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P.7. Factoring Polynomials
In the previous section we explained how to perform basic arithmetic operations on polynomials, i.e. addition, subtraction, multiplication and division. Another important arithmetic operation, especially with polynomials, is factoring. Just as factoring a number means expressing it as the product of two or more smaller numbers, factoring a polynomial means expressing it as the product of two or more polynomials of smaller degree. Factoring is difficult in general, especially for polynomials of large degree, but there are a few simple methods which are very useful for factoring polynomials of small degree.
The easiest polynomial factors to look for are linear. A linear factor of a polynomial p(x) is a factor of the form ax + b, where a and b are constants. Finding linear factors amounts to finding zeros, also known as roots, of the polynomial p(x), i.e. values r of x such that p(r) = 0. If r is a root of the polynomial p(x), then x - r is a linear factor of p(x).
The simplest nontrivial case of factoring is finding two linear factors of a quadratic polynomial. In this case, factoring is performed by trial and error. Suppose we have a quadratic polynomial of the form p(x) = x^{2 }+ Ax + B, where A and B are positive numbers. We want to express p(x) as the product (x + a)(x + b). Multiplying out these terms, we find p(x) = x^{2} + (a+b)x + ab. Equating coefficients, we find a+b = A and ab = B. Thus, factoring p(x) amounts to finding two numbers a and b whose sum is A and whose product is B.
Example 1: Factor the polynomial x^{2 }+ 7x + 10.
Solution: We need to find two numbers a and b whose sum is 7 and whose product is 10. One easily finds by trial and error the solution a=2 and b=5. Thus we have the factorization x^{2 } + 7x + 10 = (x + 2)(x + 5).
Another case to consider is quadratic polynomials of the form p(x) = x^{2} + Ax - B, where once again, A and B are positive constants. This time we look for a factorization of the form p(x) = (x + a)(x - b), where a and b are positive. Multiplying out factors and equating coefficients as before, we find A = a-b and B = ab. Thus, we need to find two numbers a and b whose difference is A and whose product is B.
Example 2: Factor the polynomial x^{2 }+ x - 12.
Solution: We need to find two numbers a and b whose difference is 1 and whose product is 12. One easily finds by trial and error the solution a=4 and b=3. Thus we have the factorization x^{2} + x - 12 = (x + 4)(x - 3).
The other two cases, namely factoring polynomials of the form x^{2 }- Ax + B or of the form x^{2} - Ax - B are similar to the two above. The only difference is that the constant coefficients of the linear factors are now the negative of what they would be if the linear term of the quadratic polynomial were positive. A couple examples will illustrate how to deal with these cases. First consider p(x) = x^{2 }- 7x + 10. This time we look for a factorization of the form p(x)=(x - a)(x - b), where once again a and b are positive numbers whose sum is 7 and whose product is 10. Since once again we have the solution a=2 and b=5, this time we find p(x) =(x - 2)(x - 5).
What about p(x) = x^{2 }- x - 12? This time we look for a factorization of the form p(x) = (x -a)(x + b), where once again a and b are positive numbers with a > b whose difference is 1 and whose product is 12. Once again we have a=4 and b=3, so the factorization is p(x) = (x - 4)(x + 3).
What if the linear term of p(x) is zero? In other words, how do we factor a polynomial of the form p(x) = x^{2} + B? This case is easy to deal with if we consider the following identity:
- (P.7.1) x^{2 }- a^{2} = (x + a)(x - a).
Thus if B is the negative of a square, then the constant terms of the linear factors are the two square roots of |B|. What if B is positive? In this case, it is not possible to factor p(x) since negative numbers have no real square roots.
Example 3: Factor the polynomial x^{2 }- 9.
Solution: Since 9 = 3^{2}, we see that x^{2} - 9 = x^{2} - 3^{2} = (x + 3)(x - 3), using formula (P.7.1) with a=3.
The following identities is also useful for factoring a special type of quadratic polynomial, namely the square of a linear polynomial:
- (P.7.2a) x^{2} + 2ax + a^{2} = (x + a)^{2}.
- (P.7.2b) x^{2} - 2ax + a^{2} = (x - a)^{2}.
Example 4: Factor the polynomial x^{2} + 10x + 25.
Solution: Noting that 25 = 5^{2} and 10 = 2*5, we see using (P.7.2a) that x^{2} + 10x + 25 = (x + 5)^{2}.
What about polynomials of higher degree? In general, these are difficult to factor, but as in the quadratic case, there are a few cases which are easy to deal with. Here are some identities for factoring some special high degree polynomials:
- (P.7.3a) x^{3} - a^{3} = (x - a)(x^{2} + ax + a^{2}).
- (P.7.3b) x^{3} + a^{3} = (x - a)(x^{2} - ax + a^{2}).
- (P.7.4a) x^{3} + 3ax^{2 }+ 3a^{2}x + a^{3} = (x + a)^{3}.
- (P.7.4b) x^{3 }- 3ax^{2 }+ 3a^{2}x - a^{3} = (x - a)^{3}.
- (P.7.5) x^{4 }- a^{4} = (x - a)(x^{3} + ax^{2 }+ a^{2}x + a^{3}) = (x - a)(x + a)(x^{2} + a^{2}).
- (P.7.6a) x^{4} + 4ax^{3} + 6a^{2}x^{2} + 4a^{3}x + a^{4} = (x + a)^{4}.
- (P.7.6b) x^{4} - 4ax^{3} + 6a^{2}x^{2} - 4a^{3}x + a^{4} = (x - a)^{4}.
Example 5: Factor x^{3} - 27.
Solution: Using (P.7.3a) with a=3, we find x^{3} - 27= (x - 3)(x^{2} + 3x + 9).
Example 6: Factor x^{3} - 6x^{2} + 12x + 8.
Solution: Using (P.7.4b) with a=2, we find x^{3} - 6x^{2} + 12x + 8 = (x - 2)^{3}.
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