Problem 23 - Algebra and Trigonometry

Question

Factor each trinomial, or state that the trinomial is prime. $$3 x^{2}-x-2$$

Step-by-Step Solution

Answer

The factored form of the trinomial is \((3x + 2)(x - 1)\)

Step 1: Identify the coefficients and constant

Given the trinomial \(3x^{2}-x-2\), it can be seen that it takes the general form of \(ax^{2}+bx+c\) where \(a=3\), \(b=-1\), and \(c=-2\)

Step 2: Factor \(ac\)

Compute the product of the coefficients \(a\) and \(c\), that is \(3*(-2) = -6\).

Step 3: Find two numbers whose product equals to factor \(ac\) and whose sum equals \(b\)

Find two numbers that multiply to -6 (the result obtained from step 2) and add up to -1 (the coefficient \(b\)). These numbers are -3 and 2

Step 4: Rewrite the trinomial

The original trinomial \(3x^{2}-x-2\) can now be rewritten by breaking down the middle term into -3x and 2x. Thus, \(3x^{2}-x-2 = 3x^{2} - 3x + 2x - 2\)

Step 5: Factor by grouping

Regroup the terms and factor out the greatest common factor (GCF) from each group. Therefore, the new expression can be rewritten as \(3x(x - 1) + 2(x - 1)\). Then, factor out the common binomial \((x - 1)\) to obtain the final factors, which are \(3x + 2\) and \(x - 1\). Thus the factored form of the trinomial is \((3x + 2)(x - 1)\)