SOLUTION: I am having trouble with a question, It is find the illegal values of c in the multiplication statement <PRE> c^2-3c-10 c^2-c-2 --------- * ---------- c^2+5c-14 c^2-2c-15

Question 6289: I am having trouble with a question, It is find the illegal values of c in the multiplication statement

c^2-3c-10 c^2-c-2 --------- * ---------- c^2+5c-14 c^2-2c-15

I think the answer is
c=-7, c=-3, c=2, and c=5
Am I right, Can someone let me know?
Answer by ichudov(507) (Show Source): You can put this solution on YOUR website!
the illegal values is the values that make the two denominators equal to zero.
c^2+5c-14=0
c^2-2c-15=0
Solve both:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=81 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 2, -7. Here's your graph:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=64 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 5, -3. Here's your graph:

So, the illegal values are 2, -7, 5, -3, you are right. Good job!
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