Factorization of Polynomials

Factorization of polynomials is the process of breaking down a polynomial expression into a product of simpler expressions (factors). This is a fundamental skill in algebra, used to solve equations, simplify expressions, and analyze the behavior of functions. We’ll focus on two types:

Quadratic Polynomials: These are polynomials of degree 2, having the general form $ax^2 + bx + c$. We’ll focus on factoring using the “splitting the middle term” method.
Cubic Polynomials: These are polynomials of degree 3, having the general form $ax^3 + bx^2 + cx + d$. We’ll use the Factor Theorem for their factorization.

Formulae

While there’s no single formula for all factorizations, the following concepts and techniques are crucial:

Splitting the Middle Term (for Quadratic): To factor $ax^2 + bx + c$, find two numbers (let’s call them ‘p’ and ‘q’) such that:
- $p + q = b$ (the coefficient of the x term)
- $p * q = a * c$ (the product of the coefficient of $x^2$ and the constant term)
Then, rewrite the middle term ($bx$) as $px + qx$ and proceed with factoring by grouping.
Factor Theorem (for Cubic and higher degree): If $p(k) = 0$ for a polynomial $p(x)$, then $(x – k)$ is a factor of $p(x)$. We find ‘k’ by trying different values (usually factors of the constant term). Once a factor is found, we can use polynomial long division or synthetic division to divide $p(x)$ by $(x – k)$ and find the remaining factor.

Examples

Let’s look at some examples to illustrate the techniques:

Example-1: Factoring a Quadratic Polynomial

Factorize $x^2 + 5x + 6$

Here, $a=1$, $b=5$, and $c=6$. We need to find two numbers that add up to 5 and multiply to 6. Those numbers are 2 and 3. So we rewrite the expression:

$x^2 + 5x + 6 = x^2 + 2x + 3x + 6$

Now, group the terms:

$(x^2 + 2x) + (3x + 6)$

Factor out the common factors:

$x(x + 2) + 3(x + 2)$

Finally, factor out the common binomial $(x + 2)$:

$(x + 2)(x + 3)$ Therefore, the factors are $(x+2)$ and $(x+3)$.

Example-2: Factoring a Cubic Polynomial using Factor Theorem

Factorize $x^3 – 7x + 6$

Let $p(x) = x^3 – 7x + 6$. We’ll try factors of the constant term (6), which are $\pm 1, \pm 2, \pm 3, \pm 6$.

Let’s try $x = 1$: $p(1) = (1)^3 – 7(1) + 6 = 1 – 7 + 6 = 0$. Therefore, $(x – 1)$ is a factor.

Now, we can divide $x^3 – 7x + 6$ by $(x-1)$ using polynomial long division or synthetic division. Using polynomial long division will give us: $x^2+x-6$

Now we have $x^3 – 7x + 6 = (x – 1)(x^2 + x – 6)$

Finally, we can factor the quadratic expression $(x^2 + x – 6)$. We need two numbers that add up to 1 and multiply to -6, which are 3 and -2:

$x^2 + x – 6 = (x + 3)(x – 2)$

Therefore, $x^3 – 7x + 6 = (x – 1)(x + 3)(x – 2)$

Theorem with Proof

Factor Theorem:

If $p(k) = 0$ for a polynomial $p(x)$, then $(x – k)$ is a factor of $p(x)$.

Proof:

Let $p(x)$ be a polynomial. By the Division Algorithm, we can write:

$p(x) = (x – k) * q(x) + r$

where $q(x)$ is the quotient and $r$ is the remainder. Since $(x – k)$ has degree 1, the remainder $r$ must be a constant.

Now, let’s substitute $x = k$:

$p(k) = (k – k) * q(k) + r$

$p(k) = 0 * q(k) + r$

$p(k) = r$

If $p(k) = 0$, then $r = 0$. This means:

$p(x) = (x – k) * q(x) + 0$

$p(x) = (x – k) * q(x)$

Therefore, $(x – k)$ is a factor of $p(x)$.

Common mistakes by students

Incorrectly Splitting the Middle Term: Students often struggle to find the correct pair of numbers that satisfy both the sum and product conditions. Careful consideration and practice are key.
Forgetting to Factor Out Completely: Students might factor out a common factor in one step but miss the opportunity to factor further at a later step.
Using the wrong factors in the Factor Theorem: Trying random numbers without thinking about the factors of the constant term, which leads to more time spent on the problem.
Mixing up the signs: Carelessness with positive and negative signs can lead to errors, particularly in quadratic equations.
Not recognizing common factors overlooking the possibility of taking a common factor before using other methods.

Real Life Application

Factorization is used extensively in many fields:

Engineering: Designing structures, circuits, and systems often involves solving polynomial equations to model various physical phenomena.
Computer Graphics: Polynomials are used to represent curves and surfaces, which are essential for creating 3D models and animations. Factorization is needed to analyze and manipulate these models.
Finance: Financial modeling and investment analysis often utilize polynomial equations to calculate returns, analyze risks, and predict market trends.
Physics: Describing the motion of an object in space.

Fun Fact

The Fundamental Theorem of Algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This means that a polynomial of degree ‘n’ will have exactly ‘n’ roots (counting multiplicities) in the complex number system. Factorization helps us find those roots!