Question

How do you factor \[2{{x}^{2}}+40x+200?\]

Answer 1

Hint: We are given \[2{{x}^{2}}+40x+200\] and we are asked to find the factored 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 or 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 answer:
We are given \[2{{x}^{2}}+40x+200\] and 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 2, 40 and 200, we can see that 2 is the common term to each of these terms. So we can separate 2 out of this and our equation will become
\[2{{x}^{2}}+40x+200=2\left( {{x}^{2}}+20x+100 \right)......\left( i \right)\]
Now we will use the middle term to split. In the middle term split, we will 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 \[{{x}^{2}}+20x+100.\] We have a = 1, b = 20 and c = 100. So, we use these values to find two terms which helps us in splitting the middle term. Now we can see that
\[a\times c=1\times 100=100\]
We can see that there are two terms 10 and 10 such that \[10\times 10=100\left[ \text{same as }a\times c \right]\] and 10 + 10 = 20. So we use this to split the middle term. So, \[{{x}^{2}}+20x+100\] becomes
\[{{x}^{2}}+\left( 10+10 \right)x+100\]
Opening the brackets, we get,
\[\Rightarrow {{x}^{2}}+10x+10x+100\]
We take common in the first 2 terms and the last 2 terms. So we get,
\[\Rightarrow x\left( x+10 \right)+10\left( x+10 \right)\]
As (x + 10) is the same, so we get,
\[\Rightarrow {{x}^{2}}+20x+100x=\left( x+10 \right)\left( x+10 \right)......\left( ii \right)\]
Using (i) and (ii)
So, the factor form of \[2{{x}^{2}}+40x+200\] is \[2\left( x+10 \right)\left( x+10 \right).\]

Note: While finding the middle term using the factor of \[a\times c\] we need to keep in mind that when the sign of ‘a’ and ‘c’ are the same then ‘b’ is obtained by addition only, if the sign of ‘a’ and ‘c’ is different then ‘b’ can be obtained using only subtraction. So, as we have a = 1 and c = 100 have the same sign, so ‘b’ is obtained as 10 + 10 = 20 by addition of 10 and 10. We can always cross-check that product of
\[2\left( x+10 \right)\left( x+10 \right)=2\left[ x\left( x+10 \right)+10\left( x+10 \right) \right]\]
On simplifying we get,
\[\Rightarrow 2\left( x+10 \right)\left( x+10 \right)=2\left( {{x}^{2}}+10x+10x+100 \right)\]
Adding the like terms, we get,
\[\Rightarrow 2\left( x+10 \right)\left( x+10 \right)=2\left( {{x}^{2}}+20x+100 \right)\]
\[\Rightarrow 2\left( x+10 \right)\left( x+10 \right)=2{{x}^{2}}+40x+200\]
So our factors are correct.