Asked by Michael
Solve graphically (x-2)^3(x+4)(3-x)^2(x+1)^4 <0
Answers
Answered by
Steve
You know where the roots are.
You know that if the root is of odd order, the graph crosses the x-axis there.
And, if the root is of even order, the graph is just tangent there.
So, knowing what you do about the general shape of polynomials, since this one is of even order,
The roots are -4,-1,2,3
y > 0 for -∞ < x < -4
y < 0 for -4 < x < -1
y < 0 for -1 < x < 2
y > 0 for 2 < x < 3
y > 0 for 3 < x < ∞
Verify this at
http://www.wolframalpha.com/input/?i=(x-2)%5E3(x%2B4)(3-x)%5E2(x%2B1)%5E4+,+-2+%3C+x+%3C+10%2F3
You know that if the root is of odd order, the graph crosses the x-axis there.
And, if the root is of even order, the graph is just tangent there.
So, knowing what you do about the general shape of polynomials, since this one is of even order,
The roots are -4,-1,2,3
y > 0 for -∞ < x < -4
y < 0 for -4 < x < -1
y < 0 for -1 < x < 2
y > 0 for 2 < x < 3
y > 0 for 3 < x < ∞
Verify this at
http://www.wolframalpha.com/input/?i=(x-2)%5E3(x%2B4)(3-x)%5E2(x%2B1)%5E4+,+-2+%3C+x+%3C+10%2F3
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