Math worksheet on The Remainder Theorem with seven problems for students to solve.

Here are the step-by-step solutions for each problem on the worksheet using the Remainder Theorem.

The Rule: To find the remainder when a polynomial is divided by $(x - c)$, you simply plug in the value $c$ into the polynomial.
* If dividing by $(x + 1)$, plug in $x = -1$.
* If dividing by $(x - 4)$, plug in $x = 4$.
* If dividing by $(2x + 3)$, solve $2x + 3 = 0$ to get $x = -3/2$, and plug that in.

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1) Find the remainder when $x^3 - 2x^2 - x - 2$ is divided by $x + 1$.

* Step 1: Set the divisor to zero: $x + 1 = 0 \rightarrow x = -1$.
* Step 2: Substitute $-1$ into the expression:
$$(-1)^3 - 2(-1)^2 - (-1) - 2$$
* Step 3: Calculate powers:
$$-1 - 2(1) + 1 - 2$$
* Step 4: Simplify:
$$-1 - 2 + 1 - 2 = -4$$

2) If $f(x) = x^3 + 3x - 4$. Find the remainder when $f(x)$ is divided by $x - 4$.

* Step 1: Set the divisor to zero: $x - 4 = 0 \rightarrow x = 4$.
* Step 2: Substitute $4$ into $f(x)$:
$$f(4) = (4)^3 + 3(4) - 4$$
* Step 3: Calculate:
$$64 + 12 - 4$$
* Step 4: Add/Subtract:
$$76 - 4 = 72$$

3) Find the remainder when $x^3 + 3x - 4$ is divided by $x + 1$.

* Step 1: Set the divisor to zero: $x + 1 = 0 \rightarrow x = -1$.
* Step 2: Substitute $-1$ into the expression:
$$(-1)^3 + 3(-1) - 4$$
* Step 3: Calculate:
$$-1 - 3 - 4$$
* Step 4: Add:
$$-8$$

4) Given that $f(x) = 6x^3 - 3x^2 - 17x + 7$, divide $f(x)$ by $2x + 3$.

* Step 1: Set the divisor to zero: $2x + 3 = 0 \rightarrow 2x = -3 \rightarrow x = -\frac{3}{2}$.
* Step 2: Substitute $-\frac{3}{2}$ into $f(x)$:
$$6(-\frac{3}{2})^3 - 3(-\frac{3}{2})^2 - 17(-\frac{3}{2}) + 7$$
* Step 3: Calculate powers:
$$6(-\frac{27}{8}) - 3(\frac{9}{4}) + \frac{51}{2} + 7$$
* Step 4: Multiply fractions:
$$-\frac{162}{8} - \frac{27}{4} + \frac{51}{2} + 7$$
Simplify $-\frac{162}{8}$ to $-\frac{81}{4}$:
$$-\frac{81}{4} - \frac{27}{4} + \frac{51}{2} + 7$$
* Step 5: Combine terms with denominator 4:
$$-\frac{108}{4} + \frac{51}{2} + 7$$
$$-27 + 25.5 + 7$$
* Step 6: Final calculation:
$$-27 + 32.5 = 5.5$$ (or $\frac{11}{2}$)

5) Find the remainder when $6x^3 + 27x^2 - 14x + 15$ is divided by $x + 5$.

* Step 1: Set the divisor to zero: $x + 5 = 0 \rightarrow x = -5$.
* Step 2: Substitute $-5$ into the expression:
$$6(-5)^3 + 27(-5)^2 - 14(-5) + 15$$
* Step 3: Calculate powers:
$$6(-125) + 27(25) + 70 + 15$$
* Step 4: Multiply:
$$-750 + 675 + 70 + 15$$
* Step 5: Add positive numbers first ($675 + 70 + 15 = 760$):
$$-750 + 760 = 10$$

6) When divided by $(x + 1)$ and $(x + 2)$, the expression $ax^2 + bx + 3$ leaves remainders 6 and 9 respectively. Find the values for $a$ and $b$.

We have two conditions to create a system of equations.

Condition 1: Divided by $(x + 1)$, remainder is 6.
* Plug in $x = -1$:
$$a(-1)^2 + b(-1) + 3 = 6$$
$$a - b + 3 = 6$$
$$a - b = 3$$ (Equation 1)

Condition 2: Divided by $(x + 2)$, remainder is 9.
* Plug in $x = -2$:
$$a(-2)^2 + b(-2) + 3 = 9$$
$$4a - 2b + 3 = 9$$
$$4a - 2b = 6$$
Divide by 2 to simplify:
$$2a - b = 3$$ (Equation 2)

Solving the System:
Subtract Equation 1 from Equation 2:
$$(2a - b) - (a - b) = 3 - 3$$
$$2a - b - a + b = 0$$
$$a = 0$$

Now substitute $a = 0$ back into Equation 1:
$$0 - b = 3$$
$$b = -3$$

So, $a = 0$ and $b = -3$.

7) Find the remainder when $x^3 + 3x^2 - 5x - 6$ is divided by $x + 2$.

* Step 1: Set the divisor to zero: $x + 2 = 0 \rightarrow x = -2$.
* Step 2: Substitute $-2$ into the expression:
$$(-2)^3 + 3(-2)^2 - 5(-2) - 6$$
* Step 3: Calculate powers:
$$-8 + 3(4) + 10 - 6$$
* Step 4: Multiply:
$$-8 + 12 + 10 - 6$$
* Step 5: Add/Subtract:
$$4 + 10 - 6 = 14 - 6 = 8$$

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Final Answer:
1) -4
2) 72
3) -8
4) 5.5 (or 11/2)
5) 10
6) a = 0, b = -3
7) 8