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:
x4+Ax3+5x2+x+D=(x2+4)(x2−x+B)+C(x+1)
Step-by-step substitution:
1. Substitute x=0:
04+A(0)3+5(0)2+0+D=(02+4)(02−0+B)+C(0+1)
D=4B+C
This gives the equation:
D=4B+C
2. Substitute x=1:
14+A(1)3+5(1)2+1+D=(12+4)(12−1+B)+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=(−12+4)((−1)2−(−1)+B)+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)((−2)2−(−2)+B)+C(−2+1)
16−8A+20−2+D=(4+4)(4+2+B)+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:
- D=4B+C
- A+D=5B+2C−7
- −A+D=5B+6
- −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=21
Now that we have B=21, substitute into equation (1):
D=4(21)+C=2+C
D=2+C
Substitute B=21 into A=B+C−7:
A=21+C−7
A=C−213
Finally, substitute A=C−213 and D=2+C into equation (4):
−8(C−213)+(2+C)=8(21)−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=6−213=−21
D=2+6=8
Final values:
- A=−21
- B=21
- C=6
- D=8
Let me know if you'd like any further clarification or more detailed explanations!
Here are 5 related questions:
- How would the approach change if we were asked to expand the equation?
- What other substitution values could be tried in this problem?
- Can this method be applied to non-quadratic polynomials as well?
- How would the system change if one of the terms were nonlinear?
- 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.