Answer:
1. x-5, x-15 - No
x+3 - Yes
2. [tex]a=k^2-2kr+r^2-p[/tex]
Step-by-step explanation:
1. If the binomial [tex]x-a[/tex] is a factor of polynomial function [tex]f(x),[/tex] then [tex]f(a)=0.[/tex]
Check all options:
A. For [tex]x-5,\ a=5[/tex] then
[tex]f(5)=5^3+5\cdot 5^2-9\cdot 5-45\\ \\=125+125-45-45\\ \\=250-90\\ \\=160\neq 0[/tex]
So, [tex]x-5[/tex] is not a factor.
B. For [tex]x+3,\ a=-3[/tex] then
[tex]f(-3)=(-3)^3+5\cdot (-3)^2-9\cdot (-3)-45\\ \\=-27+45+27-45\\ \\=0[/tex]
So, [tex]x+3[/tex] is a factor.
C. For [tex]x-15,\ a=15[/tex] then
[tex]f(15)=15^3+5\cdot 15^2-9\cdot 15-45\\ \\=3,375+1,125-135-45\\ \\=4,500-180\\ \\=4,320\neq 0[/tex]
So, [tex]x-15[/tex] is not a factor.
2. Solve for [tex]a:[/tex]
[tex]\sqrt{p+a}+r=k[/tex]
First, subtract r:
[tex]\sqrt{p+a}+r-r=k-r\\ \\\sqrt{p+a}=k-r[/tex]
Square it:
[tex](\sqrt{p+a})^2=(k-r)^2\\ \\p+a=k^2-2kr+r^2[/tex]
Subtract p:
[tex]p+a-p=k^2-2kr+r^2-p\\ \\a=k^2-2kr+r^2-p[/tex]
