Question

How do you factor the expression \[{{4}}{{{x}}^{{2}}}{{ + 4x + 1}}\] ?

Answer 1

Hint: We are given with a quadratic equation of the form \[a{x^2} + bx + c = 0\] . In order to find its factors here we will split the middle term and then take the common terms from the first two terms and last two terms separately. This separation will give us the factors of the expression. But if we observe that the discriminant if the above equation is zero so the roots will be the same.

Complete step-by-step answer:
Given that \[{{4}}{{{x}}^{{2}}}{{ + 4x + 1}}\] is the expression. It is a quadratic equation of the form \[a{x^2} + bx + c = 0\] . so very first equate it to 0.
 \[ \Rightarrow {{4}}{{{x}}^{{2}}}{{ + 4x + 1 = 0}}\]
 Now here we will factorize the middle term.
So 4x can be written as \[{{2x + 2x}}\] . Then let’s rearrange the expression,
 \[ \Rightarrow {{4}}{{{x}}^{{2}}} + {{2x + 2x + 1 = 0}}\]
Now taking \[{{2x}}\] common from first two terms and 1 common from last two terms we get,
 \[ \Rightarrow 2x\left( {{{2x + 1}}} \right){{ + 1}}\left( {{{2x + 1}}} \right) = 0\]
Now let’s rewrite the expression in factored form as,
 \[ \Rightarrow \left( {{{2x + 1}}} \right)\left( {{{2x + 1}}} \right) = 0\]
So finally we get the factors of the expression or roots of the quadratic equation as
 \[ \Rightarrow \left( {{{2x + 1}}} \right) = 0\]
Hence \[ \Rightarrow 2x = - 1\]
And at last value of x as
 \[ \Rightarrow x = \dfrac{{ - 1}}{2}\]
So factors of the expression are \[{{2x + 1}}\] and \[{{2x + 1}}\] .
This is nothing but \[ \Rightarrow {{4}}{{{x}}^{{2}}}{{ + 4x + 1 = }}{\left( {{{2x + 1}}} \right)^2}\] .
So, the correct answer is “ \[ {{4}}{{{x}}^{{2}}}{{ + 4x + 1 = }}{\left( {{{2x + 1}}} \right)^2}\] ”.

Note: Note that finding the factors means to split the middle term of the expression such that we can take common terms from the first two and last two terms respectively. Here the given expression is of the form of a quadratic equation. So we can cross check the factors by finding the roots also.
At the end we observed that the expression is a perfect square also. So simplify the solution as much as you can.