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solve without expanding by substituting value of x. Show complete working

To solve the given equation without expanding, we'll substitute different values of \( x \) to find the constants \( A \), \( B \), \( C \), and \( D \).

### The equation is:
\[
x^4 + A x^3 + 5x^2 + x + D = (x^2 + 4)\left( x^2 - x + B \right) + C(x + 1)
\]

### Step-by-step substitution:

#### 1. Substitute \( x = 0 \):

\[
0^4 + A(0)^3 + 5(0)^2 + 0 + D = (0^2 + 4)\left(0^2 - 0 + B \right) + C(0 + 1)
\]
\[
D = 4B + C
\]

This gives the equation:
\[
D = 4B + C
\]

#### 2. Substitute \( x = 1 \):

\[
1^4 + A(1)^3 + 5(1)^2 + 1 + D = (1^2 + 4)\left( 1^2 - 1 + B \right) + C(1 + 1)
\]
\[
1 + A + 5 + 1 + D = 5(1 - 1 + B) + 2C
\]
\[
7 + A + D = 5B + 2C
\]

This gives the equation:
\[
A + D = 5B + 2C - 7
\]

#### 3. Substitute \( x = -1 \):

\[
(-1)^4 + A(-1)^3 + 5(-1)^2 + (-1) + D = (-1^2 + 4)\left( (-1)^2 - (-1) + B \right) + C(-1 + 1)
\]
\[
1 - A + 5 - 1 + D = 5(1 + 1 + B) + 0
\]
\[
4 - A + D = 10 + 5B
\]

This gives the equation:
\[
-A + D = 5B + 6
\]

#### 4. Substitute \( x = -2 \):

\[
(-2)^4 + A(-2)^3 + 5(-2)^2 + (-2) + D = ((-2)^2 + 4)\left( (-2)^2 - (-2) + B \right) + C(-2 + 1)
\]
\[
16 - 8A + 20 - 2 + D = (4 + 4)\left( 4 + 2 + B \right) + C(-1)
\]
\[
34 - 8A + D = 8(6 + B) - C
\]
\[
34 - 8A + D = 48 + 8B - C
\]

This gives the equation:
\[
-8A + D = 8B - C + 14
\]

### Now we have a system of four equations:

1. \( D = 4B + C \)
2. \( A + D = 5B + 2C - 7 \)
3. \( -A + D = 5B + 6 \)
4. \( -8A + D = 8B - C + 14 \)

#### Solving this system:

From equation (1): 
\[
D = 4B + C
\]

Substitute this into equations (2), (3), and (4):

- For equation (2):
\[
A + (4B + C) = 5B + 2C - 7
\]
\[
A + 4B + C = 5B + 2C - 7
\]
\[
A = B + C - 7
\]

- For equation (3):
\[
-A + (4B + C) = 5B + 6
\]
\[
-A + 4B + C = 5B + 6
\]
Substitute \( A = B + C - 7 \):
\[
-(B + C - 7) + 4B + C = 5B + 6
\]
\[
- B - C + 7 + 4B + C = 5B + 6
\]
\[
3B + 7 = 5B + 6
\]
\[
2B = 1
\]
\[
B = \frac{1}{2}
\]

Now that we have \( B = \frac{1}{2} \), substitute into equation (1):
\[
D = 4\left(\frac{1}{2}\right) + C = 2 + C
\]
\[
D = 2 + C
\]

Substitute \( B = \frac{1}{2} \) into \( A = B + C - 7 \):
\[
A = \frac{1}{2} + C - 7
\]
\[
A = C - \frac{13}{2}
\]

Finally, substitute \( A = C - \frac{13}{2} \) and \( D = 2 + C \) into equation (4):
\[
-8(C - \frac{13}{2}) + (2 + C) = 8\left(\frac{1}{2}\right) - C + 14
\]
\[
-8C + 52 + 2 + C = 4 - C + 14
\]
\[
-7C + 54 = 18 - C
\]
\[
-7C + C = 18 - 54
\]
\[
-6C = -36
\]
\[
C = 6
\]

Now that we have \( C = 6 \), substitute back to find \( A \) and \( D \):
\[
A = 6 - \frac{13}{2} = -\frac{1}{2}
\]
\[
D = 2 + 6 = 8
\]

### Final values:
- \( A = -\frac{1}{2} \)
- \( B = \frac{1}{2} \)
- \( C = 6 \)
- \( D = 8 \)

Let me know if you'd like any further clarification or more detailed explanations!

### Here are 5 related questions:
1. How would the approach change if we were asked to expand the equation?
2. What other substitution values could be tried in this problem?
3. Can this method be applied to non-quadratic polynomials as well?
4. How would the system change if one of the terms were nonlinear?
5. Can this approach be extended to solve problems with more constants?

### Tip:
When solving for constants using substitution, aim to choose values of \( x \) that simplify the terms, like 0, 1, or -1, as this can make the calculations easier.

Given that x^4 + Ax^3 + 5x^2 + x + D = (x^2 + 4)(x^2 - x + B) + C(x + 1), find the values of constants A, B, C, and D without expanding.

Learn how to solve a polynomial equation involving constants A, B, C, and D by substituting values of x. This step-by-step solution walks through the process of forming and solving a system of linear equations. Suitable for students in Grades 10-12.

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