Chapter 2: Polynomials and Factorisation (Exercise 2.3)

Chapter 2: Polynomials and Factorisation (Exercise 2.3)

Question 1:

Find the remainder when $x^3 + 3x^2 + 3x + 1$ is divided by the following linear polynomials:

1. $x + 1$

2. $x - \frac{1}{2}$

3. $x$

4. $x + \pi$

5. $5 + 2x$

Solution:

Let $p(x) = x^3 + 3x^2 + 3x + 1$

1. Divided by $x + 1$:

   • Set $x + 1 = 0 \implies x = -1$

   • By Remainder Theorem, $\text{Remainder} = p(-1)$:

     $$p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 = -1 + 3(1) - 3 + 1 = 0$$

   • Remainder = $0$

2. Divided by $x - \frac{1}{2}$:

   • Set $x - \frac{1}{2} = 0 \implies x = \frac{1}{2}$

   • By Remainder Theorem, $\text{Remainder} = p\left(\frac{1}{2}\right)$:

     $$p\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) + 1 = \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1$$
     $$= \frac{1 + 6 + 12 + 8}{8} = \frac{27}{8}$$

   • Remainder = $\frac{27}{8}$

3. Divided by $x$:

   • Set $x = 0$

   • By Remainder Theorem, $\text{Remainder} = p(0)$:

     $$p(0) = (0)^3 + 3(0)^2 + 3(0) + 1 = 1$$

   • Remainder = $1$

4. Divided by $x + \pi$:

   • Set $x + \pi = 0 \implies x = -\pi$

   • By Remainder Theorem, $\text{Remainder} = p(-\pi)$:

     $$p(-\pi) = (-\pi)^3 + 3(-\pi)^2 + 3(-\pi) + 1 = -\pi^3 + 3\pi^2 - 3\pi + 1$$

   • Remainder = $-\pi^3 + 3\pi^2 - 3\pi + 1$

5. Divided by $5 + 2x$:

   • Set $5 + 2x = 0 \implies 2x = -5 \implies x = -\frac{5}{2}$

   • By Remainder Theorem, $\text{Remainder} = p\left(-\frac{5}{2}\right)$:

     $$p\left(-\frac{5}{2}\right) = \left(-\frac{5}{2}\right)^3 + 3\left(-\frac{5}{2}\right)^2 + 3\left(-\frac{5}{2}\right) + 1 = -\frac{125}{8} + \frac{75}{4} - \frac{15}{2} + 1$$
     $$= \frac{-125 + 150 - 60 + 8}{8} = -\frac{27}{8}$$

   • Remainder = $-\frac{27}{8}$

Question 2:

Find the remainder when $x^3 - px^2 + 6x - p$ is divided by $x - p$.

Solution:

Let $p(x) = x^3 - px^2 + 6x - p$.

Set $x - p = 0 \implies x = p$.

By Remainder Theorem:

Remainder = $5p$

Question 3:

Find the remainder when $2x^2 - 3x + 5$ is divided by $2x - 3$. Does it exactly divide the polynomial? State reason.

Solution:

Let $p(x) = 2x^2 - 3x + 5$.

Set $2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$.

By Remainder Theorem:

Reason: Since the remainder is $5$ ($\neq 0$), $2x - 3$ does not exactly divide the polynomial $2x^2 - 3x + 5$.

Question 4:

Find the remainder when $9x^3 - 3x^2 + x - 5$ is divided by $x - \frac{2}{3}$.

Solution:

Let $p(x) = 9x^3 - 3x^2 + x - 5$.

Set $x - \frac{2}{3} = 0 \implies x = \frac{2}{3}$.

By Remainder Theorem:

Remainder = $-3$

Question 5:

If the polynomials $2x^3 + ax^2 + 3x - 5$ and $x^3 + x^2 - 4x + a$ leave the same remainder when divided by $x - 2$, find the value of $a$.

Solution:

Let $f(x) = 2x^3 + ax^2 + 3x - 5$ and $g(x) = x^3 + x^2 - 4x + a$.

Set $x - 2 = 0 \implies x = 2$.

• Remainder 1 ($R_1$):

  $$R_1 = f(2) = 2(2)^3 + a(2)^2 + 3(2) - 5 = 2(8) + 4a + 6 - 5 = 16 + 4a + 1 = 4a + 17$$

• Remainder 2 ($R_2$):

  $$R_2 = g(2) = (2)^3 + (2)^2 - 4(2) + a = 8 + 4 - 8 + a = a + 4$$

Given that $R_1 = R_2$:

Question 6:

If the polynomials $x^3 + ax^2 + 5$ and $x^3 - 2x^2 + a$ are divided by $(x + 2)$ leave the same remainder, find the value of $a$.

