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If x^2 -1 is a factor of ax^3+bx^2+cx+d show tha a+c=0?
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If x^2 -1 is a factor of ax^3+bx^2+cx+d show tha a+c=0?
Proof that a*c = 0 if x^2 - 1 is a factor of ax^3 bx^2 cx d
Let's assume that x^2 - 1 is a factor of ax^3 + bx^2 + cx + d. This means that:
- x = 1 is a root of the polynomial
- x = -1 is a root of the polynomial
Using synthetic division
We can use synthetic division to write the polynomial as:
1 0 -a -a-b
1 | a b c d
| a a+b a+c
---------------
a a+b c+d a+b+c
Since x = 1 is a root of the polynomial, the remainder must be 0 when we divide by x - 1. This gives us the equation:
a + (a+b) + (c+d) + (a+b+c) = 0
Simplifying this equation, we get:
3a + 2b + 2c + d = 0
Using x = -1 as well
Similarly, we can use synthetic division to write the polynomial as:
-1 0 -a a-b
-1 | a b c d
| -a a-b -c c-d
----------------
0 b-c -c d-c+d
Since x = -1 is a root of the polynomial, the remainder must be 0 when we divide by x + 1. This gives us the equation:
b - c + d - c + d = 0
Simplifying this equation, we get:
2d - b - 2c = 0
Combining the results
Now we can solve these two equations simultaneously to get a and c in terms of b and d:
3a + 2b + 2c + d = 0
2d - b - 2c = 0
Multiplying the second equation by 3 and adding it to the first equation, we get:
9a + 7b + 4d = 0
Solving for a, we get:
a = -(7b + 4d)/9
Multiplying the second equation by 2 and subtracting it from the first equation, we get:
a + 3b - d = 0
Solving for c, we get:
c = (2d - b)/2
Proof that a*c = 0
Now we can substitute a and c in terms of b and d into the equation ac = 0:
ac = [-(7b + 4
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If x^2 -1 is a factor of ax^3+bx^2+cx+d show tha a+c=0?
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If x^2 -1 is a factor of ax^3+bx^2+cx+d show tha a+c=0?
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