Question

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

Answer 1

Hint: We will use the method of splitting the middle term to factorize the given polynomial. In order to do this, we will write the middle term as a sum of two terms. Then, we will club together like terms and take common factors out. Finally, we will get the polynomial as a product of two factors.

Complete step by step solution:
The quadratic polynomial given to us is \[{x^2} - 16\]. We have to factorize this polynomial i.e.; we have to find two factors such that the given polynomial can be expressed as a product of the two factors. To do this, we will use the method of splitting the middle term.
In the given polynomial \[{x^2} - 16\], we have to split \[0\] as a sum of two terms whose product is \[1 \times ( - 16) = - 16\]. Let us find the factors of 16 and find combinations of numbers that in addition or subtraction will give us \[0\].
We know that \[16 = 2 \times 2 \times 2 \times 2\].
We can also write this as:
 \[16 = 4 \times 4\]
We also know that \[0 = ( - 4) + 4\].
 So, the required numbers are \[ - 4\] and \[4\]. Hence, the polynomial can be written as
\[{x^2} - 16 = {x^2} - 4x + 4x - 16\]
We will club the first two terms and the last two terms together. In the first two terms the common factor is \[x\]. In the last two terms, the common factor is \[4\]. Thus,
\[ \Rightarrow {x^2} - 16 = x(x - 4) + 4(x - 4)\]
Factoring out \[(x - 4)\] from both the terms, we get

\[ \Rightarrow {x^2} - 16 = (x - 4)(x + 4)\]

Therefore, we have expressed the given polynomial as a product of two factors.

Note:
To factorize a polynomial \[a{x^2} + bx + c\] by splitting the middle term, we will write \[b\] as a sum of the two terms such that their product is \[a \times c\]. This means that we find two numbers \[p\] and \[q\] such that \[p + q = b\] and \[pq = ac\]. After finding \[p\] and \[q\], we split the middle term in the quadratic polynomial as \[px + qx\] and get desired factors by grouping the terms. Here, the terms \[p\] and \[q\] are not necessarily positive terms.