
How do you factor completely: \[2{{x}^{2}}-20x-48\]?
Answer
534.9k+ views
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 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 quadratic equation as:
\[\Rightarrow 2{{x}^{2}}-20x-48\]
We can see that the term is common in the expression therefore, on taking it out as common, we get:
$\Rightarrow 2\left( {{x}^{2}}-10x-24 \right)$
Now the term ${{x}^{2}}-10x-24$ 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:
$a=1$
$b=-10$
$c=-24$
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 $-24$ and the sum should be $-10$
We see that if $m=-12$ and $n=2$, then we get $m+n=-10$ and $m\times n=-24$
So, we will split the middle term as the addition of $-12$ and $2$ .
On substituting, we get:
$\Rightarrow 2\left( {{x}^{2}}-12x+2x-24 \right)$
Now on taking the common terms, we get:
$\Rightarrow 2\left( x\left( x-12 \right)+2\left( x-12 \right) \right)$
Now since the term $\left( x-12 \right)$ is common in both the terms, we can take it out as common and write the equation as:
$\Rightarrow 2\left( x-12 \right)\left( x+2 \right)$, which is the required factored form of the equation.
Note:
It is not necessary that all the quadratic equations would have roots which are integer numbers or real numbers therefore quadratic formula is used to solve these types of questions, the quadratic formula 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 quadratic equation as:
\[\Rightarrow 2{{x}^{2}}-20x-48\]
We can see that the term is common in the expression therefore, on taking it out as common, we get:
$\Rightarrow 2\left( {{x}^{2}}-10x-24 \right)$
Now the term ${{x}^{2}}-10x-24$ 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:
$a=1$
$b=-10$
$c=-24$
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 $-24$ and the sum should be $-10$
We see that if $m=-12$ and $n=2$, then we get $m+n=-10$ and $m\times n=-24$
So, we will split the middle term as the addition of $-12$ and $2$ .
On substituting, we get:
$\Rightarrow 2\left( {{x}^{2}}-12x+2x-24 \right)$
Now on taking the common terms, we get:
$\Rightarrow 2\left( x\left( x-12 \right)+2\left( x-12 \right) \right)$
Now since the term $\left( x-12 \right)$ is common in both the terms, we can take it out as common and write the equation as:
$\Rightarrow 2\left( x-12 \right)\left( x+2 \right)$, which is the required factored form of the equation.
Note:
It is not necessary that all the quadratic equations would have roots which are integer numbers or real numbers therefore quadratic formula is used to solve these types of questions, the quadratic formula 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.
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