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Hey can explain this question to me?

the question is: (x+4)(3x+2)(2x-3)(x+4) > 0.

U know the answer is x but I am having trouble figuring out why. Thanks!

1 Answer

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Certainly! I'd be happy to explain the question and help you understand the answer.

The given question is:
\displaystyle\sf (x+4)(3x+2)(2x-3)(x+4) > 0.

To solve this inequality, we need to determine the values of
\displaystyle\sf x for which the expression
\displaystyle\sf (x+4)(3x+2)(2x-3)(x+4) is greater than zero (
\displaystyle\sf > 0).

To find the solution, we can use the concept of interval notation and zero-product property. Here's how we can proceed step by step:

1. Begin by finding the critical values of
\displaystyle\sf x where the expression changes sign. These occur when any of the factors are equal to zero. From the given equation, we have:


\displaystyle\sf x+4= 0 \Rightarrow x=-4


\displaystyle\sf 3x+2= 0 \Rightarrow x=-(2)/(3)


\displaystyle\sf 2x-3= 0 \Rightarrow x=(3)/(2)

2. Now, we have four critical values:
\displaystyle\sf x=-4,
\displaystyle\sf x=-(2)/(3),
\displaystyle\sf x=(3)/(2), and
\displaystyle\sf x=-4 (since
\displaystyle\sf x+4= 0 yields
\displaystyle\sf x=-4 as well).

3. Plot these critical values on a number line:


\displaystyle\sf -4 \quad -(2)/(3) \quad (3)/(2) \quad -4

4. Now, we need to test the expression
\displaystyle\sf (x+4)(3x+2)(2x-3)(x+4) in the intervals created by these critical values. We will choose test points within each interval and determine if the expression is positive or negative.

For example, let's take a test point
\displaystyle\sf x=-5 from the interval
\displaystyle\sf (-\infty ,-4):


\displaystyle\sf (-5+4)(3(-5)+2)(2(-5)-3)(-5+4)=(-1)(-13)(-13)(-1)=169>0

Since the expression
\displaystyle\sf (x+4)(3x+2)(2x-3)(x+4) is positive in the interval
\displaystyle\sf (-\infty ,-4), this interval is part of the solution.

By repeating this process for the remaining intervals, you'll find that the complete solution to the given inequality is:


\displaystyle\sf x\in \left(-\infty ,-(2)/(3)\right)\cup \left((3)/(2) ,\infty \right)

In interval notation, the solution is:


\displaystyle\sf \left(-\infty ,-(2)/(3)\right)\cup \left((3)/(2) ,\infty \right)


\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}

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