Little help with polynomial identity? Suck at algebra :(

[Guest]

Hello everyone,,

find a and b for:



1 ÷ (x+2).(2x+1) ≡ a ÷ x+2 + b ÷ 2x+1

when x ≠= -2 and x ≠= -1/2



Thank you guys in advance!

[Guest]

Hello everyone,,

find a and b for:



1 ÷ (x+2).(2x+1) ≡ a ÷ x+2 + b ÷ 2x+1

when x ≠= -2 and x ≠= -1/2

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Thank you guys in advance!

Any advice guys,...

[HallsofIvy]

It's hard to understand what you are saying or what you are asking.

What you wrote is [tex]\frac{1}{(x+ 2)(2x+ 1)}= \frac{a}{x}+ 2+ \frac{b}{2x}+ 1[/tex]
but I suspect you intended [tex]frac{1}{(x+ 2)(2x+ 1)}= \frac{a}{x+ 2}+ \frac{b}{2x+ 1}[/tex].
"at x= -2 and x= -1/2".

Well, first, that is IMPOSSIBLE "at x= -2 and x= -1/2" because neither side exists at those values of x.

That is, in Calculus, referred to as "partial fractions". There are several ways to determine a and b but the basic idea is that in order to find two values, a and b, we need two equations.

I suspect you are partially remembering a specific method for determining a and b.

[tex]\frac{1}{(x+ 2)(2x+ 1)}= \frac{a}{x+ 1}+ \frac{b}{2x+ 1}[/tex]
multiply on both sides by the common denominator (x+ 2)(2x+ 1) to clear the fractions
[tex]1= a(2x+ 1)+ b(x+ 2).

Now, you said "at x= -2 and x= -1/2" and I protested that the fractions did not exist at those values of x.
But now that we have "cleared" the fractions we can set x= -2 and x= -1/2 and when we do that one of a or b is eliminated.

When x= -2, we have [tex]1= a(2(-2)+ 1)+ b(-2+ 2)= -3a[/tex] so [tex]a= -\frac{1}{3}[/tex].
When x= -1/2, we have [tex]1= a(2(-1/2)+ 1)+ b(-1/2+ 2)= (3/2)b[/tex] so [tex]b= \frac{2}{3}[/tex]

That is [tex]\frac{1}{(x+ 2)(2x+ 1)}= \frac{-\frac{1}{3}}{x+ 2}+ \frac{\frac{2}{3}}{2x+ 1}[/tex].

We can check that answer by getting the "common denominator" on the right and doing the addition:
[tex]\frac{-\frac{1}{3}(2x+ 1)+ \frac{2}{3}(x+ 2)}{(x+ 2)(2x+ 1)}= \frac{(-\frac{2}{3}x- \frac{1}{3}+ \frac{2}{3}x+ \frac{4}{3}}{(x+ 2)(2x+ 1)}= \frac{1}{(x+ 2)(2x+1)}[/tex]!