Question

2. The number 1 is a zero of the polynomial p(x)=x^3−3x^2+7x−5.

a. Write p(x) as a product of linear factors.
b. What are the solutions to the equation x^3−3x^2+7x−5=0?

Answer 1

(a) Factor will be (x-1)
`(1+2i+x)(-1+2i+x)` x = 1 , -1+2i and -1-2i

(b) Solution of the equation will be

Explanation:

We have given that 1 is the zero of the polynomial
`p(x)=x^3-3x^2+7x-5`

(a) As x is zero of the polynomial so (x-1) will completely divide the polynomial

So
`(x^3-3x^2+7x-5)/((x-1))=x^2-2x+5`

Now
`x^2-2x+5` can be factorized as
`(1+2i+x)(-1+2i+x)`

So the linear factor of polynomial
`p(x)=x^3-3x^2+7x-5` will be

(x-1)
`(1+2i+x)(-1+2i+x)`

(b) Solution of the equation will be x = 1 , -1+2i and -1-2i