Question

Factorise:
$12{x^2} - 7x + 1$

Answer 1

Hint: Here we will solve the equation by using the factoring method. Firstly we will split the middle term of the expression by using the factoring method. Then we will take numbers or variables common from two sets of terms and get our equation in product form. Finally we will equate the equation equal to zero and get our desired solution.

Complete step-by-step answer:
The equation given to us is $12{x^2} - 7x + 1$.
If we want to factorize an expression like $a{x^2} + bx + c$ we take two numbers such that their product is equal to $a.c$ and their sum is equal to $b$.
Firstly we will split the middle term of the equation by using the above technique.
Let us take one number $4$ and another number $3$ for equation $12{x^2} - 7x + 1$.
As we can see
$4 \times 3 = 12 = a \times c$
$4 - 3 = 1 = b$
So we can rewrite our equation as,
$
  12{x^2} - \left( {4x + 3x} \right) + 1 \\
   \Rightarrow 12{x^2} - 4x - 3x + 1 \\
 $
Now taking $4x$ common in first two terms and $ - 1$ common in last two terms we get,
$
   \Rightarrow 4x\left( {3x - 1} \right) - 1\left( {3x - 1} \right) \\
   \Rightarrow \left( {3x - 1} \right)\left( {4x - 1} \right) \\
 $
Now we will put the above equation equal to zero and get,
$
  \left( {3x - 1} \right)\left( {4x - 1} \right) = 0 \\
  x = \dfrac{1}{3},x = \dfrac{1}{4} \\
 $
So we got our zeroes as $x = \dfrac{1}{3},x = \dfrac{1}{4}$
Hence our factors are $\left( {x - \dfrac{1}{3}} \right)\left( {x - \dfrac{1}{4}} \right)$

So, we can factorize $12{x^2} - 7x + 1$ as $\left( {x - \dfrac{1}{3}} \right)\left( {x - \dfrac{1}{4}} \right)$

Note:
A quadratic equation is the equation having the variable with highest power as two. This method is known as factoring as we find out the factors for the equation. As the highest power of the variable is two we get two factors for the equation. We find our answer by substituting the product value equal to zero because the equation has to have at least one value equal to zero to for the equation to be equal to zero.