Question

Given $ 4{x^2} + 28x + 49 $
 Is a perfect square trinomial and how do you factor it?

Answer 1

Hint: To solve this we have to know about algebraic identity.
Quadratic equation: a polynomial equation in which the highest sum of exponents of a variable in any term is two. The value of $ x $ for which the result of the equation is zero then that value will be the root of the quadratic equation.

Complete step-by-step answer:
To solve this question. We will first factorize it:
We try to take common from the given equation $ 4{x^2} + 28x + 49 $ .
Splitting the equation. We get,
Or, $ {\left( {2x} \right)^2} + 2\times 7 \times 2x + {7^2} $
By using $ {(a + b)^2} = {a^2} + {b^2} + 2ab $
We get,
Or, $ {\left( {2x} \right)^2} + 2\times 7\times 2x + {7^2} = {(2x + 7)^2} $
As, it can write in this format hence it is a perfect square trinomial
And the factor of $ 4{x^2} + 28x + 49 $ is $ {(2x + 7)^2} $ .
So, the correct answer is “ $ {(2x + 7)^2} $ ”.

Note: To calculate it root we can do:
Hence LHS is equal to zero so, we can conclude that either,s
   $ {(2x + 7)^2} = 0 $
So, we will get,
  $ (2x + 7) = 0 $
 \[ \Rightarrow x = - \dfrac{7}{2}\]
So, roots of equation $ 4{x^2} + 28x + 49 $ is $ - \dfrac{7}{2} $
Application of the cubic equation is angle trisection and doubling the cube are two ancient problems of geometry that have been proved to not be solvable by straightedge and compass construction, because they are equivalent to solving a cubic equation.