Answer:
The complete set of roots is 5, -3, –4 + i, –4 – i
Step-by-step explanation:
The given polynomial is
[tex]f(x)=(x^2-2x-15)(x^2+8x+17)[/tex]
We equate to zero to get:
[tex](x^2-2x-15)(x^2+8x+17)=0[/tex]
[tex](x^2-5x+3x-15)(x^2+8x+17)=0[/tex]
[tex](x(x-5)+3(x-5)(x^2+8x+17)=0[/tex]
[tex](x+3)(x-5)(x^2+8x+17)=0[/tex]
This implies that:
x+3=0,or x-5=0 or [tex](x^2+8x+17)=0[/tex]
x=-3,or x=5
For [tex](x^2+8x+17)=0[/tex], we use the quadratic formula to get;
[tex]x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]
where a=1,b=8,c=17
[tex]x=\frac{-8\pm\sqrt{8^2-4(1)(17)} }{2(1)}[/tex]
[tex]x=\frac{-8\pm\sqrt{-4} }{2}[/tex]
[tex]x=-4-i[/tex] or [tex]x=-4+i[/tex]
The complete solution is 5, -3, –4 + i, –4 – i
Answer:
-5 , 3 , -4 + i, -4 - i B.is the answer on edg2020
Step-by-step explanation:
