Question

How do you factor completely: \[3{x^2} + 2x - 1\] ?

Answer 1

Hint: We have a polynomial of degree two hence it is called quadratic polynomial. Since the degree is two we will have two factors or two roots. We split the middle term of the equation and taking the common term we will get the required answer. If the factors are difficult to find or not found we use the Sridhar’s Acharya formula.

Complete step-by-step answer:
Given, \[3{x^2} + 2x - 1\]
We have the equation in form \[a{x^2} + bx + c\] , we split the middle term into a sum of two terms such that the product of these two terms will be ‘ac’ and whose sum is ‘b’.
In polynomial \[3{x^2} + 2x - 1\] we rewrite the middle term as a sum of two terms whose product is \[a \times c = 3 \times - 1 = - 3\] and whose sum is \[b = + 2\] .
The quadratic equation becomes,
 \[3{x^2} + 2x - 1\]
 \[ = 3{x^2} + 3x - 1x - 1\]
We can see that the multiple of \[3 \times - 1 = - 3\] and their sum is \[3 + ( - 1) = 2\] .
Taking ‘3x’ common in first two terms and taking ‘-1’ in reaming two terms we have
 \[ = 3x(x + 1) - 1(x + 1)\]
Taking \[(x + 1)\] common again we get,
 \[ = (3x - 1)(x + 1)\]
Thus the factors of the quadratic polynomial \[3{x^2} + 2x - 1\] are \[(3x - 1)(x + 1)\]
So, the correct answer is “ \[(3x - 1)(x + 1)\] ”.

Note: If we need to find the roots of the given quadratic polynomial we need to equate the obtained factors to zero. That is \[(3x - 1)(x + 1) = 0\] .We know from zero product property that if \[ab = 0\] then \[a = 0\] or \[b = 0\] . Using this we have, \[(3x - 1) = 0\] or \[(x + 1) = 0\] . So we get, \[x = \dfrac{1}{3}\] and \[x = - 1\] are the roots of the given quadratic polynomial. If we do this using Sridhar’s Acharya formula we will get the same answer. We follow the same procedure to solve any quadratic polynomial.