Question

How do you factor \[2{{x}^{2}}+22x+56\]?

Answer 1

Hint: First of all take 2 common from all the terms to simplify the expression. Now, use the middle term split method to factorize \[2{{x}^{2}}+22x+56\]. Split 11x into two terms such that their sum equals 11x and the product equals \[28{{x}^{2}}\]. For this process, find the prime factors of 28 and combine them in a suitable way such that the conditions are satisfied. Finally, take the common terms together and write \[2{{x}^{2}}+22x+56\] as a product of two terms.

Complete step by step solution:
Here, we have been asked to factorize the quadratic polynomial \[2{{x}^{2}}+22x+56\].
Clearly we can see that the constant 2 is common in all the terms so taking 2 common from all the terms, we get the expression as:
$\Rightarrow 2{{x}^{2}}+22x+56=2\left( {{x}^{2}}+11x+28 \right)$
Now, we need to factorize the expression $\left( {{x}^{2}}+11x+28 \right)$. Let us use the middle term split method for the factorization. In this case we have to split the middle term which is 11x into two terms such that their sum is 11x and the product is equal to the product of constant term (28) and \[{{x}^{2}}\], i.e. \[28{{x}^{2}}\]. To do this, first we need to find all the prime factors of 28.
We can write \[28=2\times 2\times 7\] as the product of its primes. Now, we have to group these factors such that our conditions of the middle terms split method are satisfied. So, we have,
(i) \[\left( 7x \right)+\left( 4x \right)=11x\]
(ii) \[\left( 7x \right)\times \left( 4x \right)=28{{x}^{2}}\]
Hence, both the conditions of the middle term split method are satisfied. So, the quadratic polynomial can be written as: -
\[\begin{align}
  & \Rightarrow 2{{x}^{2}}+22x+56=2\left( {{x}^{2}}+7x+4x+28 \right) \\
 & \Rightarrow 2{{x}^{2}}+22x+56=2\left[ x\left( x+7 \right)+4\left( x+7 \right) \right] \\
\end{align}\]
Taking (x + 7) common in the R.H.S., we get,
\[\Rightarrow 2{{x}^{2}}+22x+56=2\left( x+7 \right)\left( x+4 \right)\]

Hence, \[2\left( x+7 \right)\left( x+4 \right)\] is the factored form of the given quadratic polynomial.

Note: Here, the main motive to take 2 common from all the terms is to just reduce the calculation as it will be easier to group the prime factors and use the middle term split method. You can also use the discriminant method to solve the question. What we can do is, we will solve the given quadratic expression by substituting it equal to 0 and using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ we will find the two values of x as: x = m and x = n. Now, the product (x – m) (x – n) will be the required factored form.