Math worksheet on Remainder and Factor Theorems with practice problems.

Here are the step-by-step solutions to the problems in your worksheet.

1. Using Remainder Theorem

The Remainder Theorem states that if you divide a polynomial $f(x)$ by $(x - c)$, the remainder is $f(c)$.
* If the divisor is $(x + a)$, set $x = -a$.
* If the divisor is $(ax - b)$, set $x = \frac{b}{a}$.
* If the remainder is $0$, then $g(x)$ is a factor of $f(x)$.

i. $f(x) = x^3 + 3x^2 + 3x + 1$; $g(x) = (x + 1)$
* Set $x + 1 = 0 \Rightarrow x = -1$.
* Calculate $f(-1)$:
$$(-1)^3 + 3(-1)^2 + 3(-1) + 1$$
$$= -1 + 3(1) - 3 + 1$$
$$= -1 + 3 - 3 + 1 = 0$$
* Remainder: $0$
* Is it a factor? Yes, because the remainder is $0$.

ii. $f(x) = x^3 - 4x^2 - 2x + 20$; $g(x) = (x + 2)$
* Set $x + 2 = 0 \Rightarrow x = -2$.
* Calculate $f(-2)$:
$$(-2)^3 - 4(-2)^2 - 2(-2) + 20$$
$$= -8 - 4(4) + 4 + 20$$
$$= -8 - 16 + 4 + 20$$
$$= -24 + 24 = 0$$
* Remainder: $0$
* Is it a factor? Yes, because the remainder is $0$.

iii. $f(x) = x^3 - 3x + 1$; $g(x) = (2 - 3x)$
* Set $2 - 3x = 0 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3}$.
* Calculate $f(\frac{2}{3})$:
$$(\frac{2}{3})^3 - 3(\frac{2}{3}) + 1$$
$$= \frac{8}{27} - 2 + 1$$
$$= \frac{8}{27} - 1$$
To subtract, find a common denominator ($27$):
$$= \frac{8}{27} - \frac{27}{27} = -\frac{19}{27}$$
* Remainder: $-\frac{19}{27}$
* Is it a factor? No, because the remainder is not $0$.

iv. $f(x) = 3x^3 + x^2 - 20x + 12$; $g(x) = (3x - 2)$
* Set $3x - 2 = 0 \Rightarrow 3x = 2 \Rightarrow x = \frac{2}{3}$.
* Calculate $f(\frac{2}{3})$:
$$3(\frac{2}{3})^3 + (\frac{2}{3})^2 - 20(\frac{2}{3}) + 12$$
$$= 3(\frac{8}{27}) + \frac{4}{9} - \frac{40}{3} + 12$$
Simplify the first term ($3/27 = 1/9$):
$$= \frac{8}{9} + \frac{4}{9} - \frac{40}{3} + 12$$
Combine the ninths:
$$= \frac{12}{9} - \frac{40}{3} + 12$$
Simplify $\frac{12}{9}$ to $\frac{4}{3}$:
$$= \frac{4}{3} - \frac{40}{3} + 12$$
$$= -\frac{36}{3} + 12$$
$$= -12 + 12 = 0$$
* Remainder: $0$
* Is it a factor? Yes, because the remainder is $0$.

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2. Finding value of $k$

Problem: For what value of $k$ is $2x^4 + 3x^3 + 2kx^2 + 3x + 6$ divisible by $(x + 2)$?

* "Divisible" means the remainder is $0$.
* Set divisor $x + 2 = 0 \Rightarrow x = -2$.
* Substitute $x = -2$ into the polynomial and set equal to $0$:
$$2(-2)^4 + 3(-2)^3 + 2k(-2)^2 + 3(-2) + 6 = 0$$
$$2(16) + 3(-8) + 2k(4) - 6 + 6 = 0$$
$$32 - 24 + 8k = 0$$
$$8 + 8k = 0$$
$$8k = -8$$
$$k = -1$$

Answer: $k = -1$

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3. Finding values for $a$ and $b$

Problem: $ax^2 + bx + 3$ leaves remainder $6$ when divided by $(x + 1)$ and remainder $9$ when divided by $(x + 2)$.

Step 1: Use the first condition
* Divisor $(x + 1) \Rightarrow x = -1$. Remainder is $6$.
$$a(-1)^2 + b(-1) + 3 = 6$$
$$a - b + 3 = 6$$
$$a - b = 3 \quad \text{--- (Equation 1)}$$

Step 2: Use the second condition
* Divisor $(x + 2) \Rightarrow x = -2$. Remainder is $9$.
$$a(-2)^2 + b(-2) + 3 = 9$$
$$4a - 2b + 3 = 9$$
$$4a - 2b = 6$$
Divide the whole equation by $2$ to simplify:
$$2a - b = 3 \quad \text{--- (Equation 2)}$$

Step 3: Solve the system
From Equation 1, we know $a = b + 3$. Substitute this into Equation 2:
$$2(b + 3) - b = 3$$
$$2b + 6 - b = 3$$
$$b + 6 = 3$$
$$b = -3$$

Now find $a$:
$$a = (-3) + 3$$
$$a = 0$$

Answer: $a = 0, b = -3$

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4. Factorize using Factor Theorem

i. $x^3 - 6x^2 + 11x - 6$

1. Find a factor: Test small integers like $1, -1, 2$.
Let $x = 1$: $1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$.
So, $(x - 1)$ is a factor.
2. Divide: Divide $(x^3 - 6x^2 + 11x - 6)$ by $(x - 1)$.
Result: $x^2 - 5x + 6$.
3. Factor the quadratic: Factor $x^2 - 5x + 6$.
We need numbers that multiply to $6$ and add to $-5$. Those are $-2$ and $-3$.
So, $(x - 2)(x - 3)$.
4. Combine:
$(x - 1)(x - 2)(x - 3)$

ii. $2x^3 + 3x^2 - 17x - 30$

1. Find a factor: Test factors of $30$ divided by factors of $2$ (e.g., $\pm 1, \pm 2, \pm 3, \dots$).
Let $x = -2$:
$2(-2)^3 + 3(-2)^2 - 17(-2) - 30$
$= 2(-8) + 3(4) + 34 - 30$
$= -16 + 12 + 34 - 30$
$= -4 + 4 = 0$.
So, $(x + 2)$ is a factor.
2. Divide: Divide $(2x^3 + 3x^2 - 17x - 30)$ by $(x + 2)$.
Result: $2x^2 - x - 15$.
3. Factor the quadratic: Factor $2x^2 - x - 15$.
Multiply $a \cdot c = 2 \cdot -15 = -30$. Find factors of $-30$ that add to $b (-1)$.
Factors are $-6$ and $5$.
Rewrite middle term: $2x^2 - 6x + 5x - 15$.
Group: $2x(x - 3) + 5(x - 3)$.
Result: $(2x + 5)(x - 3)$.
4. Combine:
$(x + 2)(2x + 5)(x - 3)$

Final Answer:
1.
i. Remainder: 0; Yes, it is a factor.
ii. Remainder: 0; Yes, it is a factor.
iii. Remainder: -19/27; No, it is not a factor.
iv. Remainder: 0; Yes, it is a factor.

2. k = -1

3. a = 0, b = -3

4.
i. (x - 1)(x - 2)(x - 3)
ii. (x + 2)(x - 3)(2x + 5)