In Exercises 47-50, write the polynomial (a) as the product...

In Exercises $47-50,$ write the polynomial (a) as the product of factors that are irreducible over the rationals, (b) as the product of linear and quadratic factors that are irreducible over the reals, and (c) in completely factored form. $$ \begin{array}{l}{f(x)=x^{4}-2 x^{3}-3 x^{2}+12 x-18} \\ {\text { (Hint: One factor is } x^{2}-6 . )}\end{array} $$

In Exercises $47-50,$ write the polynomial (a) as the product of factors that are irreducible over the rationals, (b) as the product of linear and quadratic factors that are irreducible over the reals, and (c) in completely factored form.
$$
\begin{array}{l}{f(x)=x^{4}-2 x^{3}-3 x^{2}+12 x-18} \\ {\text { (Hint: One factor is } x^{2}-6 . )}\end{array}
$$

Key Concepts

Complete Factoring (over the Complex Numbers)

Complete factoring refers to the process of expressing a polynomial as a product of linear factors, which is always possible over the complex numbers according to the Fundamental Theorem of Algebra. This process involves finding all complex roots of the polynomial, which may include both real and non-real numbers, and then writing the polynomial as a product of terms corresponding to these roots.

Factor Theorem

The Factor Theorem states that if a polynomial f(x) has a root at x = r, then (x - r) is a factor of f(x). This theorem is a crucial tool in polynomial factorization because it provides a direct method to identify factors and simplify the factorization process by testing possible roots, including the use of synthetic or long division.

Irreducible Factors over the Rationals

A polynomial factor is said to be irreducible over the rationals if it cannot be factored further into polynomials of lower degree with rational coefficients. Factoring a polynomial into irreducible factors over the rationals allows one to understand its structure when only considering rational numbers as potential coefficients, making it important in abstract algebra and number theory.

Polynomial Factorization

This concept involves expressing a polynomial as a product of simpler polynomials or factors. These factors may be linear or higher degree, and the goal is to simplify the polynomial in a way that reveals its roots and structure. Techniques for polynomial factorization include grouping, applying the quadratic formula, synthetic division, and using known identities.

Factoring over the Reals

When factoring over the reals, the goal is to express the polynomial as a product of linear factors and irreducible quadratic factors, where the quadratic factors have no real roots. This type of factorization makes use of the fact that every real polynomial can be decomposed into linear and quadratic factors, as guaranteed by the Fundamental Theorem of Algebra when complex roots occur in conjugate pairs.

Transcript

00:01 Hello and welcome.

00:02 We are looking at chapter 2, section 5, problem 49.

00:07 We're asked to find the factorization to factor the following polynomial.

00:20 Now we're given the fact that x squared minus 6 is a factor.

00:25 So i'm just going to go ahead and jump right into long dividing.

00:30 So x squared goes into x to the fourth, x squared times.

00:35 So this would be x to the fourth minus 6x squared.

00:39 I'm just lining up with the like term.

00:42 So this would be 0.

00:44 This would be negative 2x cubed.

00:47 And this would be negative 3 plus 6 is positive 3.

00:53 So x squared goes into negative 2x cubed, negative 2x times.

01:00 Gives us negative 2x cubed and positive 12x.

01:08 This would be 0 plus 3x squared plus 0.

01:14 And then we can bring down that negative 18 here.

01:17 We can see that if i put our 3, that's going to cancel out and give us a remainder of 0.

01:24 So we know that this polynomial equals x squared minus 6 times x squared minus 2x plus 3.

01:35 So if we look at factoring these, let's see, two numbers that multiply to 3 and add to negative 2.

01:48 Oh, close, but that doesn't work.

01:50 So this is actually the answer to part one because it, this doesn't, if we try to factor these, we're going to either get a real number or an imaginary number, neither of which is what we want, what we want.

02:10 So this is the answer to part a.

02:13 If we go further on, we can factor x squared minus 6 to x plus root 6.

02:26 X minus root 6 using the difference of squares.

02:31 And then we won't do anything with x squared minus 2x plus 3 just yet...

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