In the trinomial, the coefficient of x^2 is 1, the coefficient of x is 6, and the stand-
alone constant is 9. We must find a number, such that adding it to itself yields 6
while squaring it yields 9. That number is 3. The binomial is therefore (x+ 3), and
we have
[(x+ 3)^2 ]^2 = 0
which simplifies to
(x+ 3)^4 = 0
(b) Here is the original equation:
(x^2 − 4 x+ 4)^3 = 0
In the trinomial, the coefficient of x^2 is 1, the coefficient of x is −4, and the con-
stant is 4. We must find a number, such that adding it to itself yields −4 while
squaring it yields 4. That number is −2. The binomial is therefore (x− 2), and
we have
[(x− 2)^2 ]^3 = 0
which simplifies to
(x− 2)^6 = 0
(c) Here is the original equation:
(16x^2 − 24 x+ 9)^4 = 0
This trinomial is the square of (4x− 3). Therefore, the original equation is equivalent to
[(4x− 3)^2 ]^4 = 0
which simplifies to
(4x− 3)^8 = 0
- In each case, we can remove the exponent from the binomial and then set it equal to 0,
obtaining a first-degree equation. The real root of the higher-degree equation is equal to
the solution of the first-degree equation. The multiplicity of the root is the value of the
exponentn to which the binomial is raised.
Chapter 26 693
