Question

How do you factorize $16{x^4} - 1 = 0$ ?

Answer 1

Hint: In this question, we are given a polynomial equation of degree 4 as the highest exponent in the given equation is 4, we know that a polynomial equation has exactly as many solutions as the degree of the equation so the given equation has four solutions. In this question, we will simply bring 1 to the other side of the equal to sign and divide both sides by 16. Then by prime factorization of $\dfrac{1}{{16}}$ ; we will write as a square of some number, then square rooting both the sides of the equation we get two numbers. Again expressing the obtained numbers as a square of some other numbers, we will get four values of x. Thus, the values obtained will be the roots of the given equation.

Complete step-by-step solution:
We are given –
$
  16{x^4} - 1 = 0 \\
   \Rightarrow {x^4} = \dfrac{1}{{16}} \\
   \Rightarrow {({x^2})^2} = {( \pm \dfrac{1}{4})^2} \\
   \Rightarrow {x^2} = \pm \dfrac{1}{4} \\
 $
Now, ${x^2} = \dfrac{1}{4}$ and ${x^2} = - \dfrac{1}{4}$
$
   \Rightarrow {x^2} = {( \pm \dfrac{1}{2})^2},\,{x^2} = {( \pm \dfrac{1}{2}\sqrt { - 1} )^2} \\
   \Rightarrow x = \pm \dfrac{1}{2},\,x = \pm \dfrac{1}{2}i \\
 $
Hence the factors of the equation $16{x^4} - 1$ are $x - \dfrac{1}{2} = 0,\,\,x + \dfrac{1}{2} = 0,\,\,x - \dfrac{1}{2}i = 0\,\,and\,\,x + \dfrac{1}{2}i = 0$ .


Note: The values of the x for which the function given as $f(x) = 16{x^4} - 1$ has a value zero are known as the solutions of the function or when we plot this function on the graph, we see the solutions of this equation are the points on which the y-coordinate is zero, thus they are simply the x-intercepts. We can also solve the given equation by using an arithmetic identity, according to which the difference of the square of one number a and the square of another number b is equal to the product of the difference of a and b and the sum of a and b, that is, ${a^2} - {b^2} = (a - b)(a + b)$ .