is there another method of solving (x+1)(x+3)(x+5)> 0 other than drawing up a chart?
Hello, furor celtica!
I believe we must make a sketch of some kind . . .Is there another method of solving $\displaystyle (x+1)(x+3)(x+5)\:>\: 0$
. . other than drawing up a chart?
The question is: .When is $\displaystyle y \:=\:(x+1)(x+3)(x+5)$ above the x-axis?
We have a cubic graph with x-intercepts: -5, -3, -1.
And we find that the graph looks like this:Then we see the solution: .$\displaystyle (\text{-}5,\text{-}3) \:\cup\: (\text{-}1,\infty)$Code:| | * * | * * | --o-----o-----o--+--- * * | * * | |
(x+ 1)(x+ 3)(x+ 5) is a polynomial, so continuous, so can change from "> 0" to "< 0" and vice versa only where it is equal to 0. That obviously happens only at x= -1, x= -3, and x= -5. We can determine which by taking one point in each interval. For example, if x= -6 which is less than -5, then (-6+1)(-6+3)(-6+5)= (-5)(-3)(-1)< 0. If we take x= -4, which is between -5 and -3, (-4+1)(-4+3)(-4+5)= (-3)(-1)(2)> 0. If we take x= -2, which is between -3 and -1, (-2+1)(-2+3)(-2+5)= (-1)(1)(3)< 0. Finally, of x= 0, which is larger than -1, (0+1)(0+3)(0+5)= (1)(3)(5)> 0. That tells us that the function is negative for x< -5, positive for -5< x< -3, negative for -3< x< -1, and, finally, positive for -1< x.
Or, rather than choosing specific values we could argue this way: x-a< 0 if x< a, x-a> 0 if x> a. (x+ 5)(x+ 3)(x+ 1)= (x-(-5))(x-(-3))(x-(-1)). If x< -5, all three factors are negative and so the entire product is negative. As we move past each of -5, -3, -1, exactly one factor changes sign so we alternate, -, +, -, +. That is really what Amer did but without the chart!