How do you factor #3y^2-10y-8=0#?

Question

Abigail Allen · Accepted Answer

(3y + 2)(y - 4)

Use the new AC Method (Socratic Search) f(y) = 3y^2 - 10y - 8 = 3(y + p)(y + q) Converted trinomial: f'(y) = y^2 - 10y - 24 = (y + p'(y + q') p' and q' have opposite sign because ac < 0. Factor pairs of (-24) --> (-2, 12)(2, -12) This sum is -10 = b. Then, p' = 2 and q' = -10. Back to f9y): p = (p')/a = 2/3 and q = (q')/a = -12/3 = - 4. Factored form --> f(y) = 3(y + 2/3)(y - 4) = (3y + 2)(y - 4)

Ethan Adkins · Answer

undefined To factor $3y^2 - 10y - 8 = 0$, you can use the quadratic formula or factorization by grouping.

Using factorization by grouping, you would first multiply the coefficient of $y^2$ (which is 3) by the constant term (which is -8), resulting in -24. Then, you need to find two numbers that multiply to -24 and add up to -10 (the coefficient of y). These numbers are -12 and 2.

Now, you split the middle term (-10y) into -12y and +2y:

$$3y^2 - 12y + 2y - 8 = 0$$

Next, factor by grouping:

$$y(3y - 12) + 2(3y - 4) = 0$$

Now, factor out the common factors:

$$y(3y - 12) + 2(3y - 4) = 0$$

Now, you can factor out the common binomial factor:

$$(y + 2)(3y - 4) = 0$$

So, the factored form of $3y^2 - 10y - 8 = 0$ is $(y + 2)(3y - 4) = 0$.