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The process of putting a quadratic into binomial factor form is called, not surprisingly, factoring Learning how to factor a quadratic takes some practice Some quadratics are easy to factor, others are difficult, and some can t be factored at all at least, not into binomials with real-number constants and coefficients Remember the general binomial factor form: (px + q)(rx + s) = 0 We must find the values p, q, r, and s We re told that they re all integers The coefficient of the x 2 term in the original equation is equal to 1 That means pr = 1 in the binomial factor form The last constant in the original equation is 15 This tells us that qs = 15 Because pr = 1, we have two possibilities: p = 1 and r = 1 p = 1 and r = 1 Because qs = 15, we have eight possibilities: q = 15 and s = 1 q = 15 and s = 1 q = 1 and s = 15 q = 1 and s = 15 q = 3 and s = 5 q = 3 and s = 5 q = 5 and s = 3 q = 5 and s = 3

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We have successfully calculated our first limit Figure 21 confirms the conclusion that our calculations gave

We can trim this down to four possibilities by eliminating duplicate scenarios That leaves us with these choices: q = 15 and s = 1 q = 15 and s = 1 q = 3 and s = 5 q = 3 and s = 5

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Now let s look at the coefficient of x in the original equation It s 2 This tells us that ps + qr = 2 If we play around with our choices for awhile, we ll see that if we let p = 1, q = 3, r = 1, and s = 5, we get ps + qr = 1 ( 5) + 3 1 = 5 + 3 = 2 That s the right coefficient for x in the original! Let s try those numbers in the binomial factors for the left side of the equation We get (px + q)(rx + s) = (x + 3)(x 5) = x 2 5x + 3x 15 = x 2 2x 15

Calculate limx 2 g( x)

That s the left side of the original quadratic Now we know that it can be written as (x + 3)(x 5) = 0

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The roots are easy to find If x = 3 or x = 5, the left side of the equation becomes equal to 0 The solution set X is therefore X = { 3,5} Let s check to be sure these solutions work in the original equation For x = 3, we have x 2 2x 15 = 0 ( 3)2 2 ( 3) 15 = 0 9 ( 6) 15 = 0 9 + 6 15 = 0 15 15 = 0 0=0 For x = 5, we have x 2 2x 15 = 0 52 2 5 15 = 0 25 10 15 = 0 15 15 = 0 0=0

There are other ways to look for the roots of a quadratic One of these methods is called completing the square

We observe that both the numerator and the denominator of the fraction defining g tend to 0 as x 2 (ie, as x tends to 2) Thus the question seems to be indeterminate However, we may factor the numerator as x 2 4 = ( x 2) ( x + 2) As long as x = 2 (and these are the only x that we examine when we calculate limx 2 ), we can then divide the denominator of the expression defining g

Perfect squares Suppose that we come across a quadratic equation whose left side breaks down into two identical factors, equivalent to a single factor multiplied by itself A polynomial of this type is called a perfect square Here are some examples:

.

The polynomial x 2 + 2x + 1 factors into (x + 1)2 The polynomial x 2 2x + 1 factors into (x 1)2 The polynomial x 2 + 4x + 4 factors into (x + 2)2 The polynomial 9x 2 + 12x + 4 factors into (3x + 2)2

The graph of the function g is shown in Figure 22 We encourage the reader to use a pocket calculator to calculate values of g for x near 2 but unequal to 2 to check the validity of our answer For example,

Quadratics built from perfect squares (by setting the right side of the equation equal to 0) are easy to solve The two roots are identical, so in effect there s only one root, done twice over

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