Problem 25 - Algebra and Trigonometry

Question

Factor each trinomial, or state that the trinomial is prime. $$3 x^{2}-25 x-28$$

Step-by-Step Solution

Answer

The factored form of the trinomial \(3x^2 - 25x - 28\) is \((3x + 7)(x - 4)\)

Step 1: Identify the coefficient and constants

The trinomial given is \(3x^2 - 25x - 28\). Here, the coefficient of \(x^2\) is 3, the coefficient of \(x\) is -25, and the constant term is -28.

Step 2: Determine the factors

Find two numbers that multiply to give a product of (3*-28) = -84 (coefficient of \(x^2\) times constant term) and those two numbers should also add up to -25 (coefficient of \(x)\). After some trial and error, we discover that -4 and 21 meet these criteria. As (-4) * 21 = -84 and -4 + 21 = 17.

Step 3: Rewrite the trinomial

Now, rewrite the middle term of the trinomial as the sum of the terms obtained in step 2: \(3x^2 - 4x + 21x - 28\). This expression is equal to the given trinomial, but rewritten in such a way as to next permit factoring by grouping.

Step 4: Factor by grouping

Group the terms and factor out common elements: \(x(3x - 4) + 7(3x - 4)\)

Step 5: Finish factoring

Notice that the expressions in parentheses are now the same, so you can factor it out, yielding the final factored form of the trinomial. Hence, \(3x^2 - 25x - 28 = (3x + 7)(x - 4)\)