Solving Quadratic Equations by Factoring

This calculator allows you to factor a quadratic equation that you provide, showing all the steps of the process. All you need to do is to provide a valid quadratic equation.

An example of a valid quadratic equation is 2x² + 5x + 1 = 0. You can also provide a quadratic equation that is not completely simplified, like for example, x² - 3/4 x + 2 = 3x - 2x², and this calculator will simplify it for you.

Once you provide a valid quadratic equation, you need to click on "Calculate", and all the steps of the process will be shown to you you.

Factoring quadratic equations is one of the methods for finding roots, but it is considered a rather "naive" method, since it is a "try and test" method, that only works well for integer and fractional roots.

How to do factoring of quadratic equations?

The process is simple, but it has limited potential results, because it only works potentially fine when the the quadratic equation has very simple roots:

What are the steps for solving quadratic equations by factoring ?

• Step 1: Identify the quadratic equation you want to solve and simplify into its form ax² + bx + c = 0
• Step 2: Investigate the coefficients a and c. If they are not integer, your changes of "guessing" the factors is nill
• Step 3: If the coefficients a and c are integer, find their integer divisors a₁, a₂, ...., and c₁, c₂,... etc. You will try to guess a solution of the equation testing the fractions of the form c_i/a_k
• Step 4: Finding roots r₁ and r₂ with this method will lead to a factorization of the form ax² + bx + c = a(x - r₁)(x - r₂) = 0

The limitation of this method is that you may not be able to guess the solutions, as the solutions may not be rational. In other words, there is not a simple formula for factoring, you rather follow a guessing process.

Now, regardless of its limitations, the solving quadratic equations with factoring is a good and quick alternative when the roots to the equation are very simple.

Why would care about factoring quadratic fractions?

Factoring plays a very important role in different contexts, and ultimately, solving a general quadratic equation relies on a sophisticated and elegant factoring process.

Often times you will use factoring within an equation not necessarily to solve the equation, but rather to group terms..

Example: Factoring Quadratic Equations

Solve the following equation by factoring \(4x^2 + 4x + 1 = 0\)

Solution:

We need to try to solve the following given quadratic equation \(\displaystyle 4x^2+4x+1=0\) by factoring.

In this case, we have that the equation we need to try to factor is \(\displaystyle 4x^2+4x+1 = 0\), which implies that corresponding coefficients are:

\[a = 4\] \[b = 4\] \[c = 1\]

Now, we need to find the integer numbers that divide \(a\) and \(c\), that will be used to construct our candidates to be factors.

The dividers of \(a = 4\) are: \(\pm 1,\pm 2,\pm 4\).

The dividers of \(c = 1\) are: \(\pm 1\).

Therefore, dividing each divider of \(c = 1\) by each divider of \(a = 4\), we find the following list of candidates to be factors:

\[\pm \frac{ 1}{ 1},\pm \frac{ 1}{ 2},\pm \frac{ 1}{ 4}\]

Now, all the candidates need to be tested to see if they are a solution. The following is obtained from testing each candidates:

\[\begin{array}{cccccccc} x & = & \displaystyle -1 &:&    & \displaystyle 4 \left(-1\right)^2+4 \left(-1\right)+1 & = & \displaystyle 1 \ne 0 \\\\ x & = & \displaystyle 1 &:&    & \displaystyle 4 \left(1\right)^2+4 \left(1\right)+1 & = & \displaystyle 9 \ne 0 \\\\ x & = & \displaystyle -\frac{1}{2} &:&    & \displaystyle 4 \left(-\frac{1}{2}\right)^2+4 \left(-\frac{1}{2}\right)+1 & = & \displaystyle 0 = 0 \\\\ x & = & \displaystyle \frac{1}{2} &:&    & \displaystyle 4 \left(\frac{1}{2}\right)^2+4 \left(\frac{1}{2}\right)+1 & = & \displaystyle 4 \ne 0 \\\\ x & = & \displaystyle -\frac{1}{4} &:&    & \displaystyle 4 \left(-\frac{1}{4}\right)^2+4 \left(-\frac{1}{4}\right)+1 & = & \displaystyle \frac{1}{4} \ne 0 \\\\ x & = & \displaystyle \frac{1}{4} &:&    & \displaystyle 4 \left(\frac{1}{4}\right)^2+4 \left(\frac{1}{4}\right)+1 & = & \displaystyle \frac{9}{4} \ne 0 \\\\ \end{array}\]

