Question

How do you factor completely: $3{{x}^{2}}+12x+7$?

Answer 1

Hint: In this question we will simplify the polynomial by taking out the common term $x$, then we will solve the simplified quadratic equation and find the factors of the equation, which are the multiplication of linear terms, which will give us the required values of $x$. Since in this question we cannot solve the equation by splitting the middle term, we will use the quadratic formula which is:
$({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, Where $({{x}_{1}},{{x}_{2}})$ are the roots of the equation and $a,b,c$ are the coefficients of the quadratic equation.

Complete step by step solution:
We have the given equation as:
$\Rightarrow 3{{x}^{2}}+12x+7$
Now we the term $3{{x}^{2}}+12x+7$ in the form of a quadratic equation which has a general form $a{{x}^{2}}+bx+c$, therefore we solve it by splitting the middle term. Now in this equation, splitting the middle term is not possible therefore, we will use the quadratic formula.
In this question we have:
$a=3$
$b=12$
$c=7$
On substituting the values in the formula, we get:
$\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-12\pm \sqrt{{{12}^{2}}-4(3)(7)}}{2(3)}$
On simplifying the root, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-12\pm \sqrt{144-84}}{2(3)}\]
On simplifying the denominator, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-12\pm \sqrt{144-84}}{6}\]
On simplifying the terms in the square root, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-12\pm \sqrt{60}}{6}\]
Now we can write the term $60$ as a product of $4\times 15$, on substituting, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-12\pm \sqrt{4\times 15}}{6}\]
On taking the term $4$ out of the square root, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-12\pm 2\sqrt{15}}{6}\]
On splitting the denominator, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-12}{6}\pm \dfrac{2\sqrt{15}}{6}\]
On simplifying the terms, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=-6\pm \dfrac{\sqrt{15}}{3}\]
Therefore, the roots of the equation are \[{{x}_{1}}=-6+\dfrac{\sqrt{15}}{3}\] and \[{{x}_{2}}=-6-\dfrac{\sqrt{15}}{3}\].
Now the equation can be written in the factorized format as $\left( x-{{r}_{1}} \right)\left( x+{{r}_{2}} \right)$.
On substituting the roots, we get:
$\Rightarrow \left( x-\left( -6+\dfrac{\sqrt{15}}{3} \right) \right)\left( x-\left( -6-\dfrac{\sqrt{15}}{3} \right) \right)$
On simplifying the brackets, we get:
$\Rightarrow \left( x-6-\dfrac{\sqrt{15}}{3} \right)\left( x-6+\dfrac{\sqrt{15}}{3} \right)$, which is the required solution.

Note:
It is to be remembered that a polynomial equation is a combination of variables and their coefficients, the equation given above is a $2nd$ degree polynomial equation.
The factors of a polynomial equation are the terms of degree $1$, which have to be multiplied among themselves so that we get the required polynomial equation. In the question above \[-6+\dfrac{\sqrt{15}}{3}\] and \[-6-\dfrac{\sqrt{15}}{3}\] are factors of the given polynomial.