Chapter 2: Polynomials and Factorisation (Exercise 2.2)

Chapter 2: Polynomials and Factorisation (Exercise 2.2)

Question 1:

Find the value of the polynomial $p(x) = x^2 - 2x + 1$ at:

1. $x = 1$

2. $x = -1$

3. $x = 0$

4. $x = 2$

5. $x = -2$

Solution:

Given polynomial: $p(x) = x^2 - 2x + 1$

1. At $x = 1$:

   $$p(1) = (1)^2 - 2(1) + 1 = 1 - 2 + 1 = 0$$

2. At $x = -1$:

   $$p(-1) = (-1)^2 - 2(-1) + 1 = 1 + 2 + 1 = 4$$

3. At $x = 0$:

   $$p(0) = (0)^2 - 2(0) + 1 = 0 - 0 + 1 = 1$$

4. At $x = 2$:

   $$p(2) = (2)^2 - 2(2) + 1 = 4 - 4 + 1 = 1$$

5. At $x = -2$:

   $$p(-2) = (-2)^2 - 2(-2) + 1 = 4 + 4 + 1 = 9$$

Question 2:

Find $p(0), p(1),$ and $p(2)$ for each of the following polynomials:

1. $p(y) = y^2 - y + 1$

2. $p(t) = 2 + t + 2t^2 - t^3$

3. $p(x) = x^3$

4. $p(x) = (x - 1)(x + 1)$

Solution:

1. $p(y) = y^2 - y + 1$:

   • $p(0) = (0)^2 - 0 + 1 = 1$

   • $p(1) = (1)^2 - 1 + 1 = 1 - 1 + 1 = 1$

   • $p(2) = (2)^2 - 2 + 1 = 4 - 2 + 1 = 3$

2. $p(t) = 2 + t + 2t^2 - t^3$:

   • $p(0) = 2 + 0 + 2(0)^2 - (0)^3 = 2$

   • $p(1) = 2 + 1 + 2(1)^2 - (1)^3 = 2 + 1 + 2 - 1 = 4$

   • $p(2) = 2 + 2 + 2(2)^2 - (2)^3 = 2 + 2 + 8 - 8 = 4$

3. $p(x) = x^3$:

   • $p(0) = (0)^3 = 0$

   • $p(1) = (1)^3 = 1$

   • $p(2) = (2)^3 = 8$

4. $p(x) = (x - 1)(x + 1)$:

   • $p(0) = (0 - 1)(0 + 1) = (-1)(1) = -1$

   • $p(1) = (1 - 1)(1 + 1) = (0)(2) = 0$

   • $p(2) = (2 - 1)(2 + 1) = (1)(3) = 3$

Question 3:

Verify whether the indicated values of $x$ are zeroes of the given polynomial:

1. $p(x) = 3x + 1, \quad x = -\frac{1}{3}$

2. $p(x) = 5x - \pi, \quad x = \frac{4}{5}$

3. $p(x) = x^2 - 1, \quad x = 1, -1$

4. $p(x) = (x + 1)(x - 2), \quad x = -1, 2$

5. $p(y) = y^2, \quad y = 0$

6. $p(x) = ax + b, \quad x = -\frac{b}{a}$

7. $p(x) = 3x^2 - 1, \quad x = -\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$

Solution:

1. $p(x) = 3x + 1$ at $x = -\frac{1}{3}$:

   $$p\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right) + 1 = -1 + 1 = 0$$

   Since $p\left(-\frac{1}{3}\right) = 0$, $x = -\frac{1}{3}$ is a zero of $p(x)$.

2. $p(x) = 5x - \pi$ at $x = \frac{4}{5}$:

   $$p\left(\frac{4}{5}\right) = 5\left(\frac{4}{5}\right) - \pi = 4 - \pi \neq 0$$

   Since $p\left(\frac{4}{5}\right) \neq 0$, $x = \frac{4}{5}$ is not a zero of $p(x)$.

3. $p(x) = x^2 - 1$ at $x = 1, -1$:

   • At $x = 1$: $p(1) = (1)^2 - 1 = 1 - 1 = 0$

   • At $x = -1$: $p(-1) = (-1)^2 - 1 = 1 - 1 = 0$

     Both $x = 1$ and $x = -1$ are zeroes of $p(x)$.