Solution:

Let $f(x) = x^3 + ax^2 + 5$ and $g(x) = x^3 - 2x^2 + a$.

Set $x + 2 = 0 \implies x = -2$.

• Remainder 1 ($R_1$):

  $$R_1 = f(-2) = (-2)^3 + a(-2)^2 + 5 = -8 + 4a + 5 = 4a - 3$$

• Remainder 2 ($R_2$):

  $$R_2 = g(-2) = (-2)^3 - 2(-2)^2 + a = -8 - 2(4) + a = -16 + a$$

Given that $R_1 = R_2$:

Question 7:

Find the remainder when $f(x) = x^4 - 3x^2 + 4$ is divided by $g(x) = x - 2$ and verify the result by actual division.

Solution:

• By Remainder Theorem:

  Set $x - 2 = 0 \implies x = 2$.

  $$\text{Remainder} = f(2) = (2)^4 - 3(2)^2 + 4 = 16 - 12 + 4 = 8$$

• Verification by Actual Long Division:

  Dividing $x^4 + 0x^3 - 3x^2 + 0x + 4$ by $x - 2$:

  • First term of quotient = $\frac{x^4}{x} = x^3 \implies x^3(x-2) = x^4 - 2x^3$. Subtracting gives $2x^3 - 3x^2$.

  • Second term of quotient = $\frac{2x^3}{x} = 2x^2 \implies 2x^2(x-2) = 2x^3 - 4x^2$. Subtracting gives $x^2 + 0x$.

  • Third term of quotient = $\frac{x^2}{x} = x \implies x(x-2) = x^2 - 2x$. Subtracting gives $2x + 4$.

  • Fourth term of quotient = $\frac{2x}{x} = 2 \implies 2(x-2) = 2x - 4$. Subtracting gives $4 - (-4) = 8$.

(Hence verified.)

Question 8:

Find the remainder when $p(x) = x^3 - 6x^2 + 14x - 3$ is divided by $g(x) = 1 - 2x$ and verify the result by long division.

Solution:

• By Remainder Theorem:

  Set $1 - 2x = 0 \implies 2x = 1 \implies x = \frac{1}{2}$.

  $$\text{Remainder} = p\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 - 6\left(\frac{1}{2}\right)^2 + 14\left(\frac{1}{2}\right) - 3$$
  $$= \frac{1}{8} - \frac{6}{4} + 7 - 3 = \frac{1}{8} - \frac{3}{2} + 4 = \frac{1 - 12 + 32}{8} = \frac{21}{8}$$

• Verification by Long Division:

  Dividing $x^3 - 6x^2 + 14x - 3$ by $-2x + 1$:

  $$\text{Quotient} = -\frac{1}{2}x^2 + \frac{11}{4}x - \frac{45}{8}, \quad \text{Remainder} = \frac{21}{8}$$

  (Hence verified.)

Question 9:

When a polynomial $2x^3 + 3x^2 + ax + b$ is divided by $(x - 2)$ leaves remainder $2$, and $(x + 2)$ leaves remainder $-2$. Find $a$ and $b$.

Solution:

Let $p(x) = 2x^3 + 3x^2 + ax + b$.

• Condition 1: When divided by $x - 2$, remainder is $2 \implies p(2) = 2$:

  $$p(2) = 2(2)^3 + 3(2)^2 + a(2) + b = 2$$
  $$16 + 12 + 2a + b = 2$$
  $$28 + 2a + b = 2 \implies 2a + b = -26 \quad \text{--- (Equation 1)}$$

• Condition 2: When divided by $x + 2$, remainder is $-2 \implies p(-2) = -2$:

  $$p(-2) = 2(-2)^3 + 3(-2)^2 + a(-2) + b = -2$$
  $$-16 + 12 - 2a + b = -2$$
  $$-4 - 2a + b = -2 \implies -2a + b = 2 \quad \text{--- (Equation 2)}$$

• Solving Equations:

  Adding Equation 1 and Equation 2:

  $$(2a + b) + (-2a + b) = -26 + 2$$
  $$2b = -24 \implies b = -12$$

  Substituting $b = -12$ into Equation 1:

  $$2a + (-12) = -26$$
  $$2a = -14 \implies a = -7$$

Answer: $a = -7, \quad b = -12$