Does the (n−1)th derivative of an n-degree polynomial give the arithmetic mean of all the roots of the polynomial?

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surgescasjag · Accepted Answer

Step 1Consider,a0xn+a1xn−1+⋯+an=0We have, sum of roots =−a1a0Now if u take the (n−1)th derivative of the expression you have,n!a0x+(n−1)!a1=0⇒x=−a1na0=−a1a0×1xThis is the arithmetic mean of the roots we have the sum of the roots divided by n.

homofilirix · Accepted Answer

Step 1f(x)=x2−5x+6This one is simple. Here, the factors are 2 and 3. So, their average is 52Also, the (n−1) or the first derivative of f(x) is2x−5=0⇒x=52Step 2(x−3)(x+2)(x+5)=x3+4x2−6x−30=f(x)Here, a.m. of roots =3−2−53=−43And f″x=6x+8=0⇒x=−43The proof,Consider, a0xn+a1xn−1+⋯+an=0When we take its (n−1)th derivative,a0xn+a1xn−1=0⇒x=−a1na0=−a1a0×1n,which is the arithmetic mean of all the roots of f(x).

Does the (n-1)th derivative of an n-degree polynomial give the

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