Problem 38 - College Algebra Essentials

Question

Factor each trinomial, or state that the trinomial is prime. $$6 x^{2}-7 x y-5 y^{2}$$

Step-by-Step Solution

Answer

The factorization of the trinomial \(6x^2-7xy-5y^2\) is \((3x-5y)(2x+y)\)

Step 1: Identify the structure of the trinomial

The given trinomial is \(6 x^{2}-7 x y-5 y^{2}\). This is a trinomial of the form \(ax^2+bxy + cy^2\).

Step 2: Look for factors

Find two numbers that multiply to \(-30\) (\(a*c=-30\)) and add to \(-7\) (coefficient of the middle term). The numbers \(-10\) and \(3\) meet these criteria because \(-10*3=-30\) and \(-10+3=-7\).

Step 3: Rewrite the trinomial

Rewrite the middle term \(-7xy\) of the trinomial as \(-10xy+3xy\). When you substitute this in the trinomial, you get: \(6x^2 - 10xy + 3xy - 5y^2\).

Step 4: Group and factor terms

Observe for common factors in each binomial group - \(6x^2 - 10xy\) and \(3xy - 5y^2\). Factor out common terms: \(2x(3x-5y) + y(3x-5y)\).

Step 5: Final factorization

Since both terms contain \((3x-5y)\), factor it out. This gives us the final factorization: \((3x-5y)(2x+y)\)