Question

x-√5 is factor of x³-3√5 x²+13x-3√5 .find zeroes

Answer 1

Since x – √5 is a factor of the cubic polynomial x3 -3√5 x2 + 13x - 3√5, let’s divide them as shown using synthetic division Hence x3 – 3√5 x2 + 13x – 3√5 = (x – √5)(x2 – 2√5x + 3) = (x – √5)[x2 – (√5 +√2) x – (√5 – √2) x + (√5 +√2)(√5 – √2)] = (x – √5)[x{x – (√5 +√2)} – (√5 – √2){ x + (√5 +√2)}] = (x – √5)[x – (√5 +√2)][x– (√5 – √2)] To get the zero of the given polynomial, we take (x – √5)[x – (√5 +√2)][x– (√5 – √2)] = 0 ⇒ (x – √5) = 0, [x – (√5 +√2)] = 0 and [x– (√5 – √2)] = 0 ⇒ x = √5, x = (√5 +√2) and x = (√5 – √2) Thus the zeroes of the given polynomial are √5, (√5 +√2) and (√5 – √2