Seeking Math help +rep pre cal questions

[John]

1. Translate the statement below to an equation using absolute value and solve:
The distance between 50 and me is at least 10.


2. Using the given zero, find all of the zeros and write a linear factorization of f(x).

1+3i is a zero of f(x)=x^4-2x^3+5x^2+10x-50

[Carrot]

1. Translate the statement below to an equation using absolute value and solve:
The distance between 50 and me is at least 10.

|x-50| ≥ 10

[John]

thanks homie

[Carrot]

No problem. I don't remember how to do #2 though D:

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[bamag]

1+3i is a zero of f(x)=x^4-2x^3+5x^2+10x-50
Since 1+3i is a root, 1-3i is also a root. ((x-1)-3i)((x-1)+3i)=(x-1)^2-(3i)^2=x^2-2x+1-9i^2=x^2-2x+10
Use long division to get x^2-5.
Find the two roots of that are +- radical 5.
Therefore, all the zeroes are 1+3i,1-3i, radical 5, negative radical 5.
Linear factorization is f(x)=(x-1-3i)(x-1+3i)(x+5^.5)(x-5^.5)

[dodowong303]

1+3i is a zero of f(x)=x^4-2x^3+5x^2+10x-50
Since 1+3i is a root, 1-3i is also a root. ((x-1)-3i)((x-1)+3i)=(x-1)^2-(3i)^2=x^2-2x+1-9i^2=x^2-2x+10
Use long division to get x^2-5.
Find the two roots of that are +- radical 5.
Therefore, all the zeroes are 1+3i,1-3i, radical 5, negative radical 5.
Linear factorization is f(x)=(x-1-3i)(x-1+3i)(x+5^.5)(x-5^.5)

it is cool, but you may give the explanation that why 1-3i is also a root.


you may add [since all coefficient are real number, therefore (a+bi)(1+3i)=constant +0i,hence a=1,b=3 ]

[bamag]

it is cool, but you may give the explanation that why 1-3i is also a root.


you may add [since all coefficient are real number, therefore (a+bi)(1+3i)=constant +0i,hence a=1,b=3 ]

It is simply the conjugate roots theorem. If a+bi is a root, then a-bi is also a root, and vise versa.

[dodowong303]

It is simply the conjugate roots theorem. If a+bi is a root, then a-bi is also a root, and vise versa.

i know but my teacher will deduct marks form missing some stupid simple explanation such as a+bi&a-bi must be pair,etc.......==