Question

How do you factor by grouping $ 4{b^2} + 4bc + {c^2} - 16 $ ?

Answer 1

Hint: To solve this problem we should know about the basic property of algebraic equations.
We have an quadratic equation or cubic equation or equation having greater power but we have to first of all simplify that equation and try to change into format of known equation then we will use algebraic equations:
  $ {a^2} - {b^2} = (a - b)(a + b) $
  $ {(a + b)^2} = {a^2} + {b^2} + 2ab $
  $ {(a - b)^2} = {a^2} + {b^2} - 2ab $

Complete step by step solution:
As given equation is $ 4{b^2} + 4bc + {c^2} - 16 $
We have to factor it by grouping.
Observing the first three terms of a given equation. We get,
  $ \left( {{{(2b)}^2} + 2.(2b).c + {c^2}} \right) - 16 $
We notice it resemble like $ {(a + b)^2} = {a^2} + {b^2} + 2ab $
So, we can rewrite it as,
  $ = {\left( {2b + c} \right)^2} - 16 $
We can rewrite it as,
  $ = {\left( {2b + c} \right)^2} - {4^2} $
By comparing it with $ {a^2} - {b^2} = (a - b)(a + b) $ . We get,
  $ = \left( {\left( {2b + c} \right) - 4} \right)\left( {\left( {2b + c} \right) + 4} \right) $
  $ = \left( {2b + c - 4} \right)\left( {2b + c + 4} \right) $
Hence, factor by grouping $ 4{b^2} + 4bc + {c^2} - 16 = \left( {2b + c - 4} \right)\left( {2b + c + 4} \right) $ .
So, the correct answer is “ $ \left( {2b + c - 4} \right)\left( {2b + c + 4} \right) $ .”.

Note: Algebraic equation is used in solving complex problems in physics, trigonometric equation, applied engineering, scientific data. When we have to find the solution of a quartic or cubic equation we simply make it equal to zero. Easily factor it by grouping we get the roots of the given equation.