
How do you factor $2{{x}^{2}}+11x+12$?
Answer
549.3k+ views
Hint: 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 ${{x}^{2}}$ 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 $2{{x}^{2}}+11x+12$.
Quadratic equation: $a{{x}^{2}}+bx+c=2{{x}^{2}}+11x+12$
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 $x+y=b=11$ and $xy=a.c=2.(12)=24$.
We see that if $x=8$ and $y=3$ , then we get $x+y=b=11$ and $xy=a.c=2.(12)=24$ .
Thus, equation: $2{{x}^{2}}+11x+12$.
Now, we use splitting the middle term method, that is we will split the middle term as the addition of 8x and 3x because $8x+3x=11x$, we get
$\begin{align}
& \Rightarrow ~2{{x}^{2}}+11x+12 \\
& \Rightarrow 2{{x}^{2}}+8x+3x+12 \\
\end{align}$
Now, take common 2x from the first two numbers and 3 from the last two numbers, we get
$\Rightarrow 2x(x+4)+3(x+4)$
As we see that (x+4) is common on both LHS and RHS of the addition sign, we get
$\Rightarrow (x+4)(2x+3)$
Therefore, $(x+4)(2x+3)$ are the factors of equation $2{{x}^{2}}+11x+12$.
Note:
While finding two numbers for factorization, use the LCM method. That is,
L.C.M of 24: $\begin{align}
& 2\left| \!{\underline {\,
24 \,}} \right. \\
& 2\left| \!{\underline {\,
12 \,}} \right. \\
& 2\left| \!{\underline {\,
6 \,}} \right. \\
& 3\left| \!{\underline {\,
3 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}$
As we see, LCM of 24= (2, 2, 2, 3)
and, if we add ${{2}^{3}}$ and 3, we get
${{2}^{3}}+3$
$8+3=11$, 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.
Complete step by step answer:
According to the question, we have to find the factors of the quadratic equation $2{{x}^{2}}+11x+12$.
Quadratic equation: $a{{x}^{2}}+bx+c=2{{x}^{2}}+11x+12$
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 $x+y=b=11$ and $xy=a.c=2.(12)=24$.
We see that if $x=8$ and $y=3$ , then we get $x+y=b=11$ and $xy=a.c=2.(12)=24$ .
Thus, equation: $2{{x}^{2}}+11x+12$.
Now, we use splitting the middle term method, that is we will split the middle term as the addition of 8x and 3x because $8x+3x=11x$, we get
$\begin{align}
& \Rightarrow ~2{{x}^{2}}+11x+12 \\
& \Rightarrow 2{{x}^{2}}+8x+3x+12 \\
\end{align}$
Now, take common 2x from the first two numbers and 3 from the last two numbers, we get
$\Rightarrow 2x(x+4)+3(x+4)$
As we see that (x+4) is common on both LHS and RHS of the addition sign, we get
$\Rightarrow (x+4)(2x+3)$
Therefore, $(x+4)(2x+3)$ are the factors of equation $2{{x}^{2}}+11x+12$.
Note:
While finding two numbers for factorization, use the LCM method. That is,
L.C.M of 24: $\begin{align}
& 2\left| \!{\underline {\,
24 \,}} \right. \\
& 2\left| \!{\underline {\,
12 \,}} \right. \\
& 2\left| \!{\underline {\,
6 \,}} \right. \\
& 3\left| \!{\underline {\,
3 \,}} \right. \\
& 1\left| \!{\underline {\,
1 \,}} \right. \\
\end{align}$
As we see, LCM of 24= (2, 2, 2, 3)
and, if we add ${{2}^{3}}$ and 3, we get
${{2}^{3}}+3$
$8+3=11$, 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.
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