Polynomials

The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Answer

We know that,

The number of points where the graph cuts the x-axis gives the number of zeroes of any polynomial p(x).

(i) In the figure,

The graph does not cuts x-axis.

Hence, there are no zeroes in this graph.

(ii) In the figure,

The graph cuts x-axis at a single point.

Hence, there is one zero in this graph.

(iii) In the figure,

The graph cuts x-axis at three points.

Hence, there are three zeroes in this graph.

(iv) In the figure,

The graph cuts x-axis at two points.

Hence, there are two zeroes in this graph.

(v) In the figure,

The graph cuts x-axis at four points.

Hence, there are four zeroes in this graph.

(vi) In the figure,

The graph cuts x-axis at three points.

Hence, there are three zeroes in this graph.

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

x² - 2x - 8

Answer

Let,

⇒ x² - 2x - 8 = 0

⇒ x² - 4x + 2x - 8 = 0

⇒ x(x - 4) + 2(x - 4) = 0

⇒ (x + 2)(x - 4) = 0

⇒ x + 2 = 0 or x - 4 = 0

⇒ x = -2 or x = 4.

Sum of zeroes = (-2) + 4 = 2 = $\dfrac{-(-2)}{1} = \dfrac{-\text{(Coefficient of x)}}{\text{(Coefficient of x}^2)}$.

Product of zeroes = -2 × 4 = -8 = $\dfrac{-8}{1} = \dfrac{\text{Constant term}}{\text{Coefficient of x}^2}$

Hence, for zero of the polynomial x² - 2x - 8, x = -2, 4.

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

4s² - 4s + 1

Answer

Let,

⇒ 4s² - 4s + 1 = 0

⇒ 4s² - 2s - 2s + 1 = 0

⇒ 2s(2s - 1) - 1(2s - 1) = 0

⇒ (2s - 1)(2s - 1) = 0

⇒ (2s - 1)= 0 or (2s - 1) = 0

⇒ 2s = 1 or 2s = 1

⇒ s = $\dfrac{1}{2}$ or s = $\dfrac{1}{2}$.

Sum of the zeroes = $\dfrac{1}{2} + \dfrac{1}{2} = 1 = \dfrac{-(-4)}{4} = \dfrac{-\text{(Coefficient of s)}}{\text{(Coefficient of s}^2)}$.

Product of zeroes = $\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4} = \dfrac{\text{Constant term}}{\text{Coefficient of s}^2}$

Hence, for zero of the polynomial 4g² - 4g + 1, g = $\dfrac{1}{2}, \dfrac{1}{2}$.

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

6x² - 3 - 7x

Answer

Let,

⇒ 6x² - 3 - 7x = 0

⇒ 6x² - 7x - 3 = 0

⇒ 6x² - 9x + 2x - 3 = 0

⇒ 3x(2x - 3) + 1(2x - 3) = 0

⇒ (3x + 1)(2x - 3) = 0

⇒ 3x + 1 = 0 or 2x - 3 = 0

⇒ 3x = -1 or 2x = 3

⇒ x = $-\dfrac{1}{3}$ or x = $\dfrac{3}{2}$

Sum of zeroes = $-\dfrac{1}{3} + \dfrac{3}{2} = \dfrac{-2 + 9}{6}= \dfrac{7}{6} = \dfrac{-(-7)}{6} = \dfrac{-\text{(Coefficient of x)}}{\text{(Coefficient of x}^2)}$

Product of zeroes = $-\dfrac{1}{3} \times \dfrac{3}{2} = -\dfrac{1}{2} = \dfrac{-3}{6} = \dfrac{\text{Constant term}}{\text{Coefficient of x}^2}$

Hence, for zero of the polynomial 6x² - 3 - 7x, x = $-\dfrac{1}{3}, \dfrac{3}{2}$.

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

4u² + 8u

Answer

Let,

⇒ 4u² + 8u = 0

⇒ 4u(u + 2) = 0

⇒ 4u = 0 or u + 2 = 0

⇒ u = 0 or u = -2.

Sum of zeroes = 0 + (-2) = -2 = $\dfrac{-8}{4} = \dfrac{-\text{(Coefficient of u)}}{\text{(Coefficient of u}^2)}$

Product of zeroes = 0 × (-2) = 0 = $\dfrac{\text{Constant term}}{\text{Coefficient of x}^2}$

Hence, for zero of the polynomial 4u² + 8u, u = 0, -2.

