Question

5. Solve the polynomial equation:

Answer 1

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.