Question

How do you factor completely ${x^2} - 16 = 0$ ?

Answer 1

Hint: Quadratic Equation: A quadratic equation is a equation which can be represented in the form of $a{x^2} + bx + c$ where $x$ is the unknown variable and a,b,c are the numbers known where $a \ne 0$.If $a = 0$ then the equation will become linear equation and will no more quadratic.
In order to determine the factors and solution of the above quadratic question using the identity $\left( {{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right)$

Complete step by step solution:
Given a quadratic equation ${x^2} - 16 = 0$ ,let it be $f(x)$
$f(x) = {x^2} - 16 = 0$
Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 1
b becomes 0
And c becomes -16
To find the quadratic factorization we’ll be writing the expression as
$f\left( x \right) = {(x)^2} - {(4)^2}$
 Consider $x$ as A and $4$ as B and Applying Identity $\left( {{A^2} - {B^2}} \right) = \left( {A - B} \right)\left( {A + B} \right)$
Now our equation becomes
$
   \Rightarrow f\left( x \right) = (x - 4)(x + 4) = 0 \\
   \Rightarrow x + 4 = 0 \\
   \Rightarrow x = - 4 \\
   \Rightarrow x - 4 = 0 \\
   \Rightarrow x = 4 \\
 $
$\therefore x = 4, - 4$
Hence, We have successfully factorized our quadratic equation
Therefore, the factors are $(x - 4)$ and $(x + 4)$ and the solution of the equation is $\therefore x = 4, - 4$.

Note: You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
Therefore, x1, x2 are root or solution to quadratic equation $a{x^2} + bx + c$
 Hence the factors will be $(x - x1)\,and\,(x - x2)\,$.