Algebra Know-It-ALL

670 Worked-Out Solutions to Exercises: Chapters 21 to 29

2. We’ve been told to factor the following quadratic, and we’ve been assured that the
   coefficients and constants are all integers.

x^2 + 10 x+ 25 = 0

Because the coefficient of x^2 is equal to 1, we know that the coefficients of x in both bino-

mials must be equal to 1. That means the factored equation looks like this:

(x+ #)(x+ #) = 0

where # represents an integer (not necessarily the same one in each case). The sum of

these integers is 10, and their product is 25. A good guess will tell us that the numbers are

both 5. Let’s see what happens if we use those numbers and multiply out:

(x+ 5)(x+ 5) = 0

x^2 + 5 x+ 5 x+ 25 = 0

x^2 + 10 x+ 25 = 0

That works, so the factored form is

(x+ 5)(x+ 5) = 0

which can also be written as

(x+ 5)^2 = 0

To solve this, we can take the square root of both sides. There is no “plus-or-minus”

ambiguity. We get

x+ 5 = 0

There is only one root here, and it is −5. The solution set is therefore {−5}.

3. We want to factor the following quadratic, and we’ve been told that the coefficients and
   constants in the binomials are all integers.

2 x^2 + 8 x− 10 = 0

The coefficient of x^2 is equal to 2. Therefore, the general form of the equation in binomial

factor form will be

(x+ #)(2x+ #) = 0

where # represents an integer (not necessarily the same one in each case). The product of

these unknown integers is −10. If we plug in integers whose product is −10 and multiply

the resulting products of binomials out, we’ll eventually come up with

(x+ 5)(2x− 2) = 0