Question

How do you write ${{x}^{2}}+2x-8$ in factored form?

Answer 1

Hint: We are given ${{x}^{2}}+2x-8$ , we are asked to find the factor form of this, to do so we will first understand the type of equation we have, once we get that we will find the greatest common factor from each term then in the remaining term be factor using the middle term.
We use $a\times b$ in such a way that its sum of difference from the ‘b’ of the equation $a{{x}^{2}}+bx+c$ .
Once we split that we will unite all factors of the equation and we get our required answer.

Complete step by step solution:
We are given ${{x}^{2}}+2x-8$, we are asked to find the factor of it.
To find the factor of the equation, we should see that as the highest power is 2 so it is or 2 degree polynomial. So it is a quadratic equation.
Now, to factor, we will first find the possible greatest common factor of all these.
In 1, 2 and -8, we can see that there is nothing in common terms which can be separated out so our equation will stay the same.
${{x}^{2}}+2x-8$
Now we will use the middle term to split.
In middle term split apply on $a{{x}^{2}}+bx+c$ , we produce ‘a’ by ‘c’ and then factor ‘ac’ in such a way that if the product is ‘ac’ while the sum or difference is made up to ‘b’.
Now, we have a middle term split on ${{x}^{2}}+2x-8$.
We have $a=1,b=2\text{ and }c=-8$ .
So, we use these values to find two terms which help us in splitting the middle term.
Now we can see that –
$a\times c=1\times -8=-8$ .
We can see that there are two terms 4 and -2.
Such that $4\times \left( -2 \right)=-8$ (same as $a\times c$ )
And $4-2=2$ .
So, we use this to split the middle term.
So,
${{x}^{2}}+2x-8$become
$\Rightarrow {{x}^{2}}+\left( 4-2 \right)x-8$
Opening brackets
${{x}^{2}}+4x-2x-8$
We take common in the first 2 terms and the last 2 terms. So we get –
$x\left( x+4 \right)-2\left( x+4 \right)$
As $x+4$ is same, so we get –
$\left( x+4 \right)\left( x-2 \right)$
So, we get –
${{x}^{2}}+2x-8=\left( x+4 \right)\left( x-2 \right)$

So, the factor form of ${{x}^{2}}+2x-8=\left( x+4 \right)\left( x-2 \right)$ .

Note: While finding the middle term using factor of $a\times c$ , we need to keep in mind that when the sign of ‘a’ and ‘c’ are same then ‘b’ is obtained by addition only, if the sign of ‘a’ and ‘c’ are different then ‘b’ can be obtained using only subtraction.
We can always cross check that –
Product of $\left( x+4 \right)\left( x-2 \right)=x\left( x-2 \right)+4\left( x-2 \right)$
By simplifying, we get –
$={{x}^{2}}-2x+4x-8$
Adding like terms we get –
$={{x}^{2}}+2x-8$
So, our factors are correct.