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?
Accepted Solution
A:
Answer:(a) Factor will be (x-1) [tex](1+2i+x)(-1+2i+x)[/tex] x = 1 , -1+2i and -1-2i(b) Solution of the equation will be Step-by-step explanation:We have given that 1 is the zero of the polynomial [tex]p(x)=x^3-3x^2+7x-5[/tex](a) As x is zero of the polynomial so (x-1) will completely divide the polynomial So [tex]\frac{x^3-3x^2+7x-5}{(x-1)}=x^2-2x+5[/tex]Now [tex]x^2-2x+5[/tex] can be factorized as [tex](1+2i+x)(-1+2i+x)[/tex]So the linear factor of polynomial [tex]p(x)=x^3-3x^2+7x-5[/tex] will be (x-1) [tex](1+2i+x)(-1+2i+x)[/tex](b) Solution of the equation will be x = 1 , -1+2i and -1-2i