Solution
We can work it the other way, multiplying the factors out to get the polynomial. Let’s rewrite the above
squared binomial as a product of sums, like this:
[x+b/(2a)] [x+b/(2a)]
When we apply the product of sums and product of quotients rules from Chap. 9, this becomes
x^2 +xb/(2a)+xb/(2a)+b^2 /(4a^2 )
We can add the two middle terms together, getting
x^2 + 2 xb/(2a)+b^2 /(4a^2 )
Canceling out the 2 in the numerator and denominator of the middle term, we obtain
x^2 +xb/a+b^2 /(4a^2 )
which can also be written as
x^2 + (b/a)x+b^2 /(4a^2 )
That’s the original polynomial.
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems.
Don’t hurry! You’ll find worked-out answers in App. C. The solutions in the appendix may
not represent the only way a problem can be figured out. If you think you can solve a particu-
lar problem in a quicker or better way than you see there, by all means try it!
- Multiply out the following equation, putting it into polynomial standard form.
(− 7 x− 5)(− 2 x+ 9) = 0
- Factor the following quadratic. Then find the roots and state the solution set. Here’s a
hint: The coefficients and constants in the factors are all integers.
x^2 + 10 x+ 25 = 0
- Factor the following quadratic. Then find the roots and state the solution set. Here’s a
hint: The coefficients and constants in the factors are all integers.
2 x^2 + 8 x− 10 = 0
- Factor the following quadratic. Then find the roots and state the solution set. Here’s a
hint: The coefficients and constants in the factors are all integers.
12 x^2 + 7 x− 10 = 0
Practice Exercises 379
