Inspired by Vieta's Derivatives
The function f ( x ) = x 5 + 6 x 4 − 1 8 x 3 − 1 0 x 2 + 4 5 x − 2 4 has only four distinct roots, each of which is real. Let the four roots be α , β , γ , and δ , in no particular order. Also let f ′ ( x ) denote the first derivative of f ( x ) . Evaluate f ′ ( α ) × f ′ ( β ) × f ′ ( γ ) × f ′ ( δ ) .
The answer is 0.
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1 solution
To clarify, repeated roots are those with multiplicity greater than 1 . For example, in P ( x ) = ( x − 6 ) 2 ( x + 2 ) , the root x = 6 has multiplicity 2 , making it a repeated root , while the root x = − 2 has multiplicity 1 and is not repeated. Graphically, this means that polynomials approach horizontal slope at repeated roots; try it out yourself!
You can also take this a step further using the derivative. If a repeated root has even multiplicity, it "bounces" off the x axis, while odd multiplicity "wiggles" across. I call this the "bounce, wiggle, cross" theorem.
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I was just wondering. What can we do it we have to find a general relation between roots of polynomial and it's derivative
Even if somebody is not familiar with this fact about repeated roots, they can easily derive it for themselves: If f ( x ) = ( x − a ) 2 g ( x ) then f ′ ( x ) = 2 ( x − a ) g ( x ) + ( x − a ) 2 g ′ ( x ) so that f ′ ( a ) = 0 .
Nice question Nice description
At first i was like holy s * , how will i solve this, but the question is very easy you have some knowledge about nature of roots , and graphs of polynomials . Nice question.
The problem can be solved in seconds if you are familiar with the fact that repeated roots of a polynomial are also roots of its first derivative, that is if α is a repeated root of a polynomial f ( x ) , then it is also a root of f ′ ( x ) .
It is given that the polynomial has 4 distinct roots, but it is of degree 5 . Therefore there must be a repeated root. Since the repeated root (let's say α ) is also a root of the derivative, i.e. f ′ ( α ) = 0 , the product is 0 .