How do you factor $ 2{x^2} + 22x + 60 $ ?

Hint: In these type of polynomials, we will solve by using splitting middle term, for a polynomial of the form $ a{x^2} + bx + c = 0 $ , rewrite the middle term as a sum of two terms whose product is $ a \cdot c = 1 \cdot 30 = 30 $ and whose sum is $ b = 11 $ , when you solve the expression we will get the required values.

Complete step-by-step solution:
The given expression is a polynomials which are algebraic expressions that are composed of two or more algebraic terms, the algebraic terms are constant, variables and exponents, there are different types of polynomials which are differentiated with the help of degree of the polynomial.
The polynomial of degree 2 is known as quadratic polynomial. The given polynomial is a quadratic polynomial.
Now given equation is $ 2{x^2} + 22x + 60 $ ,
Now taking out the 2 common we get,
 $ 2\left( {{x^2} + 11x + 30} \right) $ ,
This can be factored by splitting the middle term, now rewrite the middle term as a sum of two terms, we will get two terms. Now sum of two terms whose product is $ a \cdot c = 1 \cdot 30 = 30 $ and whose sum is $ b = 11 $ .
Now rewrite 11 as 5 and 6, then the equation becomes,
Now using distributive property, we get
 $ \Rightarrow 2\left( {{x^2} + 5x + 6x + 30} \right) $ ,
By grouping the first two terms and last two terms, we get,
 $ \Rightarrow 2\left( {\left( {{x^2} + 5x} \right) + \left( {6x + 30} \right)} \right) $ ,
Now factor out the highest common factor, we get
 $ \Rightarrow 2\left( {x\left( {x + 5} \right) + 6\left( {x + 5} \right)} \right) $ ,
Now taking common term in both, we get,
 $ \Rightarrow 2\left( {\left( {x + 5} \right)\left( {x + 6} \right)} \right) $ ,
Factorising the given polynomial we get the terms $ x + 5 $ and $ x + 6 $ ,

Factoring terms of $ 2{x^2} + 22x + 60 $ are $ x + 5 $ and $ x + 6 $ , and it is written as $ 2\left( {\left( {x + 5} \right)\left( {x + 6} \right)} \right) $ .

Note: We have several options for factoring when you are solving the polynomial equations. In a polynomial, irrespective of how many terms it has, we should always check the highest common factors first. The highest common factor is our biggest expression which would go into all our terms, and when we use H.C.F it is similar to performing the distributive property backwards. If the expression is a binomial then we must look for the differences of squares, difference of cubes, or even the sum of cubes, finally once the polynomial is factored fully, you can then use the zero property for solving the equation.