Question

How do you factor the expression \[2{{x}^{2}}+21x+49=0\]?

Answer 1

Hint: In this problem, we have to find the factor of the given expression \[2{{x}^{2}}+21x+49=0\] by factorization method. We can see that the middle term can be split into two terms as addition of 14 and 7 whose multiplication is equal to \[2\times 49=98=14\times 7\], which is the multiplication of the first term and the last term. We can then take common terms outside to get the factors.

Complete step by step solution:
We know that the given expression is,
\[2{{x}^{2}}+21x+49=0\]
We can first split the middle term to form factors.
We have to expand the middle term i.e. the x term with its coefficient in such a way that their addition is equal to the middle term i.e. 21x, and multiplication is equal to \[2\times 49=98=14\times 7\]
\[\begin{align}
  & \Rightarrow 2\times 49=98=14\times 7 \\
 & \Rightarrow 14+7=21 \\
\end{align}\]
We can apply the above step,
\[2{{x}^{2}}+14x+7x+21=0\]
We can now take the first two terms and the last two terms to take common terms outside, we get
\[\Rightarrow \left( 2{{x}^{2}}+14x \right)+\left( 7x+21 \right)\]
Now we can take the common terms outside, we get
\[\Rightarrow 2x\left( x+7 \right)+7\left( x+7 \right)\]
We can again take the common factor first then the remaining terms to make a factor, we get
\[\Rightarrow \left( 2x+7 \right)\left( x+7 \right)\]
Therefore, the factors are \[\left( 2x+7 \right)\left( x+7 \right)\].

Note: Students make mistakes while taking the common terms which when again multiplied to give the previous result. We should always concentrate on the part of the separation of the middle term, here we have all the terms positive, so we can just add the two numbers to get the coefficient of x.