Question

obtain all the zeros of x3-7x+6 if one of its zero is 1

Answer 1

hi friend,


since 1 is a zero of x³-7x+6

(x-1)is the factor of x³-7x+6

so if we divide x³-7x+6 with (x-1) then we will get x²+x-6

so we can write x³-7x+6 as (x-1)(x²+x-6)

=(x-1)(x²+3x-2x-6)

=(x-1)(x(x+3)-2(x+3))
=(x-1)(x-2)(x+3)
so the other zeros are 2 and -3


I.hope this will help u :)

Answer 2

Hi friend,

given polynomial,
x³-7x+6

One of the zero is 1,means
(x-1) is a factor of x³-7x+6

Divide x³-7x+6 by (x-1),
we get x²+x-6

x³-7x+6=(x-1)(x²+x-6)
x³-7x+6=(x-1)(x²-2x+3x-6)
x³-7x+6=(x-1)(x(x-2)+3(x-2))
x³-7x+6=(x-1)(x-2)(x+3)

The other two factors are (x-2) and (x+3)
So, the other two zeroes of the given polynomial are 2 and -3.

hope it helps