Question

How do you factor \[{x^2} + 16x + 48\]?

Answer 1

Hint: In these type 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 \[a \cdot c = 1 \cdot 48 = 48\] and whose sum is \[b = 16\], when you solve the expression we will get the required values.

Complete step-by-step solution:
Now given equation is \[{x^2} + 16x + 48\],
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 \[a \cdot c = 1 \cdot 48 = 48\] and whose sum is \[b = 16\],
Now we will find the numbers which satisfy the given data, i.e., sum of the numbers should be equal to 16 and their product should be equal to 48,
So, now rewrite the 16 as 4 and 12, whose product will be equal to 48 and their sum is 16.
Now using distributive property, we get
\[ \Rightarrow {x^2} + 4x + 12x + 48\],
By grouping the first two terms and last two terms, we get,
\[ \Rightarrow \left( {{x^2} + 4x} \right) + \left( {12x + 48} \right)\],
Now factor out the highest common factor, we get
\[ \Rightarrow x\left( {x + 4} \right) + 12\left( {x + 4} \right)\],
Now taking common term in both, we get,
\[ \Rightarrow \left( {x + 4} \right)\left( {x + 12} \right)\]
So, by factorising the polynomial we get\[\left( {x + 4} \right)\left( {x + 12} \right)\].

\[\therefore \]The factorising terms when the polynomial \[{x^2} + 16x + 48\] is factorised is equal to \[\left( {x + 4} \right)\left( {x + 12} \right)\].

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.