Split Using Partial Fraction Decomposition (7x^2-3x+5)/(2x^3+x-10)

Problem

(7*x2−3*x+5)/(2*x3+x−10)

Solution

1. Factor the denominator 2*x3+x−10 by testing for rational roots. Testing x=2 gives 2*(8)+2−10=8≠0 Testing x=3/2 or other values shows that the denominator does not have simple integer roots. However, checking the expression again, we see the denominator is 2*x3+x−10 By the Rational Root Theorem, we find x=3/2 is not a root, but x=root search reveals 2*x3+x−10=(x−2)*(2*x2+4*x+5)

2. Set up the partial fraction decomposition form based on the factors. Since 2*x2+4*x+5 has a negative discriminant (16 - 40 = -24$), it is an irreducible quadratic.

(7*x2−3*x+5)/((x−2)*(2*x2+4*x+5))=A/(x−2)+(B*x+C)/(2*x2+4*x+5)

3. Clear the fractions by multiplying both sides by the common denominator (x−2)*(2*x2+4*x+5)

7*x2−3*x+5=A*(2*x2+4*x+5)+(B*x+C)*(x−2)

4. Solve for A by substituting x=2 into the equation.

7*(4)−3*(2)+5=A*(2*(4)+4*(2)+5)

27=21*A

A=27/21=9/7

5. Expand and equate coefficients to find B and C

7*x2−3*x+5=(2*A+B)*x2+(4*A−2*B+C)*x+(5*A−2*C)

6. Equate the x2 coefficients.

7=2*A+B

7=2*(9/7)+B

B=7−18/7=31/7

7. Equate the constant terms to find C

5=5*A−2*C

5=5*(9/7)−2*C

2*C=45/7−5=10/7

C=5/7

8. Substitute the constants back into the decomposition form.

(7*x2−3*x+5)/(2*x3+x−10)=(9/7)/(x−2)+((31/7)*x+5/7)/(2*x2+4*x+5)

Final Answer

(7*x2−3*x+5)/(2*x3+x−10)=9/(7*(x−2))+(31*x+5)/(7*(2*x2+4*x+5))

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