In Exercises 103–114, factor completely. x4−5x2y2+4y4

Ch. P - Fundamental Concepts of Algebra

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Ch. P - Fundamental Concepts of Algebra

Problema 109

Capitolo 1, Problema 109

In Exercises 103–114, factor completely. x⁴−5x²y²+4y⁴

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Recognize that the given expression $ x^4 - 5x^2y^2 + 4y^4 $ is a quadratic form in terms of $ x^2 $ and $ y^2 $. Rewrite it as $ (x^2)^2 - 5(x^2)(y^2) + 4(y^2)^2 $.

Observe that this is a trinomial in the form $ a^2 - 2ab + b^2 $, which can potentially be factored as a product of two binomials. Let $ u = x^2 $ and $ v = y^2 $, so the expression becomes $ u^2 - 5uv + 4v^2 $.

Factor the trinomial $ u^2 - 5uv + 4v^2 $ by finding two numbers that multiply to $ 4 $ (the constant term) and add to $ -5 $ (the coefficient of $ uv $). These numbers are $ -4 $ and $ -1 $.

Rewrite the trinomial as $ (u - 4v)(u - v) $, substituting back $ u = x^2 $ and $ v = y^2 $. This gives $ (x^2 - 4y^2)(x^2 - y^2) $.

Notice that both $ x^2 - 4y^2 $ and $ x^2 - y^2 $ are differences of squares. Factor them further as $ (x - 2y)(x + 2y) $ and $ (x - y)(x + y) $, respectively. The fully factored form is $ (x - 2y)(x + 2y)(x - y)(x + y) $.

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Factoring Polynomials

Factoring polynomials involves rewriting a polynomial as a product of simpler polynomials or factors. This process is essential for simplifying expressions and solving equations. In the case of the given polynomial, recognizing patterns such as the difference of squares or perfect square trinomials can aid in the factoring process.

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The expression x^4−5x^2y^2+4y^4 can be viewed as a quadratic in terms of x^2. By substituting u = x^2, the polynomial transforms into a standard quadratic form, making it easier to apply factoring techniques. This approach allows for the identification of roots and factors more straightforwardly.

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The difference of squares is a factoring technique used when an expression can be written in the form a^2 - b^2, which factors into (a - b)(a + b). In the context of the given polynomial, recognizing components that fit this pattern can simplify the factoring process and lead to a complete factorization of the expression.

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