06. Factorisation (Remainder & Factor Theorems) Previous Year Questions

x² − 4 = (x − 2)(x + 2).
x³ − x² − 4x + 4 = x²(x − 1) − 4(x − 1) = (x² − 4)(x − 1) = (x − 2)(x + 2)(x − 1).
Common factors are (x − 2) and (x + 2). From given options, (x − 2) is correct. [1 Mark]

(a) By Factor Theorem, P(2) = 0 ⇒ 2(2)³ − 7(2)² + 2k − 2 = 0 ⇒ 16 − 28 + 2k − 2 = 0 ⇒ 2k = 14 ⇒ k = 7. [1.5 Marks]
(b) Polynomial is 2x³ − 7x² + 7x − 2 = (x − 2)(2x² − 3x + 1).
Factorising quadratic: 2x² − 2x − x + 1 = 2x(x − 1) − 1(x − 1) = (2x − 1)(x − 1).
Completely factorised form: (x − 2)(x − 1)(2x − 1). [1.5 Marks]

Correct Option: (d) 31
By Remainder Theorem, remainder = f(3) = 3³ + 3² − 2(3) + 1 = 27 + 9 − 6 + 1 = 31. [1 Mark]

Since (x − 1) is a factor, f(1) = 0 ⇒ 3(1) + 8(1) − 15(1) + k = 0 ⇒ k = 4. [1 Mark]
Dividing 3x³ + 8x² − 15x + 4 by (x − 1) yields quotient 3x² + 11x − 4. [1 Mark]
3x² + 11x − 4 = (3x − 1)(x + 4).
Complete factorisation: (x − 1)(3x − 1)(x + 4). [1 Mark]

Step 1 By Factor Theorem, if $(x - 2)$ is a factor of $P(x) = x^3 - kx - 12$, then $P(2) = 0$.
Step 2 Substitute $x = 2$:
$2^3 - k(2) - 12 = 0 \implies 8 - 2k - 12 = 0 \implies -2k - 4 = 0 \implies 2k = -4 \implies k = -2$.
Correct Option: (c) [1 Mark]

Step 1 Let $f(x) = 3x^3 + x^2 - ax - 81$. Since $(x - a)$ is a factor, by the Factor Theorem, $f(a) = 0$.1 Mark
Step 2 Substitute $x = a$:
$3(a)^3 + (a)^2 - a(a) - 81 = 0$
$3a^3 + a^2 - a^2 - 81 = 0$1 Mark
Step 3 Simplify algebraic terms:
$3a^3 - 81 = 0 \implies 3a^3 = 81$1 Mark
$a^3 = 27 \implies a = \sqrt[3]{27} = 3$.
Final Answer: $a = 3$.1 Mark

Let $P(x) = 2x^3 - x^2 - 13x - 6$. Factors of the constant term $-6$ are $\pm 1, \pm 2, \pm 3, \pm 6$.

Step 1: Test x = -2
$P(-2) = 2(-2)^3 - (-2)^2 - 13(-2) - 6 = 2(-8) - 4 + 26 - 6 = -16 - 4 + 26 - 6 = 0$.
By Factor Theorem, $(x + 2)$ is a factor of $P(x)$.1.5 Marks

Step 2: Polynomial Division
Dividing $2x^3 - x^2 - 13x - 6$ by $(x + 2)$:
$2x^3 - x^2 - 13x - 6 = (x + 2)(2x^2 - 5x - 3)$.1.5 Marks

Step 3: Factorize Quadratic Quotient
$2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (x - 3)(2x + 1)$.
Complete Factorisation: $(x + 2)(x - 3)(2x + 1)$.1 Mark

Step 1 Let $P(x) = 6x^3 + 17x^2 + 4x - 12$.
Test integer factors of $-12$: for $x = -2$:
$P(-2) = 6(-2)^3 + 17(-2)^2 + 4(-2) - 12 = 6(-8) + 17(4) - 8 - 12 = -48 + 68 - 20 = 0$.
Therefore, by Factor Theorem, $(x + 2)$ is a factor of $P(x)$. [1 Mark]
Step 2 Divide $P(x)$ by $(x + 2)$:
$6x^3 + 17x^2 + 4x - 12 = (x + 2)(6x^2 + 5x - 6)$. [1 Mark]
Step 3 Factorising the quotient quadratic $6x^2 + 5x - 6$:
Product $= 6 \times (-6) = -36$, Sum $= 5 \implies +9$ and $-4$.
$6x^2 + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = (2x + 3)(3x - 2)$.
Complete Factorisation: $(x + 2)(2x + 3)(3x - 2)$. [1 Mark]

Step 1 Let $k$ be added to the polynomial. Then $P(x) = 2x^3 - 3x^2 - 8x + k$.
By Remainder Theorem, when $P(x)$ is divided by $2x + 1$, remainder is $P\left(-\frac{1}{2}\right) = 10$. [1 Mark]
Step 2 Substitute $x = -\frac{1}{2}$ into $P(x)$:
$$P\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) - 8\left(-\frac{1}{2}\right) + k = -\frac{1}{4} - \frac{3}{4} + 4 + k = -1 + 4 + k = 3 + k$$ [1 Mark]
Step 3 Equating remainder to 10:
$$3 + k = 10 \implies \mathbf{k = 7}$$
Answer: 7 must be added. [1 Mark]

