Question

How do you factor completely $5{{x}^{2}}-23x+12$ ?

Answer 1

Hint: In this question, we have to find the value of x. The equation given in the problem is in the form of the quadratic equation $a{{x}^{2}}+bx+c=0$ . So, we will apply the factoring method that is splitting the middle method, to get two values of x. First we will split the middle term of the equation in the sum of $-20x$ and $-3x$ . After that, we will take common 5x in the first two terms and -3 in the last two terms. After the necessary calculations, we get two values of x, which is our required solution.

Complete step by step solution:
According to the question, we have to find the value of x.
Thus, to solve this problem we will use the splitting the middle terms method.
The equation given to us is $5{{x}^{2}}-23x+12$ ---------- (1)
As we see know, the equation (1) is in the general form of the quadratic equation$a{{x}^{2}}+bx+c=0$ , thus on comparing both the equations, we get
$a=5\text{, }b=-23\text{, and }c=+12$
Therefore, we will apply splitting the middle term method, that is, we will express the middle term of equation $b$ in such a way that it will be the sum of the factors of $a.c$ .
So, we see that $ac=5.(12)=60$ , that is
$60=(-20).(-3)$ and $-20-3=-23$
So, we will rewrite the middle term as a sum of -20x and -3x, we get
$\Rightarrow 5{{x}^{2}}-20x-3x+12=0$
Therefore, we take 5x common in the first two terms and -3 common in the last two terms, we get
 $\Rightarrow 5x(x-4)-3(x-4)=0$
Now, we take common (x-4) in the above equation, we get
$\Rightarrow (5x-3)(x-4)=0$
So, either $5x-3=0$ ----------- (3) or
$x-4=0$ --------- (4)
Thus, first, we will solve equation (3), that is
$\Rightarrow 5x-3=0$
Now, we will add 3 on both sides in the above equation, we get
$\Rightarrow 5x-3+3=0+3$
As we know, the same terms with opposite signs cancel out, therefore, we get
$\Rightarrow 5x=+3$
Now, we will divide 5 on both sides in the above equation, we get
$\Rightarrow \dfrac{5}{5}x=\dfrac{3}{5}$
On further simplification, we get
$\Rightarrow x=\dfrac{3}{5}$
Now, we will solve equation (4), which is
$x-4=0$
Now, we will add 4 on both sides in the above equation, we get
$\Rightarrow x-4+4=0+4$
As we know, the same terms with opposite signs cancel out, we get
$\Rightarrow x=4$
Therefore, for the equation $5{{x}^{2}}-23x+12$ , the value of x is $\dfrac{3}{5}$ and $4$ . Thus, factors are $(x-\dfrac{3}{5})$ and $(x-4)$

Note:
While solving this problem, do mention all the steps properly to avoid confusion and mathematical mistakes. At the time splitting the middle term, take the least common multiple of $ac$ , that will give an accurate answer to the solution.