Master Factorisation for Rbse Solutions for Class 9 Maths Chapter 2 Exercise 2.3. Learn the Factor Theorem, splitting the middle term for quadratics, and factorising cubic polynomials. Essential guide for CBSE/RBSE students.
1. Determine which of the following polynomials has $(x + 1)$ a factor:
According to the Factor Theorem, $(x+1)$ is a factor of $P(x)$ if and only if $P(-1) = 0$.
(i) $P(x) = x^3 + x^2 + x + 1$
$$P(-1) = (-1)^3 + (-1)^2 + (-1) + 1$$
$$P(-1) = -1 + 1 – 1 + 1 = 0$$
Result: Yes, $(x+1)$ is a factor.
(ii) $P(x) = x^4 + x^3 + x^2 + x + 1$
$$P(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1$$
$$P(-1) = 1 – 1 + 1 – 1 + 1 = 1$$
Result: No, $(x+1)$ is not a factor (since $P(-1) \neq 0$).
(iii) $P(x) = x^4 + 3x^3 + 3x^2 + x + 1$
$$P(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1$$
$$P(-1) = 1 + 3(-1) + 3(1) – 1 + 1$$
$$P(-1) = 1 – 3 + 3 – 1 + 1 = 1$$
Result: No, $(x+1)$ is not a factor.
(iv) $P(x) = x^3 – x^2 – (2 + \sqrt{2})x + \sqrt{2}$
$$P(-1) = (-1)^3 – (-1)^2 – (2 + \sqrt{2})(-1) + \sqrt{2}$$
$$P(-1) = -1 – 1 + (2 + \sqrt{2}) + \sqrt{2}$$
$$P(-1) = -2 + 2 + \sqrt{2} + \sqrt{2}$$
$$P(-1) = 2\sqrt{2}$$
Result: No, $(x+1)$ is not a factor.
2. Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases:
(i) $p(x) = 2x^3 + x^2 – 2x – 1, \quad g(x) = x + 1$
The zero of $g(x)$ is $x = -1$.
$$P(-1) = 2(-1)^3 + (-1)^2 – 2(-1) – 1$$
$$P(-1) = 2(-1) + 1 + 2 – 1$$
$$P(-1) = -2 + 1 + 2 – 1 = 0$$
Result: Yes, $g(x)$ is a factor of $p(x)$.
(ii) $p(x) = x^3 + 3x^2 + 3x + 1, \quad g(x) = x + 2$
The zero of $g(x)$ is $x = -2$.
$$P(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1$$
$$P(-2) = -8 + 3(4) – 6 + 1$$
$$P(-2) = -8 + 12 – 6 + 1$$
$$P(-2) = 13 – 14 = -1$$
Result: No, $g(x)$ is not a factor of $p(x)$.
(iii) $p(x) = x^3 – 4x^2 + x + 6, \quad g(x) = x – 3$
The zero of $g(x)$ is $x = 3$.
$$P(3) = (3)^3 – 4(3)^2 + 3 + 6$$
$$P(3) = 27 – 4(9) + 9$$
$$P(3) = 27 – 36 + 9$$
$$P(3) = 36 – 36 = 0$$
Result: Yes, $g(x)$ is a factor of $p(x)$.
3. Find the value of $k$, if $(x – 1)$ is a factor of $p(x)$ in each of the following cases:
Since $(x-1)$ is a factor, we must have $P(1) = 0$.