Step 1 Let $f(x) = x^3 + x^2 - 4x - 4$.
To test whether $(x - 2)$ is a factor, evaluate $f(2)$:
$f(2) = (2)^3 + (2)^2 - 4(2) - 4 = 8 + 4 - 8 - 4 = 0$.
Since $f(2) = 0$, by the Factor Theorem, $(x - 2)$ is a factor of $f(x)$. [1 Mark]
Step 2 Factorise completely by grouping or dividing by $(x - 2)$:
$x^3 + x^2 - 4x - 4 = x^2(x + 1) - 4(x + 1) = (x + 1)(x^2 - 4)$. [1 Mark]
Step 3 Apply difference of squares $a^2 - b^2 = (a - b)(a + b)$:
$x^2 - 4 = (x - 2)(x + 2)$.
$\mathbf{\therefore x^3 + x^2 - 4x - 4 = (x - 2)(x + 2)(x + 1)}$. [1 Mark]

Step 1 Let $f(x) = x^3 + (kx + 8)x + k = x^3 + kx^2 + 8x + k$.
By the Remainder Theorem:
When divided by $(x + 1)$, remainder $R_1 = f(-1)$:
$R_1 = (-1)^3 + k(-1)^2 + 8(-1) + k = -1 + k - 8 + k = \mathbf{2k - 9}$. [1 Mark]
Step 2 When divided by $(x - 2)$, remainder $R_2 = f(2)$:
$R_2 = (2)^3 + k(2)^2 + 8(2) + k = 8 + 4k + 16 + k = \mathbf{5k + 24}$. [1 Mark]
Step 3 Use condition $R_1 + R_2 = 1$:
$(2k - 9) + (5k + 24) = 1$
$7k + 15 = 1 \implies 7k = 1 - 15 = -14 \implies \mathbf{k = -2}$.
Answer: $k = -2$. [1 Mark]

Solution By Factor Theorem, if $(x - 1)$ is a factor, then $f(1) = 0$.
$f(1) = (1)^3 + 2(1)^2 - (1) - k = 0 \implies 1 + 2 - 1 - k = 0 \implies \mathbf{k = 2}$.
Correct Answer: (d) 2 [1 Mark]

Step 1 (a) Apply Remainder Theorem for both polynomials:
Let $f(x) = mx^2 - 3x + 6$. Dividing by $(x - 2)$ gives remainder $R_1 = f(2)$:
$R_1 = m(2)^2 - 3(2) + 6 = 4m - 6 + 6 = 4m$. [1 Mark]
Let $g(x) = 3x^2 + 5x - m$. Dividing by $(x + 3)$ gives remainder $R_2 = g(-3)$:
$R_2 = 3(-3)^2 + 5(-3) - m = 27 - 15 - m = 12 - m$. [1 Mark]
Step 2 Given condition $R_1 = 2 R_2$:
$4m = 2(12 - m) \implies 4m = 24 - 2m \implies 6m = 24 \implies \mathbf{m = 4}$. [1 Mark]
Step 3 (b) Calculate remainder $R_1$ using $m = 4$:
$\mathbf{\text{Remainder } R_1 = 4m = 4(4) = 16}$. [1 Mark]

Step 1 Let $f(x) = x^3 + ax + b$.
Since $(x + 2)$ is a factor, $f(-2) = 0$:
$$(-2)^3 + a(-2) + b = 0 \implies -8 - 2a + b = 0 \implies -2a + b = 8 \quad \text{--- (1)}$$ [1 Mark]
Step 2 Since $(x + 3)$ is a factor, $f(-3) = 0$:
$$(-3)^3 + a(-3) + b = 0 \implies -27 - 3a + b = 0 \implies -3a + b = 27 \quad \text{--- (2)}$$ [1 Mark]
Step 3 Subtract (2) from (1):
$$(-2a + b) - (-3a + b) = 8 - 27 \implies a = -19$$ Substitute $a = -19$ into (1):
$$-2(-19) + b = 8 \implies 38 + b = 8 \implies b = 8 - 38 = -30$$ [1 Mark]
Answer: $a = -19$, $b = -30$.

Step 1 Let $f(x) = 2x^3 + 3x^2 - 9x - 10$.
Testing $x = -1$:
$$f(-1) = 2(-1)^3 + 3(-1)^2 - 9(-1) - 10 = -2 + 3 + 9 - 10 = 0$$ $\therefore (x + 1)$ is a factor of $f(x)$ by Factor Theorem. [1 Mark]
Step 2 Divide $f(x)$ by $(x + 1)$:
$$2x^3 + 3x^2 - 9x - 10 = (x + 1)(2x^2 + x - 10)$$ [1 Mark]
Step 3 Factorise the quadratic expression $2x^2 + x - 10$ ($2 \times -10 = -20$, factors $5, -4$):
$$2x^2 + 5x - 4x - 10 = x(2x + 5) - 2(2x + 5) = (2x + 5)(x - 2)$$ $$\mathbf{(x + 1)(x - 2)(2x + 5)}$$ [1 Mark]
Answer: $(x + 1)(x - 2)(2x + 5)$.

Step 1 Let $k$ be subtracted. The resulting polynomial is $f(x) = 16x^3 - 8x^2 + 4x + 7 - k$.
By Factor Theorem, for $(2x + 1)$ to be a factor, $f\left(-\frac{1}{2}\right) = 0$. [1 Mark]
Step 2 Substitute $x = -\frac{1}{2}$:
$$16\left(-\frac{1}{2}\right)^3 - 8\left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right) + 7 - k = 0$$ $$16\left(-\frac{1}{8}\right) - 8\left(\frac{1}{4}\right) - 2 + 7 - k = 0$$ [1 Mark]
Step 3 Simplify:
$$-2 - 2 - 2 + 7 - k = 0 \implies 1 - k = 0 \implies \mathbf{k = 1}$$ [1 Mark]
Answer: 1 must be subtracted.