Question

How do you factor completely: \[2{{x}^{2}}-12x+18\]?

Answer 1

Hint: In this question, we have to find the factors of the given quadratic equation. We will first take out the common term $2$ from the expression and then we will solve it by using 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. We will then write the factored equation using the general format which is $\left( x+{{x}_{1}} \right)\left( x+{{x}_{2}} \right)$.

Complete step by step solution:
We have the quadratic equation as:
\[\Rightarrow 2{{x}^{2}}-12x+18\]
We can see that the term $2$ is common in the expression therefore, on taking it out as common, we get:
\[\Rightarrow 2\left( {{x}^{2}}-6x+9 \right)\]
Now the term \[{{x}^{2}}-6x+9\] is in the form of a quadratic equation therefore we will solve it by splitting the middle term.
As we know, the general form of quadratic equation is $a{{x}^{2}}+bx+c$
On comparing both the equations, we can see that:
$a=1$
$b=-6$
$c=9$
On substituting the values of $a,b$ and $c$, we get:
$\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-6\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 1 \right)\left( 9 \right)}}{2\left( 1 \right)}$
On simplifying the terms in the root and the denominator, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-6\pm \sqrt{36-36}}{2}\]
On simplifying, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-6\pm 0}{2}\]
On splitting the solutions, we get:
\[{{x}_{1}}=\dfrac{-6}{2}\] and \[{{x}_{2}}=\dfrac{-6}{2}\]
On simplifying, we get:
\[{{x}_{1}}=-3\] and \[{{x}_{2}}=-3\], which are the terms we used to split the middle term.
The factored form can be written as \[\left( x+{{x}_{1}} \right)\left( x+{{x}_{2}} \right)\] therefore the solution is:
\[\Rightarrow 2\left( x-3 \right)\left( x-3 \right)\]
On simplifying, we get:
\[\Rightarrow 2{{\left( x-3 \right)}^{2}}\], which is the required solution.

Note:
The alternate method to solve this quadratic equation is by splitting the middle term. We can solve it by finding two numbers such that the product of the two numbers is equal to the product of the coefficient of ${{x}^{2}}$ and the constant. Also, the sum of these two numbers is the coefficient of $x$ . Then, after finding these numbers, we split the middle term as the sum of those two numbers and simplify to get the required solution.
We have the quadratic equation as:
\[\Rightarrow 2{{x}^{2}}-12x+18\]
We can see that the term $2$ is common in the expression therefore, on taking it out as common, we get:
\[\Rightarrow 2\left( {{x}^{2}}-6x+9 \right)\]
Now the term \[{{x}^{2}}-6x+9\] is in the form of a quadratic equation therefore we will solve it by splitting the middle term.
As we know, the general form of quadratic equation is $a{{x}^{2}}+bx+c$
On comparing both the equations, we can see that:
$a=1$
$b=-6$
$c=9$
To factorize, we have to find two numbers $m$ and $n$ such that $m+n=b$ and $m\times n=a\times c$.
Therefore, the product should be $9$ and the sum should be $-6$.
We see that if $m$ and $n$ equal to $-3$, then we get $m+n=-6$ and $m\times n=9$
So, we will split the middle term as the addition of $-3$ and $-3$.
On substituting, we get:
\[\Rightarrow 2\left( {{x}^{2}}-3x-3x+9 \right)\]
Now on taking the common terms, we get:
\[\Rightarrow 2\left( x\left( x-3 \right)-3\left( x-3 \right) \right)\]
Now since the term \[\left( x-3 \right)\] is common in both the terms, we can take it out as common and write the equation as:
\[\Rightarrow 2\left( x-3 \right)\left( x-3 \right)\]
On simplifying the expression, we get:
\[\Rightarrow 2{{\left( x-3 \right)}^{2}}\], which is the required factored form of the equation.