Question

Solve the following quadratic equation by factorization method $3{{x}^{2}}-23x+40=0$.

Answer 1

Hint: Here we will apply the factorization method in which we convert the middle term of the equation of the form $a{{x}^{2}}+bx+c=0$ is converted into two different terms in such a way that after multiplying these terms we get the same number which we get after multiplying a and c.

Complete step-by-step answer:
We will consider the equation $3{{x}^{2}}-23x+40=0...(i)$. Now we use the factorization method here in order to find the value of the equation (i). For that we need to focus on the first and the last terms or we can say the coefficient of ${{x}^{2}}$ and the constant only. With these coefficients we will find the factors.
Here the coefficient of the first term is 3 and the constant is 40. Now we will multiply these together in such a way that after adding or subtracting its factors we get the middle term of the equation (i) which is - 23. This can be done as below.
We will first multiply the numbers 3 and 40. Therefore we get 120.
Now we will multiply 15 and 8 together. Here 15 and 8 are the factors of 120 only. After multiplying we clearly get 120.
Now we will add these two together. Therefore we get 23 only. Thus these factors will work here.
This means that if we add – 15 and – 8 together then we will get -23. So, the equation (i) can be written as $3{{x}^{2}}-15x-8x+40=0$.
Now we take out any common term between them so that we will be left with another common term. That is if we consider the equation $3{{x}^{2}}-15x-8x+40=0$ we will take the common term 3x from the first two terms only.
Therefore we get $3x\left( x-5 \right)-8x+40=0$.
Now we will also make another two terms in such a way that we will have another term there which is (x – 5).
That is if we consider $3x\left( x-5 \right)-8x+40=0$ then we will take the common term – 8 out from the other two terms.
Therefore we get $3x\left( x-5 \right)-8\left( x-5 \right)=0$.
Now we clearly get (x – 5) as a common term in the equation. This was needed here for the factorization. Thus we get
$\begin{align}
  & 3x\left( x-5 \right)-8\left( x-5 \right)=0 \\
 & \Rightarrow \left( 3x-8 \right)\left( x-5 \right)=0 \\
\end{align}$
Hence, the values of x are 5 and $\dfrac{8}{3}$.

Note: After we will first multiply the numbers 3 and 40 we get 120. But we will check that we will get - 23 after subtracting these together. Since, we are getting 37 after subtracting 3 and 40. Thus these factors will not work here. Now we will multiply 3 and 20 which results into 60 not equal to 120. Here 20 is one of the factors of 120. Therefore, these will also not work here. This question can also be cross checked by hit and trial method and also by the square root formula.