Question

Factorise the given equation:
(i)\[12{x^2} - 7x + 1\]
(ii)\[2{x^2} + 7x + 3\]
(iii)\[6{x^2} + 5x - 6\]
(iv)\[3{x^2} - x - 4\]

Answer 1

Hint: According to the question, which is in the form \[a{x^2} + bx + c\] . Find out the product which is ac and sum is b. Then use the splitting the middle term to solve the equation.

Complete step-by-step answer:
(i)\[12{x^2} - 7x + 1\]
Here, by using splitting the middle term method we can calculate the factors of the equation \[12{x^2} - 7x + 1\]
In this method we will find out two numbers whose sum is \[ - 7\] and product is \[12\].
So, we are getting the two numbers which are \[ - 3\] and \[ - 4\] .
 \[12{x^2} - 7x + 1\]
 \[12{x^2} - 3x - 4x + 1\]
Taking out common in the pairs of 2 we get,
 \[3x\left( {4x - 1} \right) - 1\left( {4x - 1} \right)\]
Taking 2 same factors one time we get,
\[\left( {4x - 1} \right)\left( {3x - 1} \right)\]

(ii)\[2{x^2} + 7x + 3\]
Here, by using splitting the middle term method we can calculate the factors of the equation \[2{x^2} + 7x + 3\]
 In this method we will find out two numbers whose sum is \[7\] and product is \[6\].
 So, we are getting the two numbers which are \[6\] and \[1\] .
\[2{x^2} + 7x + 3\]
\[2{x^2} + 6x + x + 3\]
Taking out common in the pairs of 2 we get,
\[2x\left( {x + 3} \right) + 1\left( {x + 3} \right)\]
Taking 2 same factors one time we get,
\[\left( {x + 3} \right)\left( {2x + 1} \right)\]

(iii)\[6{x^2} + 5x - 6\]
Here, by using splitting the middle term method we can calculate the factors of the equation \[6{x^2} + 5x - 6\]
In this method we will find out two numbers whose sum is \[5\] and product is \[ - 36\].
 So, we are getting the two numbers which are \[ - 4\] and \[9\] .
 \[6{x^2} + 5x - 6\]
\[6{x^2} - 4x + 9x - 6\]
Taking out common in the pairs of 2 we get,
\[2x\left( {3x - 2} \right) + 3\left( {3x - 2} \right)\]
Taking 2 same factors one time we get,
 \[\left( {2x + 3} \right)\left( {3x - 2} \right)\]

(iv)\[3{x^2} - x - 4\]
Here, by using splitting the middle term method we can calculate the factors of the equation \[3{x^2} - x - 4\]
 In this method we will find out two numbers whose sum is \[ - 1\] and product is \[ - 12\].
So, we are getting the two numbers which are \[3\] and \[ - 4\] .
 \[3{x^2} - x - 4\]
 \[3{x^2} + 3x - 4x - 4\]
Taking out common in the pairs of 2 we get,
\[3x\left( {x + 1} \right) - 4\left( {x + 1} \right)\]
Taking 2 same factors one time we get,
\[\left( {x + 1} \right)\left( {3x - 4} \right)\]

Note: To solve these types of questions, we can solve the equation either by splitting the middle term method or by using the formula that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . Take the common pairs very carefully and hence factorise the equation.