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

t² - 15

Answer

Let,

⇒ t² - 15 = 0

⇒ $(t - \sqrt{15})(t + \sqrt{15}) = 0$

⇒ $(t - \sqrt{15}) \text{ or } (t + \sqrt{15}) = 0$

⇒ $t = \sqrt{15} \text{ or } -\sqrt{15}$

Sum of zeroes = $\sqrt{15} + (-\sqrt{15}) = 0 = \dfrac{-\text{(Coefficient of t)}}{\text{(Coefficient of t}^2)}$

Product of zeroes = $-\sqrt{15} \times \sqrt{15} = -15 = \dfrac{\text{Constant term}}{\text{Coefficient of t}^2}$

Hence, for zero of the polynomial t² - 15, t = $-\sqrt{15}, \sqrt{15}$.

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

3x² - x - 4

Answer

Let,

⇒ 3x² - x - 4 = 0

⇒ 3x² - 4x + 3x - 4 = 0

⇒ x(3x - 4) + 1(3x - 4) = 0

⇒ (x + 1)(3x - 4) = 0

⇒ x + 1 = 0 or 3x - 4 = 0

⇒ x = -1 or 3x = 4

⇒ x = -1 or x = $\dfrac{4}{3}$.

Sum of zeroes = $-1 + \dfrac{4}{3} = \dfrac{-3 + 4}{3} = \dfrac{1}{3} = \dfrac{-(-1)}{3} = \dfrac{-\text{(Coefficient of x)}}{\text{(Coefficient of x}^2)}$

Product of zeroes = $-1 \times \dfrac{4}{3} = -\dfrac{4}{3} = \dfrac{\text{Constant term}}{\text{Coefficient of x}^2}$.

Hence, for zero of the polynomial 3x² - x - 4 = 0, x = -1, $\dfrac{4}{3}$.

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

$\dfrac{1}{4}, -1$

Answer

For given roots,

Quadratic equation : x² - (Sum of roots)x + (Product of roots) = 0

Substituting values we get :

$\Rightarrow x^2 - \dfrac{1}{4}x + (-1) = 0 \\[1em] \Rightarrow x^2 - \dfrac{1}{4}x - 1 = 0 \\[1em] \Rightarrow \dfrac{4x^2 - x - 4}{4} = 0 \\[1em] \Rightarrow 4x^2 - x - 4 = 0.$

Hence, required quadratic equation is 4x² - x - 4 = 0.

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

$\sqrt{2}, \dfrac{1}{3}$

Answer

For given roots,

Quadratic equation : x² - (Sum of roots)x + (Product of roots) = 0

Substituting values we get :

$\Rightarrow x^2 - \sqrt{2}x + \dfrac{1}{3} = 0 \\[1em] \Rightarrow \dfrac{3x^2 - 3\sqrt{2}x + 1}{3} = 0 \\[1em] \Rightarrow 3x^2 - 3\sqrt{2}x + 1 = 0.$

Hence, required quadratic equation is $3x^2 - 3\sqrt{2}x + 1 = 0$.

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

$0, \sqrt{5}$

Answer

For given roots,

Quadratic equation : x² - (Sum of roots)x + (Product of roots) = 0

Substituting values we get :

$\Rightarrow x^2 - 0x + \sqrt{5} = 0 \\[1em] \Rightarrow x^2 - 0 + \sqrt{5} = 0 \\[1em] \Rightarrow x^2 + \sqrt{5} = 0.$

Hence, required quadratic equation is $x^2 + \sqrt{5} = 0.$

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

1, 1

Answer

For given roots,

Quadratic equation : x² - (Sum of roots)x + (Product of roots) = 0

Substituting values we get :

⇒ x² - 1x + 1 = 0

⇒ x² - x + 1 = 0.

Hence, required quadratic equation is x² - x + 1 = 0.

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

$-\dfrac{1}{4}, \dfrac{1}{4}$

(vi) 4, 1

Answer

For given roots,

Quadratic equation : x² - (Sum of roots)x + (Product of roots) = 0

Substituting values we get :

$\Rightarrow x^2 - \Big(-\dfrac{1}{4}\Big)x + \Big(\dfrac{1}{4}\Big) = 0 \\[1em] \Rightarrow x^2 + \dfrac{x}{4} + \dfrac{1}{4} = 0 \\[1em] \Rightarrow \dfrac{4x^2 + x + 1}{4} = 0 \\[1em] \Rightarrow 4x^2 + x + 1 = 0.$

Hence, required quadratic equation is 4x² + x + 1 = 0.

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

4, 1

Answer

(vi) For given roots,

Quadratic equation : x² - (Sum of roots)x + (Product of roots) = 0

Substituting values we get :

⇒ x² - 4x + 1 = 0

Hence, required quadratic equation is x² - 4x + 1 = 0.