Question: Let $f(x) = 4x^2 - 7x + 1$.
When $f(x)$ is divided by $x + 2$ the remainder is:
When $f(x)$ is divided by $x + 1$ the remainder is:
When $f(x)$ is divided by $x$ the remainder is:
When $f(x)$ is divided by $x - 1$ the remainder is:
When $f(x)$ is divided by $x - 2$ the remainder is:
1. **State the problem:** We need to find the remainders when the polynomial $f(x) = 4x^2 - 7x + 1$ is divided by various linear divisors.
2. **Recall the Remainder Theorem:** When a polynomial $f(x)$ is divided by $x - a$, the remainder is $f(a)$.
3. **Apply the theorem for each divisor:**
- For divisor $x + 2$, rewrite as $x - (-2)$, so $a = -2$.
Calculate:
$$f(-2) = 4(-2)^2 - 7(-2) + 1 = 4(4) + 14 + 1 = 16 + 14 + 1 = 31$$
- For divisor $x + 1$, rewrite as $x - (-1)$, so $a = -1$.
Calculate:
$$f(-1) = 4(-1)^2 - 7(-1) + 1 = 4(1) + 7 + 1 = 4 + 7 + 1 = 12$$
- For divisor $x$, rewrite as $x - 0$, so $a = 0$.
Calculate:
$$f(0) = 4(0)^2 - 7(0) + 1 = 0 - 0 + 1 = 1$$
- For divisor $x - 1$, $a = 1$.
Calculate:
$$f(1) = 4(1)^2 - 7(1) + 1 = 4 - 7 + 1 = -2$$
- For divisor $x - 2$, $a = 2$.
Calculate:
$$f(2) = 4(2)^2 - 7(2) + 1 = 4(4) - 14 + 1 = 16 - 14 + 1 = 3$$
4. **Final answers:**
- Remainder when divided by $x + 2$ is $31$.
- Remainder when divided by $x + 1$ is $12$.
- Remainder when divided by $x$ is $1$.
- Remainder when divided by $x - 1$ is $-2$.
- Remainder when divided by $x - 2$ is $3$.