Question

I need your help ASAP! Only Correct Answers Please And Thank you.

Answer 1

1. x-5, x-15 - No

x+3 - Yes

2.
`a=k^2-2kr+r^2-p`

Explanation:

1. If the binomial
`x-a` is a factor of polynomial function
`f(x),` then
`f(a)=0.`

Check all options:

A. For
`x-5,\ a=5` then

`f(5)=5^3+5\cdot 5^2-9\cdot 5-45\\ \\=125+125-45-45\\ \\=250-90\\ \\=160\\eq 0`

So,
`x-5` is not a factor.

B. For
`x+3,\ a=-3` then

`f(-3)=(-3)^3+5\cdot (-3)^2-9\cdot (-3)-45\\ \\=-27+45+27-45\\ \\=0`

So,
`x+3` is a factor.

C. For
`x-15,\ a=15` then

`f(15)=15^3+5\cdot 15^2-9\cdot 15-45\\ \\=3,375+1,125-135-45\\ \\=4,500-180\\ \\=4,320\\eq 0`

So,
`x-15` is not a factor.

2. Solve for
`a:`

`√(p+a)+r=k`

First, subtract r:

`√(p+a)+r-r=k-r\\ \\√(p+a)=k-r`

Square it:

`(√(p+a))^2=(k-r)^2\\ \\p+a=k^2-2kr+r^2`

Subtract p:

`p+a-p=k^2-2kr+r^2-p\\ \\a=k^2-2kr+r^2-p`