x^4-19x^2+9

Question

factoring homework - stuck on how to solve

Answer 1

If the coefficient can be irrational , then you can factor this problem by completing the square.

x^4-19x^2+9

= (x^4-6x^2+9) - 13x^2

= (x^2-3)^2 - [sqrt(13)x]^2

= (x^2 -3 + sqrt(13)x) ( x^2 -3 - sqrt(13)x)

Answer 2

usually when you have an x⁴ and an x², you set a value u = x² and therefore u² = x⁴, then factor as "normal"

but once I do that I don't get something that is factorable. I don't know what two numbers would get you a 9 when multiplied and a 19 when added.

Maybe the trick with setting the u value will help on other problems.

Answer 3

I believe this is a prime trinomial. You can tell because the only factors for +9 would be +/-1, +/-3, +/-9. None of the combinations of these positive or negative will give us -19x^2 in the middle term if we try to factor out. You can also check this by using the quadratic formula. If you get two integers it can be factored. If you get something with a radical then it is prime.

[-(-19) +/- √-19^2 - 4(1)(9)]/2(1) yields (19 +/- √325)/2 which must remain partially radical so the trinomial is prime.