Rebecca M. answered • 07/11/19
Math tutor for 4 yrs at a charter high school; 1.5 yrs at Job Corp
To find the solution of the equation or value for x the polynomial must be factored. Before factoring the polynomial must be set to zero as has been done here. The equation must be set to zero due to the zero property of multiplication which states if one factor is equal to zero then the product is equal to zero. Then each factor can be set to zero to solve for the value of x.
The "ac" method is the easiest to use when the first coefficient does not equal 1. The "ac" method assigns the value of each coefficient to a letter. Thus the first coefficient is a, the second is b, and the third is c. In this case a = 3, b = 5 and c = -2. Two integers must be found that have a product equal to ac and have a sum of b. The product of ac is -6. Two integers which have a product of ac and a sum of 5 are -1 and 6.
Now this information is used to split the b term for factoring by groups. The new equation with the split b is 3 x^2 +6x-x-2. The equation is then divided into two binomials: (3x^2+6x) + (-x-2); each to be facotred with a common binomial factor. The factoring of the first is 3x( x+2) and the second is -1(x+2). The factor x+ 2 is common. Putting the other two parts together the complete factoring becomes: (3x-1) (x+2). Now each factor must be set to 0 for solving for x. Thus 3x-1 = 0 is solved by adding 1 to both sides and then dividing both sides by 3; x = 1/3. The other factor is solved x + 2 = 0 by subtracting 2 from both sides and x = -2.
When c is negative I prefer to think of a difference for b than a sum since a negative integer times a positive is a negative number. Also I remember that the larger of the two integers in this case has the same sign as b. The order of the two middle terms which add to b does not matter. As (3x^2-x) + (6x-2) will factor as x(3x-1) and 2(3x-1). Combining terms will yield (x+2) (3x-1) as before.
