Solve (x-1)(x+4)(3x+4)<0 algebraically using interval chart and showing your full work.
Don't know how you were taught to use the interval chart, but the way I do this:
critical values are x = 1, -4, and -4/3
So on a number line place those points.
look at the part x < -4, the farthest point to the left.
pick any x, say x = -5, and test it,
(-6)(-1)(-11) < 0 ?, Yes, since the result is negative
look at x between -4 and -4/3 , lets pick x = -2
(-3)(2)(-2) < 0 , NO, since the result is positive
look at x between -4/3 and 1, lets pick x = 0
(-1)(4)(4) < 0 ? , Yes, we get a negative
look at x> 1, say x = 2
(1)(6)(10) < 0 , NO, we get a positive
so we have:
-4/3 < x < 1 OR x < -4
I will let you put that into the current interval notation
Of course we could have just looked at a quick graph of
y = (x-1)(x+4)(3x+4), which has x intercepts of -4, -4/3, 1
with a leading +3x^3 term, so it goes to ∞ as x goes to ∞
and goes to -∞ as x goes to -∞
and knowing the general shape of a cubic with 3 intercepts would
give us the above result