Question

How do you factor \[4{x^2} - 16\] ?

Answer 1

Hint: We can solve this using algebraic identities. We use the identity \[{a^2} - {b^2} = (a - b)(a + b)\] to solve the given problem. We can see that 4 and 16 are perfect squares. We can convert the given problem into \[{a^2} - {b^2}\] , since the square of 2 is 4 and square of 4 is 16.

Complete step-by-step answer:
Given, \[4{x^2} - 16\]
We can rewrite it as \[ = {2^2}.{x^2} - {4^2}\]
 \[ = {(2x)^2} - {4^2}\] .
That is it is in the form \[{a^2} - {b^2}\] , where \[a = 2x\] and \[b = 4\] .
We have the formula \[{a^2} - {b^2} = (a - b)(a + b)\] .
Then above becomes,
 \[ = (2x - 4)(2x + 4)\] . These are the factors of the \[4{x^2} - 16\] .
We can find the root of the polynomial by equating the obtained factors to zero. That is
 \[ \Rightarrow (2x - 4)(2x + 4) = 0\]
Using the zero product principle, that is of ab=0 then a=0 or b=0.using this we get,
 \[ \Rightarrow 2x - 4 = 0\] and \[2x + 4 = 0\] .
 \[ \Rightarrow 2x = 4\] and \[2x = - 4\] .
 \[ \Rightarrow x = \dfrac{4}{2}\] and \[x = - \dfrac{4}{2}\] .
 \[ \Rightarrow x = 2\] and \[x = - 2\]
This is the roots of the given polynomial.
We know that on the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis that is the roots are simply the x- intercept.
So, the correct answer is “(2x - 4)(2x + 4)”.

Note: Since the given equation is a polynomial. The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. Here the degree is 2. Hence it is called a quadratic equation. (We know the quadratic equation is of the form \[a{x^2} + bx + c = 0\] , in our problem coefficient of ‘x’ is zero) Hence, we have two roots or two factors.