Question

Can someone explain to me how to do this:

Answer 1

each time i try to solve that I get a fraction

Answer 2

those are separate questions, right ?

Answer 3

the one that starts from "Determine the value"
and the other that starts from "The graph of a 4\(^{th}\) degree"

Answer 4

yes they are, sorry

Answer 5

ok, even if instead of "1" on top of the x-5 you had an unknown (say B) you would be able to solve for A and B

you need a partial fraction expansion for this

Answer 6

I've never seen something like this

Answer 7

when do I will learn this?

Answer 8

well If im correct this is part algebra and Trig

Answer 9

yes, algebra

Answer 10

so I think it will be next year

Answer 11

4th degree polynomial 4 roots only 2 real showing

Answer 12

\(\color{black}{ \displaystyle \frac{ A }{x-4} +\frac{ B }{x-5}=\frac{ -x+6 }{x^2-9x+20} }\)

you see that (x-4)(x_5) is the same as \(x^2-9x+20\) but factored, (right?)
multiply every term by the common denominator \(x^2-9x+20\), and you get:

\(\color{black}{ \displaystyle \frac{ A }{x-4} +\frac{ B }{x-5}=\frac{ -x+6 }{x^2-9x+20} }\)

\(\color{black}{ \displaystyle \frac{ A(x-5)(x-4) }{x-4} +\frac{ B(x-5)(x-4) }{x-5}=\frac{ (-x+6)(x-5)(x-4) }{x^2-9x+20} }\)

\(\color{black}{ \displaystyle A(x-5) +B(x-4) =-x+6}\)

plug in x=4 and find the A
plug in x=5 and find the B

Answer 13

in this case, you are already given the B, so you have

\(\color{black}{ \displaystyle A(x-5) +1(x-4) =-x+6}\)
\(\color{black}{ \displaystyle A(x-5) +x+4=-x+6}\)
\(\color{black}{ \displaystyle A(x-5) +x=-x+2}\)
\(\color{black}{ \displaystyle A(x-5) =-2x+2}\)

Answer 14

it is still true for all x values, so you can plug any x value.

try x=4

Answer 15

I miss those days partial fractions are great

Answer 16

that would give you 6

Answer 17

I did it incorrectly

Answer 18

A(x-5)+1(x-4)=-x+6
A(x-5)+x-4=-x+6
A(x-5)=-2x+10

now plug in x=6