Chapter 2: Polynomials and Factorisation (Exercise 2.4)

Chapter 2: Polynomials and Factorisation (Exercise 2.4)

Question 1:

Determine which of the following polynomials has $(x + 1)$ as a factor:

1. $x^3 + x^2 + x + 1$

2. $x^4 + x^3 + x^2 + x + 1$

3. $x^4 + 3x^3 + 3x^2 + x + 1$

4. $x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$

Solution:

By Factor Theorem, $(x + 1)$ is a factor of $p(x)$ if $p(-1) = 0$.

1. $p(x) = x^3 + x^2 + x + 1$:

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

   Since $p(-1) = 0$, $(x + 1)$ is a factor.

2. $p(x) = x^4 + x^3 + x^2 + x + 1$:

   $$p(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1 - 1 + 1 - 1 + 1 = 1 \neq 0$$

   Since $p(-1) \neq 0$, $(x + 1)$ is not a factor.

3. $p(x) = x^4 + 3x^3 + 3x^2 + x + 1$:

   $$p(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1 = 1 - 3 + 3 - 1 + 1 = 1 \neq 0$$

   Since $p(-1) \neq 0$, $(x + 1)$ is not a factor.

4. $p(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$:

   $$p(-1) = (-1)^3 - (-1)^2 - (2 + \sqrt{2})(-1) + \sqrt{2}$$
   $$= -1 - 1 + (2 + \sqrt{2}) + \sqrt{2} = -2 + 2 + \sqrt{2} + \sqrt{2} = 2\sqrt{2} \neq 0$$

   Since $p(-1) \neq 0$, $(x + 1)$ is not a factor.

Question 2:

Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each case:

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

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

3. $p(x) = x^3 - 4x^2 + x + 6, \quad g(x) = x - 3$

Solution:

1. For $g(x) = x + 1 \implies x = -1$:

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

   Since $p(-1) = 0$, $g(x)$ is a factor of $p(x)$.

2. For $g(x) = x + 2 \implies x = -2$:

   $$p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1 = -8 + 12 - 6 + 1 = -1 \neq 0$$

   Since $p(-2) \neq 0$, $g(x)$ is not a factor of $p(x)$.

3. For $g(x) = x - 3 \implies x = 3$:

   $$p(3) = (3)^3 - 4(3)^2 + 3 + 6 = 27 - 36 + 3 + 6 = 0$$

   Since $p(3) = 0$, $g(x)$ is a factor of $p(x)$.

Question 3:

Find the value of $k$, if $x - 1$ is a factor of $p(x)$ in each case:

1. $p(x) = x^2 + x + k$

2. $p(x) = 2x^2 + kx + \sqrt{2}$

3. $p(x) = kx^2 - \sqrt{2}x + 1$

4. $p(x) = kx^2 - 3x + k$

Solution:

If $(x - 1)$ is a factor of $p(x)$, then $p(1) = 0$.

1. $p(1) = (1)^2 + 1 + k = 0 \implies 2 + k = 0 \implies k = -2$

2. $p(1) = 2(1)^2 + k(1) + \sqrt{2} = 0 \implies 2 + k + \sqrt{2} = 0 \implies k = -(2 + \sqrt{2})$

3. $p(1) = k(1)^2 - \sqrt{2}(1) + 1 = 0 \implies k - \sqrt{2} + 1 = 0 \implies k = \sqrt{2} - 1$

4. $p(1) = k(1)^2 - 3(1) + k = 0 \implies k - 3 + k = 0 \implies 2k = 3 \implies k = \frac{3}{2}$

Question 4:

Factorise the following quadratic polynomials:

1. $12x^2 - 7x + 1$

2. $2x^2 + 7x + 3$

3. $6x^2 + 5x - 6$

4. $3x^2 - x - 4$

Solution:

1. $12x^2 - 7x + 1$:

   • Product = $12 \times 1 = 12$, Sum = $-7$. Factors are $-4$ and $-3$.

     $$= 12x^2 - 4x - 3x + 1 = 4x(3x - 1) - 1(3x - 1) = (3x - 1)(4x - 1)$$

2. $2x^2 + 7x + 3$:

   • Product = $2 \times 3 = 6$, Sum = $7$. Factors are $6$ and $1$.

     $$= 2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)$$

3. $6x^2 + 5x - 6$:

   • Product = $6 \times (-6) = -36$, Sum = $5$. Factors are $9$ and $-4$.

     $$= 6x^2 + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = (2x + 3)(3x - 2)$$

4. $3x^2 - x - 4$:

   • Product = $3 \times (-4) = -12$, Sum = $-1$. Factors are $-4$ and $3$.

     $$= 3x^2 - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (3x - 4)(x + 1)$$

Question 5:

Factorise the following cubic polynomials:

1. $x^3 - 2x^2 - x + 2$

2. $x^3 - 3x^2 - 9x - 5$

3. $x^3 + 13x^2 + 32x + 20$

4. $2y^3 + y^2 - 2y - 1$

Solution:

1. $x^3 - 2x^2 - x + 2$:

   Grouping terms in pairs:

   $$= x^2(x - 2) - 1(x - 2) = (x - 2)(x^2 - 1) = (x - 2)(x - 1)(x + 1)$$

2. $x^3 - 3x^2 - 9x - 5$:

   • Factors of constant term $-5$ are $\pm 1, \pm 5$.

   • Test $x = -1$: $p(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 = -1 - 3 + 9 - 5 = 0$. So, $(x + 1)$ is a factor.

   • Divide $x^3 - 3x^2 - 9x - 5$ by $(x + 1)$:

     $$x^3 - 3x^2 - 9x - 5 = (x + 1)(x^2 - 4x - 5)$$

   • Factorise $x^2 - 4x - 5 = x^2 - 5x + x - 5 = x(x - 5) + 1(x - 5) = (x - 5)(x + 1)$.

   • Final Answer: $(x + 1)(x + 1)(x - 5)$

3. $x^3 + 13x^2 + 32x + 20$:

   • Test $x = -1$: $p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20 = -1 + 13 - 32 + 20 = 0$. So, $(x + 1)$ is a factor.

   • Divide $x^3 + 13x^2 + 32x + 20$ by $(x + 1)$:

     $$x^3 + 13x^2 + 32x + 20 = (x + 1)(x^2 + 12x + 20)$$

   • Factorise $x^2 + 12x + 20 = x^2 + 10x + 2x + 20 = (x + 10)(x + 2)$.

   • Final Answer: $(x + 1)(x + 2)(x + 10)$

4. $2y^3 + y^2 - 2y - 1$:

   Grouping terms in pairs:

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