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• e^(2x) + 2 + e^(-2x) = (e^x)^2 + 2 + (e^(-x))^2 = (e^x)^2 + 2(e^x)(e^(-x)) + (e^(-x))^2 = (e^x + e^(-x))^2

  in Factor into two groups? e^(2x)+2+e^(-2x)? Comment by alwbsok June 2021
• This is a quadratic in x^2 (rather than in x, like you may be used to). We factorise the same way: x^4 - 5x^2 - 6 = (x^2 - 6)(x^2 + 1) Now, we have some differences of two squares: = (x^2 - (sqrt(6))^2)(x^2 - i^2) = (x - sqrt(6))(x + sqrt…

  in how do you factor x^4-5x^2-6? Comment by alwbsok June 2021
• This is a quadratic in x^2 (rather than in x, like you may be used to). We factorise the same way: x^4 - 5x^2 - 6 = (x^2 - 6)(x^2 + 1) Now, we have some differences of two squares: = (x^2 - (sqrt(6))^2)(x^2 - i^2) = (x - sqrt(6))(x + sqrt…

  in how do you factor x^4-5x^2-6? Comment by alwbsok May 2021
• This is a quadratic in x^2 (rather than in x, like you may be used to). We factorise the same way: x^4 - 5x^2 - 6 = (x^2 - 6)(x^2 + 1) Now, we have some differences of two squares: = (x^2 - (sqrt(6))^2)(x^2 - i^2) = (x - sqrt(6))(x + sqrt…

  in how do you factor x^4-5x^2-6? Comment by alwbsok May 2021
• This is a quadratic in x^2 (rather than in x, like you may be used to). We factorise the same way: x^4 - 5x^2 - 6 = (x^2 - 6)(x^2 + 1) Now, we have some differences of two squares: = (x^2 - (sqrt(6))^2)(x^2 - i^2) = (x - sqrt(6))(x + sqrt…

  in how do you factor x^4-5x^2-6? Comment by alwbsok May 2021
• This is a quadratic in x^2 (rather than in x, like you may be used to). We factorise the same way: x^4 - 5x^2 - 6 = (x^2 - 6)(x^2 + 1) Now, we have some differences of two squares: = (x^2 - (sqrt(6))^2)(x^2 - i^2) = (x - sqrt(6))(x + sqrt…

  in how do you factor x^4-5x^2-6? Comment by alwbsok May 2021
• e^(2x) + 2 + e^(-2x) = (e^x)^2 + 2 + (e^(-x))^2 = (e^x)^2 + 2(e^x)(e^(-x)) + (e^(-x))^2 = (e^x + e^(-x))^2

  in Factor into two groups? e^(2x)+2+e^(-2x)? Comment by alwbsok May 2021