Question

How do you find the completely factored ${{x}^{2}}+5x+6$?

Answer 1

Hint: Now we are given with a quadratic equation. Now to find the roots of the quadratic equation we will solve the equation by completing the square method. Hence we will add and subtract the term ${{\left( \dfrac{b}{2a} \right)}^{2}}$ and then simplify the equation using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ we will simplify the obtained equation and take square root to eliminate the square. Now we will simplify the equation to find the value of x which is nothing but the roots of the equation. Hence we will find the corresponding factors.

Complete step by step solution:
Now consider the given expression ${{x}^{2}}+5x+6$.
Now the given expression is a quadratic expression in x.
Now we will first find the roots of the given expression.
To find the roots of the expression we have ${{x}^{2}}+5x+6=0$ .
Now we will use the completing square method to find the roots of the expression.
Now to use this method we need the coefficient if ${{x}^{2}}$ must be one.
Here we already have the coefficient as 1.
Now we will add and subtract the term ${{\left( \dfrac{b}{2a} \right)}^{2}}={{\left( \dfrac{5}{2} \right)}^{2}}=\dfrac{25}{4}$ on both sides.
Hence we get,
$\Rightarrow {{x}^{2}}+5x+\dfrac{25}{4}-\dfrac{25}{4}+6=0$
Now we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Hence we get,
$\Rightarrow {{\left( x+\dfrac{5}{2} \right)}^{2}}-\dfrac{25}{4}+6=0$
Now simplifying the above equation we get,
$\begin{align}
  & \Rightarrow {{\left( x+\dfrac{5}{2} \right)}^{2}}=\dfrac{25}{4}-6 \\
 & \Rightarrow {{\left( x+\dfrac{5}{2} \right)}^{2}}=\dfrac{25-24}{4} \\
 & \Rightarrow {{\left( x+\dfrac{5}{2} \right)}^{2}}=\dfrac{1}{4} \\
\end{align}$
Now taking square root on both sides we get,
$\Rightarrow \left( x+\dfrac{5}{2} \right)=\dfrac{\pm 1}{2}$
Now rearranging the terms we get,
$x=\dfrac{-5}{2}\pm \dfrac{1}{2}$
Hence we get the roots of the equation are $\dfrac{-5+1}{2}$ and $\dfrac{-5-1}{2}$ .
Hence the roots of the equation are – 2 and – 3
Hence we have the factors of the expression as $\left( x-\left( -3 \right) \right)\left( x-\left( 2 \right) \right)$
The factors of the given equation are $\left( x+3 \right)\left( x+2 \right)$

Note: Now note that to find the roots of the quadratic equation of the form $a{{x}^{2}}+bx+c$ we can directly use the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and hence substitute the values of a, b and c. hence we get the solution of the quadratic equation.