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

Question

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

Kennedy Adams · Accepted Answer

The answer is #x in ]-1/2, 4[#

Let #f(x)=3(4-x)(2x+1)#

We build a sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-1/2##color(white)(aaaa)##4##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##2x+1##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##4-x##color(white)(aaaaaaaa)##+##color(white)(aaaa)##+##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-#

Therefore,

#f(x)>0# when #x in ]-1/2, 4[#

Kennedy Adams · Answer

undefined To solve the inequality $3(4 - x)(2x + 1) > 0$, first, find the critical points by setting each factor equal to zero: $4 - x = 0$ gives $x = 4$ $2x + 1 = 0$ gives $x = -\frac{1}{2}$ These critical points divide the number line into three intervals: $(-∞, -\frac{1}{2})$, $(-1/2, 4)$, and $(4, ∞)$. Choose test points within each interval and evaluate the expression to determine the sign of $3(4 - x)(2x + 1)$: For $x = -1$, $3(4 - (-1))(2(-1) + 1) = 3(5)(-1) = -15 < 0$, so this interval is negative. For $x = 0$, $3(4 - 0)(2(0) + 1) = 3(4)(1) = 12 > 0$, so this interval is positive. For $x = 5$, $3(4 - 5)(2(5) + 1) = 3(-1)(11) = -33 < 0$, so this interval is negative. Thus, the solution to the inequality is $x \in (-\frac{1}{2}, 4)$.

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

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