To solve an equation is to find the set of values of the unknowns satisfying the equation. While we can easily solve equations of lower degrees, it is not easy to solve equations of higher degrees. However, by factorization, we can rewrite a high degree polynomial into a product of low degree polynomials. The method of solving equations be factorization is based on the following theorem:
Zero product property Let and be real or complex valued. If , then or . Proof Either or holds true. If , then " or " holds true. If , then exists. Multiply it to both sides of :
.
Then " or " also holds true.QED
Therefore, when we solve a polynomial equation, we can first factorize the polynomial and form smaller equations equating the factors with zero.
Example 1 Solve . Solution
or or
or or
or
In the previous chapter, we have seen the use of unit roots in determining the divisibility of a polynomial. In fact we can similarly factorize some polynomials by considering the properties of unit roots.
Example 2 Factorize . Anaylsis The indices in the expression are 8, 6, 4, 2, 0. When they are divided by 5, the remainders are 3, 1, 4, 2, 0. Therefore, when is replaced by a non-real fifth root of unity, the expression equals zero. So the expression is divisible by . Answer
If we allow complex coefficients in the factors, any polynomials can be factorized as a product of linear factors; if we allow any real coefficients, then any polynomials with real coefficients can be factorized as a product of factors of at most the second degree.
Example 3 Factorize
(a) (three factors)
(b) (two factors) Analysis We can check that both (a) and (b) is zero when is replaced by a non-real cube root of unity. However, we have to further find another factor for (a). For this purpose, we first let so the expression becomes , which also equals zero when is replaced by a non-real cube root of unity. Solution (a) . So,
Using the cube root of unity, we can derive the formula for cubic equations.
Let be a non-real cube root of unity, then:
, .
Then we can show that:
.
Therefore, the roots of the cubic equation in :
are:
, , .
Now consider the cubic equation:
.
We may let and . So . Therefore, and are the roots of the following equation:
.
The roots of this equation are:
.
We let be any of these roots, and be the other root. Then:
,
.
(In fact we may also take other non-real cube roots, as long as the relation holds. However, we need not consider the non-real cube roots because we have considered them through the specification of and .)
Example 4 Solve . Solution In the example, , . So: