Chapter 2: Polynomials
Exercise 2.1
Q1
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.
Solution:
The number of zeroes of the polynomial \( p(x) \) is the number of points where the graph of \( y = p(x) \) intersects the x-axis.
- (i) The graph does not intersect the x-axis. Therefore, the number of zeroes is 0.
- (ii) The graph intersects the x-axis at one point. Therefore, the number of zeroes is 1.
- (iii) The graph intersects the x-axis at three points. Therefore, the number of zeroes is 3.
- (iv) The graph intersects the x-axis at two points. Therefore, the number of zeroes is 2.
- (v) The graph intersects the x-axis at four points. Therefore, the number of zeroes is 4.
- (vi) The graph intersects (touches/cuts) the x-axis at three points. Therefore, the number of zeroes is 3.
Exercise 2.2
Q1
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) \( 3x^2 - x - 4 \)
(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:
(i) \( x^2 - 2x - 8 \)
Factorising the polynomial:
\[ x^2 - 2x - 8 = x^2 - 4x + 2x - 8 \]
\[ = x(x - 4) + 2(x - 4) = (x - 4)(x + 2) \]
Zeroes are \( x = 4 \) and \( x = -2 \). Let \( \alpha = 4 \) and \( \beta = -2 \).
Verification:
Sum of zeroes \( = 4 + (-2) = 2 \)
\[ \frac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2} = \frac{-(-2)}{1} = 2 \]
Product of zeroes \( = 4 \times (-2) = -8 \)
\[ \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{-8}{1} = -8 \]
(ii) \( 4s^2 - 4s + 1 \)
Factorising the polynomial:
\[ 4s^2 - 4s + 1 = (2s - 1)^2 \]
Zeroes are \( s = \frac{1}{2}, \frac{1}{2} \). Let \( \alpha = \frac{1}{2} \) and \( \beta = \frac{1}{2} \).
Verification:
Sum of zeroes \( = \frac{1}{2} + \frac{1}{2} = 1 \)
\[ \frac{-(\text{Coefficient of } s)}{\text{Coefficient of } s^2} = \frac{-(-4)}{4} = 1 \]
Product of zeroes \( = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]
\[ \frac{\text{Constant term}}{\text{Coefficient of } s^2} = \frac{1}{4} \]
(iii) \( 6x^2 - 3 - 7x \)
Rearranging: \( 6x^2 - 7x - 3 \)
\[ 6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3) \]
Zeroes are \( x = -\frac{1}{3} \) and \( x = \frac{3}{2} \).
Verification:
Sum of zeroes \( = -\frac{1}{3} + \frac{3}{2} = \frac{-2 + 9}{6} = \frac{7}{6} \]
\[ \frac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2} = \frac{-(-7)}{6} = \frac{7}{6} \]
Product of zeroes \( = \left(-\frac{1}{3}\right) \times \left(\frac{3}{2}\right) = -\frac{3}{6} = -\frac{1}{2} \]
\[ \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{-3}{6} = -\frac{1}{2} \]
(iv) \( 4u^2 + 8u \)
\[ 4u^2 + 8u = 4u(u + 2) \]
Zeroes are \( u = 0 \) and \( u = -2 \).
Verification:
Sum of zeroes \( = 0 + (-2) = -2 \]
\[ \frac{-(\text{Coefficient of } u)}{\text{Coefficient of } u^2} = \frac{-8}{4} = -2 \]
Product of zeroes \( = 0 \times (-2) = 0 \]
\[ \frac{\text{Constant term}}{\text{Coefficient of } u^2} = \frac{0}{4} = 0 \]
(v) \( t^2 - 15 \)
\[ t^2 - 15 = (t - \sqrt{15})(t + \sqrt{15}) \]
Zeroes are \( t = \sqrt{15} \) and \( t = -\sqrt{15} \).
Verification:
Sum of zeroes \( = \sqrt{15} + (-\sqrt{15}) = 0 \]
\[ \frac{-(\text{Coefficient of } t)}{\text{Coefficient of } t^2} = \frac{-0}{1} = 0 \]
Product of zeroes \( = \sqrt{15} \times (-\sqrt{15}) = -15 \]
\[ \frac{\text{Constant term}}{\text{Coefficient of } t^2} = \frac{-15}{1} = -15 \]
(vi) \( 3x^2 - x - 4 \)
\[ 3x^2 - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (x + 1)(3x - 4) \]
Zeroes are \( x = -1 \) and \( x = \frac{4}{3} \).
Verification:
Sum of zeroes \( = -1 + \frac{4}{3} = \frac{1}{3} \]
\[ \frac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2} = \frac{-(-1)}{3} = \frac{1}{3} \]
Product of zeroes \( = -1 \times \frac{4}{3} = -\frac{4}{3} \]
\[ \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{-4}{3} \]
Q2
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \( \frac{1}{4}, -1 \)
(ii) \( \sqrt{2}, \frac{1}{3} \)
(iii) \( 0, \sqrt{5} \)
(iv) \( 1, 1 \)
(v) \( -\frac{1}{4}, \frac{1}{4} \)
(vi) \( 4, 1 \)
(i) \( \frac{1}{4}, -1 \)
(ii) \( \sqrt{2}, \frac{1}{3} \)
(iii) \( 0, \sqrt{5} \)
(iv) \( 1, 1 \)
(v) \( -\frac{1}{4}, \frac{1}{4} \)
(vi) \( 4, 1 \)
Solution:
A quadratic polynomial is given by \( k[x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})] \).
(i) Sum = \( \frac{1}{4} \), Product = \( -1 \)
\[ x^2 - \frac{1}{4}x + (-1) = x^2 - \frac{1}{4}x - 1 \]
Multiplying by 4 to remove the fraction: \( 4x^2 - x - 4 \)
(ii) Sum = \( \sqrt{2} \), Product = \( \frac{1}{3} \)
\[ x^2 - \sqrt{2}x + \frac{1}{3} \]
Multiplying by 3: \( 3x^2 - 3\sqrt{2}x + 1 \)
(iii) Sum = \( 0 \), Product = \( \sqrt{5} \)
\[ x^2 - 0x + \sqrt{5} \]
Polynomial: \( x^2 + \sqrt{5} \)
(iv) Sum = \( 1 \), Product = \( 1 \)
\[ x^2 - 1x + 1 \]
Polynomial: \( x^2 - x + 1 \)
(v) Sum = \( -\frac{1}{4} \), Product = \( \frac{1}{4} \)
\[ x^2 - \left(-\frac{1}{4}\right)x + \frac{1}{4} = x^2 + \frac{1}{4}x + \frac{1}{4} \]
Multiplying by 4: \( 4x^2 + x + 1 \)
(vi) Sum = \( 4 \), Product = \( 1 \)
\[ x^2 - 4x + 1 \]
Polynomial: \( x^2 - 4x + 1 \)