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to look for the removable point of discontinuity we first going to factor both the numerator and the denominator of this function if the product of two integers is negative 24 and their sum is five three and eight seems to be those two numbers we need to find out which one is negative negative three times eight negative 24 negative three plus eight five so the factors are x minus three times x plus eight the product of two integers is negative six and the sum is a coefficient of x negative one three and two and three is negative so the factors are x minus 3 and x plus 2. so x minus 3 is the common factor and the removal point of this continuity occurs when the number is both a zero of the numerator and the denominator x minus three equal to zero so that number is x equal to three now when we cancel x minus 3 we have f of x equal to x plus 8 divided by x plus 2. so we redefine the function so that the point of discontinuity can be removed and that removable point will be f of th
