Chapter 2.5, Problem 73E

To determine

To compile: A list of all the various techniques for factoring a polynomial by giving examples and a paragraph discussing that when the use of each technique is appropriate.

Explanation of Solution

Given information: Techniques of factoring a polynomial have to be discussed.

The various methods to factor a polynomial are:

  $(a)Commonfactor:$ In this method, the distributive property of multiplication with respect to the sum is applied. According to this property, $x(a+b)=xa+xb$ .

For example, to factor the polynomial $8x+4$ , distributive property can be used:

  $8x+4=4(2x+1)$ .

When there are coefficients with common factors and a variable with different exponents, then the greatest common divisor and the variable with exponent of minor grade are taken out common. For example, $20x^2+10x^3+5x=5x(4x+2x^2+1)$ , where $5$ is the greatest common divisor and $x$ is the variable with minor grade.

  $(b)Differenceofsquares:$ It is equal to the sum of the roots of each term by its difference. It is based on the formula $a^2-b^2=(a+b)(a-b)$ .

For example, $9x^2-1=(3x+1)(3x-1)$ .

  $(c)Perfectsquaretrinomial:$ Based on the formulas:

  ${(a+b)}^2=a^2+2ab+b^{2}and{(a-b)}^2=a^2-2ab+b^2$ ,

it is equal to the square of the binomials.

For example, to factorize $4x^2-12x+9$ , the square roots of first and third terms are $\sqrt{4x^2}=2x$ and $\sqrt{9}=3$ . Comparing $4x^2-12x+9$ with ${(a-b)}^2=a^2-2ab+b^2$ ,

  $a=2,b=3$ .

So $4x^2-12x+9={(2x-3)}^2$ .

  $(d)FactorizationofTrinomialof\sec ondgrade:$ To factor a polynomial of the type $a{x}^2+bx+c$ , the trinomial is equated to zero and the equation $a{x}^2+bx+c=0$ is solved by using the formula $x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

For example to factorize $x^2-5x+6$ , equating it to zero, $x^2-5x+6=0$ .

Comparing with $a{x}^2+bx+c=0$ , $a=1,b=-5,c=6$ .

  $\begin{array}{l}{x}_{1,2}=\frac{-(-5)\pm \sqrt{25-4(1)(6)}}{2(1)}\\ =\frac{5\pm \sqrt{1}}{2}\\ =\frac{5\pm 1}{2}\\ \Rightarrow x_1=\frac{5+1}{2}=\frac{6}{2}=3\\ \&x_2=\frac{5-1}{2}=\frac{4}{2}=2\end{array}$

So, the original polynomial must be $(x-3)(x-2)$ .

Thus, $x^2-5x+6=(x-3)(x-2)$ .

  $(e)Ruffinirule:$ This rule can be applied for any polynomial having whole roots.

It is a special case of synthetic division when the divisor is a linear factor. To divide a polynomial $p(x)$ by a polynomial $q(x)$ , synthetic division is done.

For example, to divide $p(x)=2x^3+3x^2-4$ by $q(x)=x+1=x-(-1)$ ,

  $\begin{array}{lllll}-1\hfill & 2\hfill & 3\hfill & 0\hfill & -4\hfill \\ \hfill & \hfill & -2\hfill & -1\hfill & 1\hfill \\ \hfill & 2\hfill & 1\hfill & -1\hfill & -3\hfill \end{array}$

So, $\begin{array}{l}p(x)=q(x)r(x)+s\\ \Rightarrow 2x^3+3x^2-4=(2x^2+x-1)(x+1)-3\end{array}$