How do you express #(x^3 - 5x + 2) / (x^2 - 8x + 15)# in partial fractions?

Question

Elias Ackerman · Accepted Answer

#(x^3-5x+2)/(x^2-8x+15) = x+8-7/(x-3)+51/(x-5)#

#(x^3-5x+2)/(x^2-8x+15)#

#=(x^3-8x^2+15x+8x^2-64x+120+44x-118)/(x^2-8x+15)#

#=(x(x^2-8x+15)+8(x^2-8x+15)+44x-118)/(x^2-8x+15)#

#=x+8+(44x-118)/(x^2-8x+15)#

#=x+8+(44x-118)/((x-3)(x-5))#

#=x+8+A/(x-3)+B/(x-5)#

Using Heaviside's cover-up method, we find:

#A=(44(3)-118)/((3)-5) = (132-118)/(-2) = 14/(-2) = -7#

#B=(44(5)-118)/((5)-3) = (220-118)/2 = 102/2 = 51#

So:

#(x^3-5x+2)/(x^2-8x+15) = x+8-7/(x-3)+51/(x-5)#

Sadie Alexander · Answer

undefined To express $ \frac{x^3 - 5x + 2}{x^2 - 8x + 15} $ in partial fractions, we first need to factor the denominator.

The quadratic expression $ x^2 - 8x + 15 $ factors as $ (x - 3)(x - 5) $.

So, we can rewrite the expression as:

$$ \frac{x^3 - 5x + 2}{x^2 - 8x + 15} = \frac{x^3 - 5x + 2}{(x - 3)(x - 5)} $$

Now, we need to express the fraction on the right-hand side in partial fractions. Since the degree of the numerator is greater than or equal to the degree of the denominator, we start by performing polynomial long division or synthetic division to divide $ x^3 - 5x + 2 $ by $ (x - 3)(x - 5) $.

After performing the division, we obtain:

$$ \frac{x^3 - 5x + 2}{(x - 3)(x - 5)} = \frac{A}{x - 3} + \frac{B}{x - 5} $$

where $ A $ and $ B $ are constants to be determined.

Now, we multiply both sides by $ (x - 3)(x - 5) $ to clear the fractions:

$$ x^3 - 5x + 2 = A(x - 5) + B(x - 3) $$

Expanding and collecting like terms, we get:

$$ x^3 - 5x + 2 = (A + B)x - 5A - 3B $$

Comparing coefficients, we can solve for $ A $ and $ B $.

Finally, we substitute the values of $ A $ and $ B $ back into the expression to obtain the partial fraction decomposition.