1. ## Sign Charts

Dear Forum I solved the following but I need help solving is using (sign charts- critical value method) I solved the inequality but I need help solving it using sign charts.

1.Solve x2 + 11x + 18 > 0 using the Critical Value (Sign Chart) method.
what I came up with was:
x^(2)+11x+18>0

Factor the trinomial x^(2)+11x+18.
(x+9)(x+2)>0

Set each of the factors of the left-hand side of the inequality equal to 0 to find the critical points.
x+9=0_x+2=0

Since 9 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 9 from both sides.
x=-9_x+2=0

Set each of the factors of the left-hand side of the inequality equal to 0 to find the critical points.
x=-9_x+2=0

Since 2 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 2 from both sides.
x=-9_x=-2

Since this is a 'greater than 0' inequality, all intervals that make the expression positive are part of the solution.
x<-9 or x>-2

2. It's difficult to describe the process in words, so you might do better with studying some online lessons that show pictures of the process.

Once you've learned the basic process, look at the two factors you have. The factors x + 9 and x + 2 split the number line into the intervals (-infinity, -9), (-9, -2), and (-2, +infinity). From what you know of graphing, you already know that the quadratic is above the axis (that is, greater than zero) on its "ends", so you already know which two intervals you need for your solution.

But looking at the signs of the factors, you have x + 9 > 0 for x > -9, and x + 2 > 0 for x > -2. Then the two factors together are positive (that is, the quadratic is positive) where both factors are positive or else both are negative. Both are negative for x < -9 and both are positive for x > -2.

Have fun!

3. If x> a x-a is positive. If x< a, x- a is negative
Start from the left and "step over" each critical point, one at a time.

If x< -9, x is less than both -9 and -2 so both factors are negative and the product is positive.

If -9< x< -2, one of the factors (x+ 9) is now positive but the other is still negative so the product is negative.

If x> -2, x is larger than both -9 and -2 so both factors are positive and the product is positive.

The point is that exactly one of the factors changes sign at each critical point. That being the case, the sign of the product MUST change at each critical point.

(One thing to be careful about: if one factor is to an even power, it never changes sign so you just ignore that critical point.)

For example, if $\displaystyle f(x)= (x+ 3)(x+ 2)^3(x- 1)^4(x- 2)$, the critical points are -3, -2, 1, and 2 but x- 1 is to an even power so we can ignore that. If x< -3, there are 5 negative factors (counting the three in (x-2)^3 but it really doesn't matter whether you think of it as 5 or 3) so f(x) is negative. If -3< x< -2, only x+ 3 changes sign so we now have an even number of negative factors and f(x) is positive. If -2< x< 2, we have only one negative factor so f(x) is non-positive (it will be 0 at x= 1).