Factoring Polynomials

Here we talk about factoring monomials and polynomials from other polynomials, the quadratic formula, and polynomial division

The Essentials

Factoring monomials from polynomials is done by dividing a common term out and setting it in front of the polynomial:

\[ 10x + 15xy = 2(5x) + 3y(5x) \]
\[ = 5x(2 + 3y) \]

Likewise, a polynomial can be factored out as well. Consider this expression:

\[ 2x + 2 + 3xy + 3y \]

It can be written as two terms of something times \( (x+1) \):

\[ 2(x + 1) + 3y(x + 1) \]

We can treat the \( (x+1) \) the same as the 5x in the first example and factor it out:

\[(x + 1)(2 + 3y)\]

Quadratic expressions can also be factored. Consider a quadratic expression that fits in the form \( (x+a)(x+b) \) where a and b could be positive or negative numbers. Distributing gives us \( x^2 + (a + b)x + ab \). So, if we have a quadratic expression in the form \( x^2 + Cx + D \) then we can factor our expression using \( a + b = C \), \( ab = D \) if the number are nice enough. Here are some common patterns that are useful for factoring:

\[ (x + a)(x + a) = x^2 + 2ax + a^2 \]

(square of sum)

\[ (x - a)(x - a) = x^2 - 2ax + a^2 \]

(square of difference)

\[ (x + a)(x - a) = x^2 - a^2 \]

(difference of squares)

If we want to factor an expression in this form: \( ax^2 + bx + c \), where a isn’t one, we can use the quadratic equation to factor any quadratic expression:

\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]

This formula will give two answers. At these two x values, \( x_1 \) and \( x_2 \) the quadratic expression \( ax^2 + bx + c \) will be equal to zero, so we can factor the expression as: \( (x − x_1)(x − x_2) \).

Higher-order polynomials don’t have a convenient formula like the quadratic formula. Some tricks could be used to factor them. In this expression: \( y^4 + 4y^2 + 4 \), we can substitute \( x \) for \( y^2 \) and solve as a quadratic expression. Consider this expression: \( x^3 + x^2 + 2x + 2 \). If we look closely, we might notice that we can factor out a \( (x + 1) \) term and factor it as \( (x^2 + 2)(x + 1)\) .

Such tricks aren’t always easy to find (or even possible). Polynomial division (aka synthetic division or long division) can be used in these cases. Here we will not explain this process, we will show an example:

\[ \require{enclose} \begin{array}{r@{}l} x + 4 \\ \begin{array}{r@{}l} x - 2\,\enclose{longdiv}{\,x^2 + 2x + 3} \\ - (x^2 - 2x) \\ \hline 0 + 4x + 3 \\ - (4x - 8) \\ \hline 0 + 11 \\ \end{array} \end{array} \]

Here the remainder is 11, so the polynomial can be re-written like this:

\[ x^2 + 2x + 3 = (x + 4)(x - 2) + 11 \]

Example

Factor this polynomial: \(64c^4 - 32c^2 + 4\)

This polynomial fits into the square of difference form: \( (x - a)(x + a) = x^2 - 2ax + x^2 \).

\[ (8c^2)^2 - 2(8c^2 \cdot 2) + 2^2 \]

Therefore it can be factored as \( (8c^2 - 2)^2 \).

Practice

Factor these polynomials:

  1. \( 100x^2 + 80x + 16 \)
  2. \( 2x^3 + 2x^2 + 3x + 3 \)
  3. \( x^3 + 6x^2 - 9x - 14 \), where \( (x + 1) \) is a factor

Solution:

  1. \( 4(5x + 2)^2 \)
  2. \( (x + 1)(2x^2 + 3) \)
  3. \( (x + 1)(x − 2)(x + 7) \)

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