(i) $p(x) = x^2 + x + k$
$$P(1) = (1)^2 + (1) + k = 0$$
$$1 + 1 + k = 0$$
$$2 + k = 0$$
$$\mathbf{k = -2}$$
(ii) $p(x) = 2x^2 + kx + \sqrt{2}$
$$P(1) = 2(1)^2 + k(1) + \sqrt{2} = 0$$
$$2 + k + \sqrt{2} = 0$$
$$\mathbf{k = -2 – \sqrt{2}}$$
(iii) $p(x) = kx^2 – \sqrt{2}x + 1$
$$P(1) = k(1)^2 – \sqrt{2}(1) + 1 = 0$$
$$k – \sqrt{2} + 1 = 0$$
$$\mathbf{k = \sqrt{2} – 1}$$
(iv) $p(x) = kx^2 – 3x + k$
$$P(1) = k(1)^2 – 3(1) + k = 0$$
$$k – 3 + k = 0$$
$$2k – 3 = 0$$
$$2k = 3$$
$$\mathbf{k = \frac{3}{2}}$$
4. Factorise (Splitting the Middle Term):
We look for two numbers whose product is $a \cdot c$ and whose sum is $b$.
(i) $12x^2 – 7x + 1$
Here, $a=12, b=-7, c=1$. Product $a \cdot c = 12$. Sum $= -7$.
Numbers: $-3$ and $-4$.
$$12x^2 – 3x – 4x + 1$$
$$3x(4x – 1) – 1(4x – 1)$$
$$\mathbf{(4x – 1)(3x – 1)}$$
(ii) $2x^2 + 7x + 3$
Here, $a=2, b=7, c=3$. Product $a \cdot c = 6$. Sum $= 7$.
Numbers: $6$ and $1$.
$$2x^2 + 6x + x + 3$$
$$2x(x + 3) + 1(x + 3)$$
$$\mathbf{(x + 3)(2x + 1)}$$
(iii) $6x^2 + 5x – 6$
Here, $a=6, b=5, c=-6$. Product $a \cdot c = -36$. Sum $= 5$.
Numbers: $9$ and $-4$.
$$6x^2 + 9x – 4x – 6$$
$$3x(2x + 3) – 2(2x + 3)$$
$$\mathbf{(2x + 3)(3x – 2)}$$
(iv) $3x^2 – x – 4$
Here, $a=3, b=-1, c=-4$. Product $a \cdot c = -12$. Sum $= -1$.
Numbers: $-4$ and $3$.
$$3x^2 – 4x + 3x – 4$$
$$x(3x – 4) + 1(3x – 4)$$
$$\mathbf{(3x – 4)(x + 1)}$$
5. Factorise (Factor Theorem and Grouping):
For cubic polynomials, we first use the Factor Theorem to find one root, and then use polynomial division or grouping to find the remaining quadratic factor.
(i) $x^3 – 2x^2 – x + 2$
By grouping:
$$x^2(x – 2) – 1(x – 2)$$
$$(x – 2)(x^2 – 1)$$
Using $a^2 – b^2 = (a-b)(a+b)$ on $x^2 – 1$:
$$\mathbf{(x – 2)(x – 1)(x + 1)}$$
(ii) $x^3 – 3x^2 – 9x – 5$
The possible factors are divisors of $-5$: $\pm 1, \pm 5$.
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 $P(x)$ by $(x+1)$ or use coefficient comparison to get the quadratic factor $x^2 – 4x – 5$.
Factorise $x^2 – 4x – 5$: $(x-5)(x+1)$.
$$\mathbf{(x + 1)(x + 1)(x – 5) \quad \text{or} \quad (x + 1)^2 (x – 5)}$$
(iii) $x^3 + 13x^2 + 32x + 20$
Possible factors are divisors of $20$: $\pm 1, \pm 2, \pm 4, \dots$
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 $P(x)$ by $(x+1)$ to get the quadratic factor $x^2 + 12x + 20$.
Factorise $x^2 + 12x + 20$ (product 20, sum 12): $(x+10)(x+2)$.
$$\mathbf{(x + 1)(x + 2)(x + 10)}$$
(iv) $2y^3 + y^2 – 2y – 1$
By grouping:
$$y^2(2y + 1) – 1(2y + 1)$$
$$(2y + 1)(y^2 – 1)$$
Using $a^2 – b^2 = (a-b)(a+b)$ on $y^2 – 1$:
$$\mathbf{(2y + 1)(y – 1)(y + 1)}$$