How to simplify using partial fraction. 5/(x^2-5x+6) ?
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calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We have to express the given expression 5/(x^2-5x+6) as partial fractions.
5/(x^2-5x+6)
=> 5 / ( x^2 - 3x - 2x + 6)
=> 5/ ( x(x - 3) - 2( x - 3))
=> 5/ (x - 2)(x - 3)
=> A / (x - 2) + B/ (x - 3)
=> [A(x - 3) + B(x - 2)]/ (x - 2)(x -3)
Ax - 3A + Bx - 2B = 5
=> Ax + Bx - 3A - 2B = 5
Equate the coefficients of x and the numeric coefficients
=> A + B = 0 and 3A + 2B = -5
=> A = -B
Substitute in 3A + 2B = -5
=> -3B + 2B = -5
=> B = 5
A = -5
We can write 5/(x^2-5x+6) = 5/(x - 3) - 5/(x - 2).
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
We need to rewrite the fraction 5/(x^2-5x+6).
First we will simplify the denominator.
==> 5/(x^2-5x+6)= 5/(x-2)(x-3)
==> 5/(x-2)(x-3)= A/(x-2) + B/(x-3)
Now we will simplify by multiplying both sides by (x-2)(x-3)
==> 5 = A(x-3) + B(x-2)
==> 5 = Ax - 3A + Bx -2B
==> 5= (A+B)x + (-3A-2B)
==> A+B = 0 ==> A = -B
==> -3A -2B = 5
==> -3(-B) - 2B = 5
==> 3B - 2B = 5
==> B= 5
==> A = -5
==> 5/(x^2 -5x+6) = -5/(x-2) + 5/(x-3)
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