find the illegal values of C in the multiplication statement c^2-3c-10/c^2+5c-14 . c^2-c-2/c^2-2c-15
I will assume you mean
(c^2-3c-10)/(c^2+5c-14) (c^2-c-2)/(c^2-2c-15)
= (c-5)(c+2)/((c+7)(c-2)) (c-2)(c+1)/((c-5)(c+3))
= (c+2)(c+1)/((c+7)(c+3)) c ≠ -7,-3, 2, 5
To find the illegal values of C, we need to identify the values that make the denominators of either fraction equal to zero.
For the first fraction, the denominator is c^2 + 5c - 14. To find the values that make it equal to zero, we can factorize it:
c^2 + 5c - 14 = (c - 2)(c +7)
Setting each factor equal to zero, we get:
c - 2 = 0 or c + 7 = 0
Solving these equations, we find:
c = 2 or c = -7
So, c cannot be equal to 2 or -7 for the first fraction.
For the second fraction, the denominator is c^2 - 2c - 15. To find the values that make it equal to zero, we can factorize it:
c^2 - 2c - 15 = (c - 5)(c + 3)
Setting each factor equal to zero, we get:
c - 5 = 0 or c + 3 = 0
Solving these equations, we find:
c = 5 or c = -3
So, c cannot be equal to 5 or -3 for the second fraction.
Therefore, the illegal values of C are 2, -7, 5, and -3.
To find the illegal values of C in the given multiplication statement, we need to identify the values of C that would make the denominators of either fraction equal to zero. This is because division by zero is undefined and therefore illegal.
Let's examine the denominators of each fraction separately and find the values of C that would make them equal to zero.
For the first fraction, c^2 + 5c - 14:
To determine when this expression equals zero, we can factorize it or use the quadratic formula.
c^2 + 5c - 14 = 0
Using factorization:
(c - 2)(c + 7) = 0
Setting each factor to zero, we find:
c - 2 = 0 or c + 7 = 0
Solving these equations, we get:
c = 2 or c = -7
For the second fraction, c^2 - 2c - 15:
Again, we need to find the values of C that make this expression equal to zero.
c^2 - 2c - 15 = 0
Using factorization:
(c - 5)(c + 3) = 0
Setting each factor to zero, we find:
c - 5 = 0 or c + 3 = 0
Solving these equations, we get:
c = 5 or c = -3
So, the illegal values of C in the given multiplication statement are C=2, C=-7, C=5, and C=-3, as those are the values that make the denominators equal to zero.