what number can I substitute to X so that x(cube)-7x(squared)+7x-6 will be equal to zero?

Answer 1

x^3 - 7x^2 + 7x - 6 = 0
x^3 - x^2 - 6x^2 + 6x + x -6 = 0
x^2(x-1) - 6x(x-1) + (x-6) =0
(x-1)(x^2 - 6x) + (x-6) =0
(x-1)(x-6)(x) + (x-6) =0
(x-6)(x^2 - x +1 )=0
so we have x- 6 =0
which gives x= 6

and x^2 - x + 1=0
which gives solution in imaginary number (i= sqaure root of -1)

x= (1+isqrt3)/2
x= (1-isqrt3)/2

Answer 2

x3 - 7x2 + 7x - 6 = 0
Put x = 6
216 - 252 + 42 - 6 = 0
so (x - 6) is the one root of the equation

Divide x3 - 7x2 + 7x - 6 = 0 by (x - 6)

x2(x - 6) - x (x - 6) + 1 (x - 6) = 0
(x - 6) (x2 - x + 1) = 0
(x - 6) (x2 - x + 1) = 0
x = 0 and x2 – x + 1 = 0
for x2 – x + 1 = 0
x = 1 ± sqrt(1-4 ) / 2*1
x = 1 ± sqrt(-3) / 2
x = (1 ± 3i) / 2
this cubic equation has one real root x = 6 and other two roots are imaginary i..e x = (1 ± 3i) / 2

Answer 3

x3 - 7x2 + 7x - 6 = 0
(writing mxn for m* x to the power n)
or (x+2)(x2 -2x -3) = 0
or (x+2)(x+1)(x-3) = 0
so you can substitute any of the values
-2, -1 or 3

Answer 4

you factor the polynomial
x(cube)-7x(sqr)-6

you want to try something of the form (ax+b) where a is a factor of 1 (since it's one x(cube)) and b is going to be a factor of (-6)

so it equals (x-6)[(x(sqr)-x+1)]
which implies x must = 6 for this to equal to zero.

(since the solution to the quadratic would consists of imaginary number. if you need this, you can simply solve it by using the quadratic equation)

Answer 5

there is only one real root
p(6)=0
p(x)=(x-6)(x^2-x+1)=0

x1=6

the complex roots are
x2=(1-i*sqrt(3))/2

x3=(1+i*sqrt(3))/2