Denominator of a rational expression cannot be 0 because if it will be 0 then the expression will be undefined. Hence, we need to exclude the values of x at which the denominator will be 0. So,
x²-2x-15=0
We can solve this equation by factoring the trinomial on the left side of the equation.
For that we need to find the two multiples of the constant term -15 so that their addition will result the coefficient of -2.
So, -15= -5*3 and -5+3=-2.
So, we can replace -2x with -5x+3x in the above equation.
x²-5x+3x-15=0
(x²-5x)+(3x-15)=0 Making the group of two terms.
x(x - 5) + 3(x - 5) =0 Taking out the common factor from each terms.
(x - 5) (x + 3) =0 Taking out the common factor (x-5) from both the terms.
So, x - 5 =0 and x + 3 =0 Set up each factor equal to 0.
So, x=5 and x=-3.
Hence, the excluded values are 3) -3 and 5.
