Question

How do you factor $ {x^3} - 3{x^2} + 4x - 12 = 0 $ ?

Answer 1

Hint: Factorization: Mathematics, factorization or factoring is defined as the breaking or decomposition of an entity (for example a number, a matrix, or a polynomial) into a product of another entity, or factors, which when multiplied together give the original number or a matrix, etc.
Cubic equation: the equation in which the highest power of the given polynomial is three then it is the cubic equation.

Complete step by step solution:
As we have $ {x^3} - 3{x^2} + 4x - 12 = 0 $
So let,
Step 1: Take $ {x^2} $ common form first two terms. We get,
  $ {x^2}(x - 3) + 4x - 12 = 0 $
Step 2: Take $ 4 $ as a common form in the last two terms. We get,
  $ {x^2}(x - 3) + 4(x - 3) = 0 $
Step 3:
Take $ (x - 3) $ common form both forms. We get,
  $ (x - 3)({x^2} + 4) = 0 $
So, be can write factor of $ {x^3} - 3{x^2} + 4x - 12 = 0 $ is $ (x - 3)({x^2} + 4) = 0 $
We can further factorize it by using complex number;
We can write,
  $ (x - 3)({x^2} + 4) = 0 $ as $ (x - 3)({x^2} - ( - 4)) = 0 $ .
As we know $ \sqrt { - 4} = 2i, - 2i $
So, $ (x - 3)({x^2} - ( - 4)) = (x - 3)(x - 2i)(x + 2i) $
Roots of the equation will be $ (x - 3)(x - 2i)(x + 2i) = 0 $
From above:
  $ x = 3,2i, - 2i $
So, the correct answer is “ $ x = 3,2i, - 2i $ ”.

Note: In mathematics, a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a symbol called the imaginary unit. ‘i’ is an imaginary number, it was called an imaginary number by René Descartes.