Question

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!

Answer 1

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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