Question

How do you factor completely: $2{{x}^{2}}-x-10$

Answer 1

Hint: In this question, we have to find the factors of the given quadratic equation. 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:
 $\Rightarrow 2{{x}^{2}}-x-10\to (1)$
Now the term 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 $a{{x}^{2}}+bx+c\to (2)$
On comparing equations $(1)$ and $(2)$, we get:
$a=2$
$b=-1$
$c=-10$
To factorize, we have to find two numbers $p$ and $q$ such that $p+q=b$ and $p\times q=a\times c$.
Therefore, the product should be $-20$ and the sum should be $-1$
We see that if $p=-5$ and $q=4$, then we get $p+q=-1$ and \[p\times q=-20\]
So, we will split the middle term as the addition of $-5$ and $4$. On substituting, we get:
 $\Rightarrow 2{{x}^{2}}-5x+4x-10$
Now on taking the common terms, we get:
$\Rightarrow x\left( 2x-5 \right)+2\left( 2x-5 \right)$
Now since the term $\left( 2x-5 \right)$ is common in both the terms, we can take it out as common and write the equation as:
$\Rightarrow \left( x+2 \right)\left( 2x-5 \right)$, which is the required factored form of the equation.

Note:
It is to be remembered that when doing a factoring sum the alternate method can be used which is called as completing the square method. In this method the expression is converted into the form of ${{a}^{2}}\pm 2ab+{{b}^{2}}$ where in the term $a$, the term regarding $x$ is present and in term $b$, the term regarding the constant is present. Thereafter the equation is simplified as ${{\left( a\pm b \right)}^{2}}$.