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Factor the polyomial completely using any method 6a^4+13a^2-5 Medal given for answer
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let x = a^2 6a^4+13a^2-5 substituting: 6x^2 + 13x - 5 (6x + 3)(6x + 10 )/6 3(2x + 1)2(3x + 5)/6 (2x + 1)(3x + 5) substituting back: (2a^2 + 1)(2a^2 + 5) Ans. (2a^2 + 1)(2a^2 + 5)
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sorry, small problem, correcting...
let x = a^2 6a^4+13a^2-5 substituting: 6x^2 + 13x - 5 (6x - 2)(6x + 15 )/6 2(3x - 1)3(2x + 5)/6 (3x - 1)(2x + 5) substituting back: (3a^2 - 1)(2a^2 + 5) ( sqrt(3)a - 1)(sqrt(3)a + 1)(2a^2 + 5) Ans. ( sqrt(3)*a - 1)(sqrt(3)*a + 1)(2a^2 + 5)
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