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How do you factor #6x^2+5x-4#?

Answer 1

(2x - 1)( 3x + 4)

Use the new AC Method (Socratic Search) #y = 6x^2 + 5x - 4 =# 6(x + p)(x + q) Converted trinomial: #y' = x^2 + 5x - 24 =# (x + p')(x + q'). p' and q' have opposite signs because ac < 0. factor pairs of (ac = - 24) --> (-3, 8). This sum is (5 = b). Then p' = -3 and q' = 8. Back to trinomial y --> #p = (p')/a = -3/6 = -1/2#, and #q = (q')/a = 8/6 = 4/3#. Factored form #y = 6(x - 1/2)(x + 4/3) = (2x - 1)(3x + 4)#

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Answer 2

Genesis Allen

To factor the expression (6x^2 + 5x - 4), you can use the factoring by grouping method. First, multiply the coefficient of (x^2) (6) by the constant term (-4) to get -24. Then, find two numbers that multiply to -24 and add up to the coefficient of (x) term (5). These numbers are 8 and -3. Rewrite the middle term using these numbers as coefficients:

[6x^2 + 8x - 3x - 4]

Now, group the terms:

[(6x^2 + 8x) + (-3x - 4)]

Factor out the greatest common factor from each group:

[2x(3x + 4) - 1(3x + 4)]

Notice that both groups have a common factor of (3x + 4). Factor it out:

[(2x - 1)(3x + 4)]

So, (6x^2 + 5x - 4) factors into ((2x - 1)(3x + 4)).

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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.