroes and the coefficients of the polynomial can be derived from Vieta's formulas.

3. Vieta's Formulas

For a polynomial of degree $n$, the relationships can be stated as follows:

  • If the polynomial is:

    • $$P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$$

  • The sum of the zeroes, $r_1 + r_2 + \ldots + r_n$, is given by:

    • $$r_1 + r_2 + \ldots + r_n = -\frac{a_{n-1}}{a_n}$$

  • The sum of the products of the zeroes taken two at a time, $r_1r_2 + r_1r_3 + \ldots$, is given by:

    • $$r_1r_2 + r_1r_3 + \ldots = \frac{a_{n-2}}{a_n}$$

  • Similarly, for the product of the zeroes (for even and odd degrees):
    • If $n$ is odd: $$r_1r_2 \ldots r_n = -\frac{a_0}{a_n}$$
    • If $n$ is even: $$r_1r_2 \ldots r_n = \frac{a_0}{a_n}$$
4. Example

Consider the polynomial:

$$P(x) = 2x^3 - 3x^2 + 4x - 5$$

Here, the coefficients are $a_3 = 2$, $a_2 = -3$, $a_1 = 4$, and $a_0 = -5$. According to Vieta's formulas:

  • The sum of the zeroes ($r_1 + r_2 + r_3$) is:

    • $$r_1 + r_2 + r_3 = -\frac{-3}{2} = \frac{3}{2}$$

  • The sum of the products of the zeroes taken two at a time ($r_1r_2 + r_2r_3 + r_3r_1$) is:

    • $$r_1r_2 + r_2r_3 + r_3r_1 = \frac{4}{2} = 2$$

  • The product of the zeroes ($r_1r_2r_3$) is:

    • $$r_1r_2r_3 = -\frac{-5}{2} = \frac{5}{2}$$

5. Importance of Zeroes and Coefficients

Understanding the relationship between zeroes and coefficients is beneficial for several reasons:

  • It allows for quick calculations of zeroes without solving the polynomial directly.
  • It aids in graphing polynomials by providing information about their behavior.
  • It helps in determining the nature of the roots, such as whether they are real or complex.
6. Additional Examples

Here are more examples illustrating the relationships:

  • Example 1: For the polynomial $P(x) = x^2 - 5x + 6$, the coefficients are $1$, $-5$, and $6$. The zeroes are $r_1 = 2$ and $r_2 = 3$.

    • Sum of zeroes: $$2 + 3 = 5 = -\frac{-5}{1}$$

    Product of zeroes: $$2 \cdot 3 = 6 = \frac{6}{1}$$

  • Example 2: For $P(x) = x^3 - 4x^2 + 5x - 2$, where $a_3 = 1$, $a_2 = -4$, $a_1 = 5$, and $a_0 = -2$:

    • Sum of zeroes: $$r_1 + r_2 + r_3 = 4 = -\frac{-4}{1}$$

    Product of zeroes: $$r_1r_2r_3 = 2 = -\frac{-2}{1}$$

7. Practice Problems

To deepen your understanding, attempt the following problems:

  • 1. Find the sum and product of the zeroes for $P(x) = 3x^2 - 12x + 9$.
  • 2. Determine the relationships for the polynomial $P(x) = 2x^3 + 3x^2 - 8x - 4$.
  • 3. Given the zeroes of $P(x) = x^2 - 4$, verify the relationships with the coefficients.

Engaging with these examples and practice problems will reinforce the concepts of relationships between zeroes and coefficients of a polynomial.

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) $x^2 –2x–8 $
(ii) $4s^2 –4s+1 $
(iii) $6x^2 –3–7x $
(iv) $4u^2 +8u $
(v) $t^2 –15 $
(vi) $3x^2 – x– 4$

Solution:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^2 - 2x - 8$ The given quadratic polynomial is $x^2 - 2x - 8$. To find the zeroes, we use the factorization method: $x^2 - 2x - 8 = (x - 4)(x + 2)$. So, the zeroes are $x = 4$ and $x = -2$. Verification of the relationship between the zeroes and coefficients: The sum of the zeroes = $4 + (-2) = 2$. The product of the zeroes = $4 \times (-2) = -8$. From the equation $x^2 - 2x - 8$, the sum of the zeroes should be $-(-2) = 2$ and the product should be $\frac{-8}{1} = -8$. Hence, the relationship is verified.

(ii) $4s^2 - 4s + 1$ The given quadratic polynomial is $4s^2 - 4s + 1$. To find the zeroes, we use the quadratic formula: $s = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(1)}}{2(4)}$ $s = \frac{4 \pm \sqrt{16 - 16}}{8}$ $s = \frac{4 \pm \sqrt{0}}{8}$ $s = \frac{4 \pm 0}{8}$ $s = \frac{4}{8}$ $s = \frac{1}{2}$. So, the zeroes are $s = \frac{1}{2}$ (repeated root). Verification of the relationship between the zeroes and coefficients: The sum of the zeroes = $\frac{1}{2} + \frac{1}{2} = 1$. The product of the zeroes = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. From the equation $4s^2 - 4s + 1$, the sum of the zeroes should be $\frac{-(-4)}{4} = 1$ and the product should be $\frac{1}{4}$. Hence, the relationship is verified.

