Solution,
Given; x^3-7x^2-x+7=0
Using factor theorem,
we have,
i)x^3-7x^2-x+7=0
let's check whether 1 is the factor of given equation or not by replacing the value of x by one..
or, 1^3-7.1^2-1+7=0
or, 1-7-1+7=0(.°. Cancellation of +1&-1 and +7&-7 in equation)
.°.0=0
which proves that 1 is one of the factor of x
therefore(x-1) is a factor of x^3-7x^2-x+7=0.
Now,
Expanding the x^3-7x^2-x+7=0 to get (x-1) as a factor...
ii)x^3-7x^2-x+7=0
or,x^3-x^2-6x^2+6x-7x+7
or,x^2(x-1) - 6x(x-1) - 7(x-1) = 0
or,(x-1) (x^2-6x-7)w = 0
or,,(x-1) (x^2-(7-1)x-7) =0
or,,(x-1) (x^2-7x + x -7) =0
or,(x-1) [x(x-7)+1(x-7)] = 0
or,(x-1) (x+1) (x-7) = 0
Either,
i) x - 1=0
.°.x = 1
And,
ii)x + 1 = 0
.°. x = -1
Or,
iii) x - 7 = 0
.°.x =7
Hence, the required value of x are -1 ,+1 & 7.
