Question

If the zeroes of the polynomial p(x)=x^3-3x^2+x+1 are a-b,a,a+b find a and b

Answer 1

Given,

${p(x) = x^3-3x^2+x+1}$

From this equation,

a = 1, b = -3, c = 1, d = -1

The roots are a - b, a , a+ b

Sum of roots = ${\huge{\frac{-b}{a}}}$

==> a - b + a + a+ b = ${\huge{\frac{-(-3)}{1}}}$

==> 3a = 3

==> a = 1........(1)

Product of roots = ${\huge{\frac{-d}{a}}}$

==> (a - b)(a)(a + b) = ${\huge{\frac{-(1)}{1}}}$

==> (1-b)(1)(1+b) = -1 [From (1)]

==> ${1-b^2 = -1}$

==> ${b^2 = 2}$

==> ${b = +-\sqrt{2}}$

${\huge{\orange{\boxed{a = 1 ; b = +-\sqrt{2}}}}}$