(iii) $6x^2 - 3 - 7x$ The given quadratic polynomial is $6x^2 - 7x - 3$. To find the zeroes, we use the quadratic formula: $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(-3)}}{2(6)}$ $x = \frac{7 \pm \sqrt{49 + 72}}{12}$ $x = \frac{7 \pm \sqrt{121}}{12}$ $x = \frac{7 \pm 11}{12}$. So, the two zeroes are: $x = \frac{7 + 11}{12} = \frac{18}{12} = \frac{3}{2}$ and $x = \frac{7 - 11}{12} = \frac{-4}{12} = \frac{-1}{3}$. Verification of the relationship between the zeroes and coefficients: The sum of the zeroes = $\frac{3}{2} + \frac{-1}{3} = \frac{9}{6} - \frac{2}{6} = \frac{7}{6}$. The product of the zeroes = $\frac{3}{2} \times \frac{-1}{3} = \frac{-3}{6} = \frac{-1}{2}$. From the equation $6x^2 - 7x - 3$, the sum of the zeroes should be $\frac{-(-7)}{6} = \frac{7}{6}$ and the product should be $\frac{-3}{6} = \frac{-1}{2}$. Hence, the relationship is verified.

(iv) $4u^2 + 8u$ The given quadratic polynomial is $4u^2 + 8u$. Factor out the common term: $4u^2 + 8u = 4u(u + 2)$. So, the zeroes are $u = 0$ and $u = -2$. Verification of the relationship between the zeroes and coefficients: The sum of the zeroes = $0 + (-2) = -2$. The product of the zeroes = $0 \times (-2) = 0$. From the equation $4u^2 + 8u$, the sum of the zeroes should be $\frac{-8}{4} = -2$ and the product should be $\frac{0}{4} = 0$. Hence, the relationship is verified.

(v) $t^2 - 15$ The given quadratic polynomial is $t^2 - 15$. To find the zeroes, we use the quadratic formula: $t = \frac{-0 \pm \sqrt{0^2 - 4(1)(-15)}}{2(1)}$ $t = \frac{0 \pm \sqrt{60}}{2}$ $t = \frac{\pm \sqrt{60}}{2}$ $t = \frac{\pm 2\sqrt{15}}{2}$ $t = \pm \sqrt{15}$. So, the zeroes are $t = \sqrt{15}$ and $t = -\sqrt{15}$. Verification of the relationship between the zeroes and coefficients: The sum of the zeroes = $\sqrt{15} + (-\sqrt{15}) = 0$. The product of the zeroes = $\sqrt{15} \times (-\sqrt{15}) = -15$. From the equation $t^2 - 15$, the sum of the zeroes should be $0$ and the product should be $\frac{-15}{1} = -15$. Hence, the relationship is verified.

(vi) $3x^2 - x - 4$ The given quadratic polynomial is $3x^2 - x - 4$. To find the zeroes, we use the quadratic formula: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-4)}}{2(3)}$ $x = \frac{1 \pm \sqrt{1 + 48}}{6}$ $x = \frac{1 \pm \sqrt{49}}{6}$ $x = \frac{1 \pm 7}{6}$. So, the two zeroes are: $x = \frac{1 + 7}{6} = \frac{8}{6} = \frac{4}{3}$ and $x = \frac{1 - 7}{6} = \frac{-6}{6} = -1$. Verification of the relationship between the zeroes and coefficients: The sum of the zeroes = $\frac{4}{3} + (-1) = \frac{4}{3} - \frac{3}{3} = \frac{1}{3}$. The product of the zeroes = $\frac{4}{3} \times (-1) = \frac{-4}{3}$. From the equation $3x^2 - x - 4$, the sum of the zeroes should be $\frac{-(-1)}{3} = \frac{1}{3}$ and the product should be $\frac{-4}{3}$. Hence, the relationship is verified.

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Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) $\dfrac{1}{4}$, -1
(ii) $\sqrt{2}, \dfrac{1}{3}$
(iii) 0, $\sqrt{5}$
(iv) 1,1
(v) $-\dfrac{1}{4}, \dfrac{1}{4}$
(vi) 4, 1

Solution:

Find a quadratic polynomial for the given sum and product of its zeroes:
(i) Sum = $\dfrac{1}{4}$, Product = -1

Quadratic polynomial: $x^2 - \left( \dfrac{1}{4} \right)x - 1$

(ii) Sum = $\sqrt{2}$, Product = $\dfrac{1}{3}$

Quadratic polynomial: $x^2 - \sqrt{2}x + \dfrac{1}{3}$

(iii) Sum = 0, Product = $\sqrt{5}$

Quadratic polynomial: $x^2 + \sqrt{5}$

(iv) Sum = 1, Product = 1

Quadratic polynomial: $x^2 - x + 1$

(v) Sum = $-\dfrac{1}{4}$, Product = $\dfrac{1}{4}$

Quadratic polynomial: $x^2 + \dfrac{1}{4}x + \dfrac{1}{4}$

(vi) Sum = 4, Product = 1

Quadratic polynomial: $x^2 - 4x + 1$

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