Is $ 9{x^2} + 42x + 49 $ a perfect square trinomial and how do you factor it?

Hint: To prove the perfect square trinomial, the equation given should be represented as the perfect identity. To make the identity firstly, divide the whole equation by coefficient of $ {x^2} $ and then half the coefficient of the x and the form the identity.

Complete step-by-step answer:
The given equation is $ 9{x^2} + 42x + 49 = 0 $
The coefficient of x is $ 9 $ . Firstly, divide the whole equation with the coefficient of x.
 $ \dfrac{{9{x^2}}}{9} + \dfrac{{42x}}{9} + \dfrac{{49}}{9} = 0 $
Which gives $ {x^2} + \dfrac{{42x}}{9} + \dfrac{{49}}{9} = 0 $
Then, half the coefficient of x on both sides
 $
  {x^2} + 2x\left( {\dfrac{{42}}{9} \times \dfrac{1}{2}} \right) + \dfrac{{49}}{9} = 0 \\
  {x^2} + 2x\left( {\dfrac{7}{3}} \right) + \dfrac{{49}}{9} = 0 \\
  $
 $ \ldots \left( 1 \right) $
We know the identity
 $ {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab $
Now take $ a = x $ and $ b = \dfrac{7}{3} $ converting the values of a and b into the identity
 $ {\left( {x + \dfrac{7}{3}} \right)^2} = {\left( x \right)^2} + {\left( {\dfrac{7}{3}} \right)^2} + 2\left( x \right)\left( {\dfrac{7}{3}} \right) $
 $ = {x^2} + \dfrac{{49}}{9} + 2x\left( {\dfrac{7}{3}} \right) $
Which is same as the above equation $ \left( 1 \right) $
Hence, the given question equation is a perfect square trinomial.
Now, the factors for the equation are $ x = - \dfrac{7}{3}, - \dfrac{7}{3} $
So, the correct answer is “ $ x = - \dfrac{7}{3}, - \dfrac{7}{3} $ ”.

Note: We can find the factors of a quadratic equation $ \left( {a{x^2} + bx + c = 0} \right) $ by the formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
We can also find the factors by minimizing the middle term which is known as the factoring method. Students mainly make silly mistakes while dividing the coefficient.