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Question

Question: How do you factor \(2{{x}^{2}}+11x+12\)?...

How do you factor 2x2+11x+122{{x}^{2}}+11x+12?

Explanation

Solution

In this question, we have to find the factors of the equation. A quadratic equation is given in the problem, so we apply the middle term split method to get the factors for the same. We start solving this problem by finding two numbers such that the product of the two numbers is equal to the product of the coefficient of x2{{x}^{2}} and the constant, and the sum of these two numbers is the coefficient of x. After finding the 2 numbers, we split the middle term as the sum of those two numbers and make the necessary calculations to get the required answer to the problem.

Complete step by step answer:
According to the question, we have to find the factors of the quadratic equation 2x2+11x+122{{x}^{2}}+11x+12.
Quadratic equation: ax2+bx+c=2x2+11x+12a{{x}^{2}}+bx+c=2{{x}^{2}}+11x+12
Thus, it is given that a=2, b=11, and c=12
To factorize, we have to find two numbers x and y such that x+y=b=11x+y=b=11 and xy=a.c=2.(12)=24xy=a.c=2.(12)=24.
We see that if x=8x=8 and y=3y=3 , then we get x+y=b=11x+y=b=11 and xy=a.c=2.(12)=24xy=a.c=2.(12)=24 .
Thus, equation: 2x2+11x+122{{x}^{2}}+11x+12.
Now, we use splitting the middle term method, that is we will split the middle term as the addition of 8x and 3x because 8x+3x=11x8x+3x=11x, we get
 2x2+11x+12 2x2+8x+3x+12 \begin{aligned} & \Rightarrow ~2{{x}^{2}}+11x+12 \\\ & \Rightarrow 2{{x}^{2}}+8x+3x+12 \\\ \end{aligned}
Now, take common 2x from the first two numbers and 3 from the last two numbers, we get
2x(x+4)+3(x+4)\Rightarrow 2x(x+4)+3(x+4)
As we see that (x+4) is common on both LHS and RHS of the addition sign, we get
(x+4)(2x+3)\Rightarrow (x+4)(2x+3)
Therefore, (x+4)(2x+3)(x+4)(2x+3) are the factors of equation 2x2+11x+122{{x}^{2}}+11x+12.

Note: While finding two numbers for factorization, use the LCM method. That is,
L.C.M of 24:
2!24 2!12 2!6 3!3 1!1 \begin{aligned} & 2\left| \\!{\underline {\, 24 \,}} \right. \\\ & 2\left| \\!{\underline {\, 12 \,}} \right. \\\ & 2\left| \\!{\underline {\, 6 \,}} \right. \\\ & 3\left| \\!{\underline {\, 3 \,}} \right. \\\ & 1\left| \\!{\underline {\, 1 \,}} \right. \\\ \end{aligned}
As we see, LCM of 24= (2, 2, 2, 3)
and, if we add 23{{2}^{3}} and 3, we get
23+3{{2}^{3}}+3
8+3=118+3=11, which is the coefficient of x.
You can also use the Hit and Trial method for solving this problem. Let different values of x and put that value in the function, if you get the functional value as 0, then that number is the factor of the equation.