Subtracting 5 from each side and then dividing through by 3, we get x=−5/3. Next, we apply the qua-
dratic formula to the trinomial factor. If we let a 2 = 16, b 2 =−56, and c= 49, then
x= [−b 2 ± (b 22 − 4 a 2 c)1/2] / (2a 2 )
= {56 ± [(−56)^2 − 4 × 16 × 49]1/2} / (2 × 16)
= [56 ± (3,136 − 3,136)1/2] / 32
= 56/32
= 7/4
In this case, 7/4 is the only real root of the quadratic we obtain by setting the trinomial factor equal to 0.
The original cubic therefore has two real roots:
x=−5/3 or x= 7/4and the real solution set X is {−5/3, 7/4}. The root x= 7/4 has multiplicity 2. In Practice Exercise 7 at the
end of this chapter, you’ll check these roots.
Here’s another challenge!
Find the real roots of the following cubic equation using the approach described in this section, and state
the real solution set X.
(− 2 x+ 11)(2x^2 − 2 x+ 5) = 0Solution
Our first step, as in the previous “challenge,” is to create a first-degree equation by setting the binomial
factor equal to 0. That gives us
− 2 x+ 11 = 0When we subtract 11 from each side and then divide the entire equation through by −2, we get the solu-
tionx=−11/(−2), which is equal to 11/2. Then we apply the quadratic formula to the trinomial factor. If
we let a 2 = 2, b 2 =−2, and c= 5, we have
x= [−b 2 ± (b 22 − 4 a 2 c)1/2] / (2a 2 )
= {2 ± [(−2)^2 − 4 × 2 × 5]1/2} / (2 × 2)
= [2 ± (4 − 40)1/2] / 4
= [2 ± (−36)1/2] / 4
We can stop right here, because the discriminant is negative. The quadratic we obtain by setting the
trinomial factor equal to 0 has no real roots (although it has two complex roots). The original cubic
therefore has only one real root, x= 11/2. The real solution set X contains only that single root, so
X= {11/2}.
Binomial Times Trinomial 421