Question

How do you factor completely ${x^2} - 16x + 64$ ?

Answer 1

To order to determine the factors of the above quadratic question use the Splitting up the middle

Complete step by step solution:
Given a quadratic equation ${x^2} - 16x + 64$ ,

let it be $f(x)$
$f(x) = {x^2} - 16x + 64$

Comparing the equation with the standard Quadratic equation $a{x^2} + bx + c$
a becomes 1
b becomes -16
And c becomes 64

To find the quadratic factorization we’ll use splitting up the middle term method
So first calculate the product of coefficient of ${x^2}$ and the constant term which comes to be
$= 64 \times 1 = 64$

Now the second Step is to find the 2 factors of the number 64 such that the whether addition or subtraction of those numbers is equal to the middle term or coefficient of x and the product of those factors results in the value of constant .

So if we factorize 64 ,the answer comes to be -8 and -8 as $- 8 - 8 = - 16$ that is the middle term . and $8 \times 8 = 64$ which is perfectly equal to the constant value.

Now writing the middle term sum of the factors obtained ,so equation $f(x)$ becomes
$f(x) = {x^2} - 8x - 8x + 64$

Now taking common from the first 2 terms and last 2 terms
$f(x) = x(x - 8) - 8(x - 8)$

Finding the common binomial parenthesis, the equation becomes
$f(x) = (x - 8)(x - 8)$

Hence , We have successfully factorized our quadratic equation.

Therefore the factors are $(x - 8)$ and $(x - 8)$

Alternative:
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 $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$

x1,x2 are root to quadratic equation $a{x^2} + bx + c$

Hence the factors will be $(x - x1)\,and\,(x - x2)\,$ .

Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given quadratic equation with the standard one every time.