Question

If the zeros of a polynomial P(x)=x³-3x²+x+1 are in A.P, then find the zeros

Answer 1

Hey There,

Here, we are going to use the following concepts:

For a general cubic polynomial:
$ax^3+bx^2+cx+d$


Sum of zeros = $-\frac{b}{a}$

Product of zeros = $-\frac{d}{a}$

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Now, since this is a Cubic Polynomial, there would be Three Zeros.


We are given that the zeros are in AP.

Let the zeros be a-d, a, a+d



Here our polynomial is: $P(x)=x^3-3x^2+x+1$


Now, Sum of zeros is:

$(a-d)+a+(a+d)=-\frac{(-3)}{1} = 3 \\ \\ \implies 3a=3 \\ \\ \implies \boxed{a=1}$


Also, Product of Zeros is:

$(a-d)a(a+d)=-\frac{1}{1} \\ \\ \implies (1-d)(1+d)\times 1 = -1 \\ \\ \implies 1-d^2=-1 \\ \\ \implies d^2 = 2 \\ \\ \implies \boxed{d=\pm\sqrt{2}}$ 


In any case:

Zeros are $1-\sqrt{2} \, , 1 \, and \, 1+\sqrt{2}$


Hope it helps
Purva
Brainly Community