Question

How do you know when to use factoring or quadratic formula?

Answer 1

Step-1: Describe when to use factoring and when to use the quadratic formula:

Factoring by splitting the linear term is a method of solving quadratic equations.

It is convenient when the solutions of the given quadratic equation are all integers or rational numbers.

The quadratic formula is a more powerful but slightly more complicated method than factoring.

The quadratic formula gives a direct expression of the solutions $x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$ of the quadratic equation $A{x}^2+Bx+C=0;A\ne 0$ in terms of the coefficients $A,B,\text{and}C$.

If the given quadratic polynomial $A{x}^2+Bx+C$ can be factored easily then one should use factoring for solving the quadratic equation.

Sometimes, factoring into simple factors may be impossible depending on the values of the coefficients $A,B,\text{and}C$.

There exist quadratic equations with simple-looking coefficients but still, they are impossible to break into simple factors.

There exist quadratic equations which have only irrational or complex number solutions.

For these cases using the quadratic formula arrives at the solution faster.

Step-2: Illustrate the solution of a particular quadratic equation by factoring.

For example, solve $5x^2-11x-12=0$ by factoring.

Find the product of the quadratic term $5x^2$ and the constant term $-12$, $5x^2\times \left(-12\right)=-60x^2$ .

Through a hit and trial process, find two factors of $-60x^2$ that add up to give the linear term $-11x$.

The two factors that satisfy these conditions are $-15x$ and $4x$, because their sum is equal to the linear term $-11x$ and their product is equal to $-60x^2$.

Rewrite the term $-11x$ in the expression $5x^2-11x-12$ as $-15x+4x$.

Then factoring by grouping the terms:

$\begin{array}{rcl}5x^2-11x-12& =& 5x^2-15x+4x-12\\ & =& \left(5x^2-15x\right)+\left(4x-12\right)\\ & =& 5x\left(x-3\right)+4\left(x-3\right)\\ & =& \left(x-3\right)\left(5x+4\right)\end{array}$

Substitute the factored form of $5x^2-11x-12$ in the equation $5x^2-11x-12=0$, and solve for $x$:

$\begin{array}{rrcl}& 5x^2-11x-12& =& 0\\ \Rightarrow & \left(x-3\right)\left(5x+4\right)& =& 0\\ \Rightarrow & x& =& 3,\frac{-4}{5}\end{array}$

Thus, $5x^2-11x-12=0$ is solved using factoring to get solutions $\begin{array}{rcl}x& =& 3,\frac{-4}{5}\end{array}$.

Step-3: Illustrate the solution of a particular quadratic equation by Quadratic Formula.

As an example, solve $5x^2-11x-12=0$ by using Quadratic Formula.

Compare the equation $5x^2-11x-12=0$ with $A{x}^2+Bx+C=0;A\ne 0$ to determine $A=5,B=-11,\text{and}C=-12$.

Substitute these values in the Quadratic formula $x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$ and simplify to obtain the solutions.

$\begin{array}{rcl}x& =& \frac{-B\pm \sqrt{B^2-4AC}}{2A}\\ & \Rightarrow & x=\frac{-\left(-11\right)\pm \sqrt{{\left(-11\right)}^2-4\left(5\right)\left(-12\right)}}{2\left(5\right)}\\ & \Rightarrow & x=\frac{11\pm \sqrt{121+240}}{10}\\ & \Rightarrow & x=\frac{11\pm \sqrt{361}}{10}\\ & \Rightarrow & x=\frac{11\pm \sqrt{{\left(19\right)}^2}}{10}\\ & \Rightarrow & x=\frac{11\pm 19}{10}\\ & \Rightarrow & x=\frac{11+19}{10},\frac{11-19}{10}\\ & \Rightarrow & x=\frac{30}{10},\frac{-8}{10}\\ & \Rightarrow & x=3,-\frac{4}{5}\end{array}$

Thus, $5x^2-11x-12=0$ is solved using the quadratic formula to get the solutions $\begin{array}{rcl}x& =& 3,\frac{-4}{5}\end{array}$.

Hence, if the quadratic polynomial $A{x}^2+Bx+C$ can be factored easily then the factoring method is a quicker way to solve the quadratic equation $A{x}^2+Bx+C=0;A\ne 0$ otherwise the quadratic formula can be used to arrive at the solution.