How do you factor $ {x^2} - 4x + 24 = 0 $ ?

Hint: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. The roots of an equation are the points on the x- axis that is the roots are simply the x-intercepts.

Factoring the equation and a special formula called completing the square method are the ways to find out the roots of a quadratic equation. In the given question, we have to find the factors of the equation which can be done by any of the mentioned methods.

Complete step by step answer:
The degree of the equation $ {x^2} - 4x + 24 = 0 $ is 2, so the number of roots of the given equation is 2.

On comparing the given equation with the standard quadratic equation – $ a{x^2} + bx + c = 0 $
We get - $ a = 1,\,b = - 4,\,c = 24 $

For factorization, the standard equation is rewritten as $ a{x^2} + {b_1}x + {b_2}x + c = 0 $
In the given question, we have to find the value of $ {b_1} $ and $ {b_2} $ by hit and trial method such that $ {b_1} \times {b_2} = 24 $ (both the numbers should be either positive or negative so that their product is 24) and $ {b_1} + {b_2} = - 4 $

But we are not able to find the values of $ {b_1} $ and $ {b_2} $ by hit and trial method, so we find the roots by completing the square method.
 $
x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(1)(24)} }}{{2(1)}} \\
x = \dfrac{{4 \pm \sqrt {16 - 96} }}{2} \\
\Rightarrow x = \dfrac{{4 \pm \sqrt { - 80} }}{2} = \dfrac{{4 \pm 4\sqrt { - 5} }}{2} \\
\Rightarrow x = 2 + 10i,\,x = 2 - 10i \\
 $

Hence the factors of $ {x^2} - 4x + 24 = 0 $ are $ x - 2 - 10i = 0 $ and $ x - 2 + 10i = 0 $

Note:If we solve an equation by factorization then the condition to form factors is that we have to express b as a sum of two numbers such that their product is equal to the product of $ a $ and $ c $ that is $ a
\times c = {b_1} \times {b_2} $ . $ \sqrt { - 1} $ is an imaginary number and is called iota, denoted by
 $ i $ , thus the two roots of the given equation are imaginary.