Imagine that we were presented with the above equation for the first time, and we had
never seen it in factored form. We could factor x out, getting
x(x^2 + 14 x+ 48) = 0We could try to break up the trinomial part of this equation into binomial factors. With a
few trials and errors, we’d find them. We would notice that the stand-alone constants must
add up to 14 and multiply to 48; it wouldn’t take us long to guess that they are 6 and 8. Once
we had the three binomial factors of the cubic, finding the real roots would be easy, as the
next example will show. Even if we couldn’t factor the trinomial, we could attack it with the
quadratic formula. Later, we’ll see an example of that tactic.
What are the real roots?
To see how we can solve a binomial-factor cubic with real coefficients and a real constant, let’s
find the real roots of
(3x+ 2)(5x+ 6)(− 7 x− 1) = 0To find the first real root, we solve the equation created by setting the first binomial factor
equal to 0. That equation is
3 x+ 2 = 0Subtracting 2 from each side and then dividing through by 3, we get x=−2/3. To find the
second real root, we do the same thing with the second binomial factor. We have
5 x+ 6 = 0Subtracting 6 from each side and then dividing through by 5, we get x=−6/5. To find the
third real root, we solve the equation
− 7 x− 1 = 0Adding 1 to each side and then dividing through by −7, we get x= 1/(−7)=−1/7. The real
roots of the original cubic are therefore
x=−2/3 or x=−6/5 or x=−1/7and the solution set X is {−2/3,−6/5,−1/7}. In Practice Exercises 3 and 4 at the end of this
chapter, you’ll multiply out the factors of the original cubic, and then test these roots in the
resulting equation.
Are you confused?
You might wonder, “Why not divide the following equation through by x in an attempt to simplify and
solve it?”
x^3 + 14 x^2 + 48 x= 0Three Binomial Factors 417