If the function ax^5+ bx^3+c has five roots which of a, b and c cannot be zero?

Asked on by dhopkins895

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

let f(x) = ax^5 + bx^3 + c

Given that f(x) has 5 roots:

If c = 0  ==> f(x) = x^3(ax^2 + b)

==> f(x) will have 3 roots oly, then c can not be 0.

If b = 0 ==> f(x) = ax^5 + c ==> x = (c/a)^1/5 is the only zero.

Then b can not be zero.

if a = 0 ==> f(x) = bx^3  +c ==> f(x) will be 3 gegree funcion and wil not have 5 roots. then a can not be 0 .

We conclude that:

if f(x) has 5 roots, then a,b, c, can not be zero.

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

For a function f(x) which has five roots, the highest power of x should be 5. We are given f(x) = ax^5 + bx^3 + c.

So this has to have the highest power of x as 5, therefore 'a' cannot be zero else the expression will have a highest power of 3.

Also c cannot be zero, because if it is taken to be zero, the expression is reduced to ax^5 + bx^3= x^3(ax^2+b), this can have a maximum of three roots including 0.

If b is zero the expression is ax^5+c. Now equating ax^5+c=0,

x^5 = c/a . Now this also cannot yield 5 values of x.

So neither of  a , b or c can be zero if the expression has 5 roots.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

f(x) = ax^5+bx^3+c  has 5 roots.

To determine which of a, b and c could be zero.

Solution:

If a = 0, then f(x) = bx^3+c becomes a 3rd degree polnomial which could have 3 roots. So it cannot have 5 roots.

So a cannot be zero.

If c = 0, then f(x) = ax^5+bx^3 = 0 becomes x^3(ax^2+b) = so x =0 or 2 roots from ax^2+b = 0 whic contradicts f(x) has 5 roots.

If b= 0, then ax^5 +c = 0 can have 5 roots real or complex.

So and c cannot be zero. b could be zero under the given condition.

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