4. $p(x) = (x + 1)(x - 2)$ at $x = -1, 2$:

   • At $x = -1$: $p(-1) = (-1 + 1)(-1 - 2) = (0)(-3) = 0$

   • At $x = 2$: $p(2) = (2 + 1)(2 - 2) = (3)(0) = 0$

     Both $x = -1$ and $x = 2$ are zeroes of $p(x)$.

5. $p(y) = y^2$ at $y = 0$:

   $$p(0) = (0)^2 = 0$$

   $y = 0$ is a zero of $p(y)$.

6. $p(x) = ax + b$ at $x = -\frac{b}{a}$:

   $$p\left(-\frac{b}{a}\right) = a\left(-\frac{b}{a}\right) + b = -b + b = 0$$

   $x = -\frac{b}{a}$ is a zero of $p(x)$.

7. $p(x) = 3x^2 - 1$ at $x = -\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$:

   • At $x = -\frac{1}{\sqrt{3}}$:

     $$p\left(-\frac{1}{\sqrt{3}}\right) = 3\left(-\frac{1}{\sqrt{3}}\right)^2 - 1 = 3\left(\frac{1}{3}\right) - 1 = 1 - 1 = 0$$

     So, $x = -\frac{1}{\sqrt{3}}$ is a zero of $p(x)$.

   • At $x = \frac{2}{\sqrt{3}}$:

     $$p\left(\frac{2}{\sqrt{3}}\right) = 3\left(\frac{2}{\sqrt{3}}\right)^2 - 1 = 3\left(\frac{4}{3}\right) - 1 = 4 - 1 = 3 \neq 0$$

     So, $x = \frac{2}{\sqrt{3}}$ is not a zero of $p(x)$.

Question 4:

Find the zero of the polynomial in each of the following cases:

1. $p(x) = x + 5$

2. $p(x) = x - 5$

3. $p(x) = 2x + 5$

4. $p(x) = 3x - 2$

5. $p(x) = 3x$

6. $p(x) = ax, \quad a \neq 0$

7. $p(x) = ax + b, \quad a \neq 0, a, b \in \mathbb{R}$

Solution:

To find the zero of $p(x)$, set $p(x) = 0$:

1. $x + 5 = 0 \implies x = -5$ $\rightarrow$ Zero = $-5$

2. $x - 5 = 0 \implies x = 5$ $\rightarrow$ Zero = $5$

3. $2x + 5 = 0 \implies 2x = -5 \implies x = -\frac{5}{2}$ $\rightarrow$ Zero = $-\frac{5}{2}$

4. $3x - 2 = 0 \implies 3x = 2 \implies x = \frac{2}{3}$ $\rightarrow$ Zero = $\frac{2}{3}$

5. $3x = 0 \implies x = 0$ $\rightarrow$ Zero = $0$

6. $ax = 0 \implies x = \frac{0}{a} = 0$ $\rightarrow$ Zero = $0$

7. $ax + b = 0 \implies ax = -b \implies x = -\frac{b}{a}$ $\rightarrow$ Zero = $-\frac{b}{a}$

Question 5:

If $2$ is a zero of the polynomial $p(x) = 2x^2 - 3x + a$, find the value of $a$.

Solution:

Since $2$ is a zero of $p(x)$, $p(2) = 0$.

Substitute $x = 2$ into $p(x)$:

Question 6:

If $0$ and $1$ are zeroes of the polynomial $f(x) = 2x^3 - 3x^2 + ax + b$, find the values of $a$ and $b$.

Solution:

Since $0$ and $1$ are zeroes of $f(x)$, $f(0) = 0$ and $f(1) = 0$.

• Step 1: Using $f(0) = 0$:

  $$f(0) = 2(0)^3 - 3(0)^2 + a(0) + b = 0 \implies b = 0$$

• Step 2: Using $f(1) = 0$ and $b = 0$:

  $$f(1) = 2(1)^3 - 3(1)^2 + a(1) + 0 = 0$$
  $$2 - 3 + a = 0$$
  $$-1 + a = 0 \implies a = 1$$

Answer: $a = 1, \quad b = 0$