Algebra Know-It-ALL

Multiplying to confirm, we get

(x+ 5)(2x− 2) = 0

2 x^2 − 2 x+ 10 x− 10 = 0

2 x^2 + 8 x− 10 = 0

The roots are found by solving these two first-degree equations:

x+ 5 = 0

and

2 x− 2 = 0

The solution in the first case is easily seen to be x=−5, and in the second case the solu-

tion is x= 1. The roots of the quadratic are −5 and 1. The solution set is {−5, 1}.

4. We want to morph the left side of the following quadratic into a product of binomials
   whose coefficients and constants in the binomials are all integers.

12 x^2 + 7 x− 10 = 0

In this case, the coefficient of x^2 is equal to 12. That means the general binomial factor

form will look like one of these:

(x+ #)(12x+ #) = 0

(2x+ #)(6x+ #) = 0

(4x+ #)(3x+ #) = 0

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

unknown integers is −10. As before, we can start plugging in integers whose product is −10,

multiply the resulting product of binomials out on every attempt, and see what we get. There

are lots of choices here, and the process could take time. Eventually we’ll come up with

(4x+ 5)(3x− 2) = 0

When we multiply this out, we find

(4x+ 5)(3x− 2) = 0

12 x^2 + (− 8 x)+ 15 x+ (−10)= 0

12 x^2 + 7 x− 10 = 0

To find the roots, we must solve

4 x+ 5 = 0

Chapter 22 671