So, only one of the candidates, \(x = \displaystyle -\frac{1}{2}\) turns out to be a root, so then the we have that the given quadratic equation can be factored as \( 4 \left(x+\frac{1}{2}\right)^2 = 0\).

Example: Solving quadratic equations by factoring

Solve the following quadratic equation by factoring \(x^2 + 5x + 6 = 0\)

Solution: We need to try to factor \(\displaystyle x^2+5x+6 = 0\), so then corresponding coefficients are:

\[a = 1\] \[b = 5\] \[c = 6\]

Now, we need to find the integer numbers that divide \(a\) and \(c\), that will be used to construct our candidates to be factors.

The dividers of \(a = 1\) are: \(\pm 1\).

The dividers of \(c = 6\) are: \(\pm 1,\pm 2,\pm 3,\pm 6\).

Therefore, dividing each divider of \(c = 6\) by each divider of \(a = 1\), we find the following list of candidates to be factors:

\[\pm \frac{ 1}{ 1},\pm \frac{ 2}{ 1},\pm \frac{ 3}{ 1},\pm \frac{ 6}{ 1}\]

Now, all the candidates need to be tested to see if they are a solution. The following is obtained from testing each candidates:

\[\begin{array}{cccccccc} x & = & \displaystyle -1 &:&    & \displaystyle 1 \left(-1\right)^2+5 \left(-1\right)+6 & = & \displaystyle 2 \ne 0 \\\\ x & = & \displaystyle 1 &:&    & \displaystyle 1 \left(1\right)^2+5 \left(1\right)+6 & = & \displaystyle 12 \ne 0 \\\\ x & = & \displaystyle -2 &:&    & \displaystyle 1 \left(-2\right)^2+5 \left(-2\right)+6 & = & \displaystyle 0 = 0 \\\\ x & = & \displaystyle 2 &:&    & \displaystyle 1 \left(2\right)^2+5 \left(2\right)+6 & = & \displaystyle 20 \ne 0 \\\\ x & = & \displaystyle -3 &:&    & \displaystyle 1 \left(-3\right)^2+5 \left(-3\right)+6 & = & \displaystyle 0 = 0 \\\\ x & = & \displaystyle 3 &:&    & \displaystyle 1 \left(3\right)^2+5 \left(3\right)+6 & = & \displaystyle 30 \ne 0 \\\\ x & = & \displaystyle -6 &:&    & \displaystyle 1 \left(-6\right)^2+5 \left(-6\right)+6 & = & \displaystyle 12 \ne 0 \\\\ x & = & \displaystyle 6 &:&    & \displaystyle 1 \left(6\right)^2+5 \left(6\right)+6 & = & \displaystyle 72 \ne 0 \\\\ \end{array}\]

So, two of the candidates turn out to be roots, \(x_1 = \displaystyle -2\) and \(x = \displaystyle -3\), so then the we have found our solutions, and we can factor the given equation as \( \displaystyle \left(x+2\right)\left(x+3\right) = 0\).

Other useful quadratic calculators

The quadratic formula is really one of the most important ones in basic Algebra, and it has applications in many contexts. You may want to calculate a quadratic equation, you may want to express it in vertex form, there are lots of possibilities.

There are may elements that are all tied together, like the quadratic equation discriminant, or the axis of symmetry of a parabola. All those elements are tightly related and play an important role together.