Math Problem Statement

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

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.

Solution

To solve the given equation without expanding, we'll substitute different values of xx to find the constants AA, BB, CC, and DD.

The equation is:

x4+Ax3+5x2+x+D=(x2+4)(x2x+B)+C(x+1)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=0x = 0:

04+A(0)3+5(0)2+0+D=(02+4)(020+B)+C(0+1)0^4 + A(0)^3 + 5(0)^2 + 0 + D = (0^2 + 4)\left(0^2 - 0 + B \right) + C(0 + 1) D=4B+CD = 4B + C

This gives the equation: D=4B+CD = 4B + C

2. Substitute x=1x = 1:

14+A(1)3+5(1)2+1+D=(12+4)(121+B)+C(1+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(11+B)+2C1 + A + 5 + 1 + D = 5(1 - 1 + B) + 2C 7+A+D=5B+2C7 + A + D = 5B + 2C

This gives the equation: A+D=5B+2C7A + D = 5B + 2C - 7

3. Substitute x=1x = -1:

(1)4+A(1)3+5(1)2+(1)+D=(12+4)((1)2(1)+B)+C(1+1)(-1)^4 + A(-1)^3 + 5(-1)^2 + (-1) + D = (-1^2 + 4)\left( (-1)^2 - (-1) + B \right) + C(-1 + 1) 1A+51+D=5(1+1+B)+01 - A + 5 - 1 + D = 5(1 + 1 + B) + 0 4A+D=10+5B4 - A + D = 10 + 5B

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

4. Substitute x=2x = -2:

(2)4+A(2)3+5(2)2+(2)+D=((2)2+4)((2)2(2)+B)+C(2+1)(-2)^4 + A(-2)^3 + 5(-2)^2 + (-2) + D = ((-2)^2 + 4)\left( (-2)^2 - (-2) + B \right) + C(-2 + 1) 168A+202+D=(4+4)(4+2+B)+C(1)16 - 8A + 20 - 2 + D = (4 + 4)\left( 4 + 2 + B \right) + C(-1) 348A+D=8(6+B)C34 - 8A + D = 8(6 + B) - C 348A+D=48+8BC34 - 8A + D = 48 + 8B - C

This gives the equation: 8A+D=8BC+14-8A + D = 8B - C + 14

Now we have a system of four equations:

  1. D=4B+CD = 4B + C
  2. A+D=5B+2C7A + D = 5B + 2C - 7
  3. A+D=5B+6-A + D = 5B + 6
  4. 8A+D=8BC+14-8A + D = 8B - C + 14

Solving this system:

From equation (1): D=4B+CD = 4B + C

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

  • For equation (2): A+(4B+C)=5B+2C7A + (4B + C) = 5B + 2C - 7 A+4B+C=5B+2C7A + 4B + C = 5B + 2C - 7 A=B+C7A = B + C - 7

  • For equation (3): A+(4B+C)=5B+6-A + (4B + C) = 5B + 6 A+4B+C=5B+6-A + 4B + C = 5B + 6 Substitute A=B+C7A = B + C - 7: (B+C7)+4B+C=5B+6-(B + C - 7) + 4B + C = 5B + 6 BC+7+4B+C=5B+6- B - C + 7 + 4B + C = 5B + 6 3B+7=5B+63B + 7 = 5B + 6 2B=12B = 1 B=12B = \frac{1}{2}

Now that we have B=12B = \frac{1}{2}, substitute into equation (1): D=4(12)+C=2+CD = 4\left(\frac{1}{2}\right) + C = 2 + C D=2+CD = 2 + C

Substitute B=12B = \frac{1}{2} into A=B+C7A = B + C - 7: A=12+C7A = \frac{1}{2} + C - 7 A=C132A = C - \frac{13}{2}

Finally, substitute A=C132A = C - \frac{13}{2} and D=2+CD = 2 + C into equation (4): 8(C132)+(2+C)=8(12)C+14-8(C - \frac{13}{2}) + (2 + C) = 8\left(\frac{1}{2}\right) - C + 14 8C+52+2+C=4C+14-8C + 52 + 2 + C = 4 - C + 14 7C+54=18C-7C + 54 = 18 - C 7C+C=1854-7C + C = 18 - 54 6C=36-6C = -36 C=6C = 6

Now that we have C=6C = 6, substitute back to find AA and DD: A=6132=12A = 6 - \frac{13}{2} = -\frac{1}{2} D=2+6=8D = 2 + 6 = 8

Final values:

  • A=12A = -\frac{1}{2}
  • B=12B = \frac{1}{2}
  • C=6C = 6
  • D=8D = 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 xx that simplify the terms, like 0, 1, or -1, as this can make the calculations easier.

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Math Problem Analysis

Mathematical Concepts

Polynomials
Substitution
System of Equations

Formulas

Polynomial equation
Substitution method

Theorems

Substitution and solving systems of linear equations

Suitable Grade Level

Grades 10-12