Question

How do you factor completely: \[2{{x}^{2}}+20={{x}^{2}}+9x\]?

Answer 1

Hint: In this question we have to find the factors of the given polynomial. We will first transfer the terms from the right-hand side to the left-hand side and simplify the expression to get a quadratic equation in the form of $a{{x}^{2}}+bx+c$. We will then solve it by splitting the middle term to get the factors for the same. We start solving this problem 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.

Complete step by step solution:
We have the expression as:
\[\Rightarrow 2{{x}^{2}}+20={{x}^{2}}+9x\]
On transferring the terms from the right-hand side to the left-hand side, we get:
\[\Rightarrow 2{{x}^{2}}+20-{{x}^{2}}-9x=0\]
On simplifying the expression, we get:
\[\Rightarrow {{x}^{2}}+20-9x=0\]
On rearranging the terms in the equation, we get:
\[\Rightarrow {{x}^{2}}-9x+20=0\]
Now we have the above expression in the form of a quadratic equation in the form of $a{{x}^{2}}+bx+c$
On comparing both the equations, we can see that:
$a=1$
$b=-9$
$c=20$
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 $20$ and the sum should be $-9$
We see that if $m=-4$ and $n=-5$, then we get $m+n=-9$ and $m\times n=20$
So, we will split the middle term as the addition of $-4$ and $-5$.
On substituting, we get:
\[\Rightarrow {{x}^{2}}-4x-5x+20=0\]
Now on taking the common terms, we get:
$\Rightarrow x\left( x-4 \right)-5\left( x-4 \right)=0$
Now since the term $\left( x-4 \right)$ is common in both the terms, we can take it out as common and write the equation as:
$\Rightarrow \left( x-4 \right)\left( x-5 \right)=0$, which is the required factored form of the equation.

Note:
The alternate method to solve the quadratic equation is 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.
On substituting the values of $a,b$ and $c$, we get:
$\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-9\pm \sqrt{{{\left( -9 \right)}^{2}}-4\left( 1 \right)\left( 20 \right)}}{2\left( 1 \right)}$
On simplifying the terms in the root and the denominator, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-9\pm \sqrt{81-80}}{2}\]
On simplifying, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-9\pm 1}{2}\]
On splitting the solutions, we get:
\[{{x}_{1}}=\dfrac{-9+1}{2}\] and \[{{x}_{2}}=\dfrac{-9-1}{2}\]
On simplifying, we get:
\[{{x}_{1}}=-4\] and ${{x}_{2}}=-5$, 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:
$\Rightarrow \left( x-4 \right)\left( x-5 \right)$ which is the required solution.