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Is it possible to factor #y=2x^2+4x-30#? If so, what are the factors?

Answer 1

y = 2(x - 3)(x + 5)

I use the new AC Method to factor trinomials (Socratic Search). #y = 2x^2 + 4x - 30 =# 2(x + p)(x + q) (1) Converted trinomial: #y' = x^2 + 4x - 60 = #(x + p')(x + q') p' and q' have opposite signs because ac < 0. Compose factor pairs of (ac = -60) --> (-4, 15)(-6, 10). This sum is (10 - 6 = 4 = b). Then p' = -6 and q' = 10. Back to original trinomial (1): #p = (p')/a = -6/2 = -3# and #q = (q')/a = 10/2 = 5# Factored form: y = 2(x - 3)(x + 5)

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Answer 2

Nathan Axel

Yes, it is possible to factor the quadratic equation y = 2x^2 + 4x - 30. The factors are (2x - 6)(x + 5).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.