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Chapter 0: Problem 38

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

Short Answer

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

Step by step solution

01

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 \(ax^2+bxy + cy^2\).
02

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 \(-10+3=-7\).
03

Rewrite the trinomial

Rewrite the middle term \(-7xy\) of the trinomial as \(-10xy+3xy\). When you substitute this in the trinomial, you get: \(6x^2 - 10xy + 3xy - 5y^2\).
04

Group and factor terms

Observe for common factors in each binomial group - \(6x^2 - 10xy\) and \(3xy - 5y^2\). Factor out common terms: \(2x(3x-5y) + y(3x-5y)\).
05

Final factorization

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

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