# Solving Quadratic Equations I: Factoring (Grouping)

Sometimes, quadratic equations can be solved by factorizing, which is also called grouping. The solve by factoring process is usually consists of three major steps:
All terms should be moved to one side of the equation using addition or subtraction. Rewrite the equation so that the left side of the equation is set equal to 0. For example, if the original equation is $x^2 = 5x - 3$, it should be rewritten as $x^2 - 5x + 3 = 0$.
The equation should be factored completely. It will have two factors.
Each factor should be set equal to zero. Since factors are of first power, they are easy to be solved.
All of the solutions should be combined to obtain the full solution set for the original equation.
The catch of this method is the finding of factors. Except some trivial cases, factoring quadratic equations is not easier than using other methods to solve the quadratic equations. Here are a few tips to help you factorize a quadratic equations:

### 1) Use polynomial identities:

$(a+b)^2 = a^2 + 2ab + b^2$
$x^2 + (a+b)x + ab = (x + a)(x + b)$
$a^2 - b^2 = (a+b)(a-b)$
Example 1:
Solve the quadratic equation  $x^2+2x+1=0$.
Comparing with $(a+b)^2 = a^2 + 2ab + b^2$, we see that substituting $a=x$ and $b=1$ we have,
$x^2+2x+1=0 \to (x+1)^2=0 \to (x+1)=0 \to x=-1.$
Example 2:
Solve the quadratic equation  $y^2+3y+2=0$.
To use the identity $x^2 + (a+b)x + ab = (x + a)(x + b)$ we need to find $a$ and $b$ such that their summation is $3$ and multiplication equals to latex $2$. It is not that hard to "guess" that $a=1$ and $b=2$ satisfy the condition. Therefore,
$y^2+3y+2=0 \to (y+1)(y+2)=0 \to \left\{ \begin{array}{l} y+1=0 \\ y+2=0 \end{array} \right. \to \left\{ \begin{array}{l} y=-1 \\ y=-2 \end{array} \right.$
Example 3:
Solve the quadratic equation $x^2-9=0.$
Assuming $a=x$ and $b=\sqrt{9}=3$, the identity $a^2 - b^2 = (a+b)(a-b)$ can be used to factor this equation. We have,
$x^2-9=0 \to x^2-3^2=0 \to (x+3)(x-3)=0\to \left\{ \begin{array}{l} x+3=0 \\ x-3=0 \end{array} \right. \to \left\{ \begin{array}{l} x=-3 \\ x=3 \end{array} \right.$

### 2) Guess one of the factors:

If coefficients are integer, it is sometimes easy to guess one of the factors.
Example 4: Solve the quadratic equation $2x^2-3x+1=0$
Since $x=1$ satisfies the equation, $x-1$ is one of the factors. We can divide the equation by $(x - 1)$ to find the other factor:
$\begin{array}{l | l} & x-1 \\ 2x^2-3x+1 &2x-1\\2x^2-2x & \\ -x+1&\\-x+1&\\ 0&\end{array}$
Therefore, the other factor is $2x-1$ and the original has two answers $x=1$ and $2x-1=0 \to x= \frac{1}{2}$.
Example 5: Solve the quadratic equation $-4x-3=x^2$
$-4x-3=x^2 \to x^2+4x+3=0$
$x=-1$ satisfies the equation; therefore, $x+1$ is one of the factors. By dividing the original equation by $(x+1)$ the other factor can be found:
$\begin{array}{l | l} & x+1 \\ x^2+4x+3 &x+3\\x^2+x & \\ 3x+3&\\3x+3&\\ 0 & \end{array}$
Thus, the other factor is $x+3$ and $x=-3$ is the other answer. Therefore, the original has two answers $x=-1$ and $x=-3$.
As explained, factoring quadratic equations does not have a straight-forward procedure to follow. Therefore, the usage of this method is restricted to some trivial equations and relies mostly on your skills. You can learn about other methods of solving quadratic equations here.

## 2 thoughts on “Solving Quadratic Equations I: Factoring (Grouping)”

1. emt training says:

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