How do you solve # 3(4 - x)(2x + 1)

Question

How do you solve # 3(4 - x)(2x + 1)<0#?

Eli Assouline · Accepted Answer

x > 4 and x < #-1/2#. In other words, x is outside #[ -1/2, 4 ]# Drop the positive factor 3. #-2 x^2 + 7 x +4 < 0#. #( x - 7/4 )^2 > 81/16# #| x - 7/4 | > 9/4#. If #| x - a | > b, x > a + b and x < a - b# Here, #x > 4 and x < -1/2#.

Hazel Anglin · Answer

undefined To solve $3(4 - x)(2x + 1) < 0$, we need to find the intervals of $x$ values where the expression is negative.

1. First, determine the critical points by setting each factor equal to zero:
   $4 - x = 0$ and $2x + 1 = 0$.

2. Solve for $x$ to find the critical points:
   $x = 4$ and $x = -\frac{1}{2}$.

3. Next, plot these critical points on a number line.

4. Test each interval created by the critical points by selecting test points within each interval and determining if the expression is positive or negative.

5. Determine the sign of the expression in each interval and note where it is negative.

6. Finally, express the solution as an interval or combination of intervals where the expression is negative.

The solution is $x < -\frac{1}{2}$ or $4 < x < 4. $

How do you solve # 3(4 - x)(2x + 1)&lt;0#?

How do you solve # 3(4 - x)(2x + 1)<0#?