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Worked solution

Divide an improper rational expression before decomposing

Express 3x3+x−4(3x+1)(x2+1)\dfrac{3x^3+x-4}{(3x+1)(x^2+1)} in partial fractions.

CJC · O Level revision · Example 15

Answer

1−43x+1−1−xx2+11-\dfrac4{3x+1}-\dfrac{1-x}{x^2+1}, for real x≠−13x\ne-\dfrac13.

Divide the equal-degree polynomials first

Answer

Quotient 11, remainder −x2−2x−5-x^2-2x-5.

Expand the denominator and include the missing numerator term

(3x+1)(x2+1)=3x3+x2+3x+1,3x3+0x2+x−4(3x+1)(x^2+1)=3x^3+x^2+3x+1,\qquad3x^3+0x^2+x-4

The source callout requires long division because numerator and denominator both have degree 3. Keep 0x² in the numerator to align coefficients.

Subtract the denominator once

(3x3+0x2+x−4)−(3x3+x2+3x+1)=−x2−2x−5,3x3+x−4(3x+1)(x2+1)=1−x2+2x+5(3x+1)(x2+1).\begin{aligned}(3x^3+0x^2+x-4)-(3x^3+x^2+3x+1)&=-x^2-2x-5,\\\frac{3x^3+x-4}{(3x+1)(x^2+1)}&=1-\frac{x^2+2x+5}{(3x+1)(x^2+1)}.\end{aligned}

This retains the complete printed division: quotient 1 and the negative quadratic remainder.

How did this part go?

Decompose the positive remainder numerator

Answer

x2+2x+5(3x+1)(x2+1)=43x+1+1−xx2+1\dfrac{x^2+2x+5}{(3x+1)(x^2+1)}=\dfrac4{3x+1}+\dfrac{1-x}{x^2+1}.

Clear the proper fraction denominator

x2+2x+5(3x+1)(x2+1)=A3x+1+Bx+Cx2+1 ⟹ x2+2x+5=A(x2+1)+(Bx+C)(3x+1)\begin{aligned}&\frac{x^2+2x+5}{(3x+1)(x^2+1)}\\ & =\frac A{3x+1}+\frac{Bx+C}{x^2+1}\ \Longrightarrow\ x^2\\ &{}+2x+5\\ & =A(x^2+1)\\ &{}+(Bx+C)(3x+1)\end{aligned}

The source decomposes the positive x²+2x+5 numerator; the outer minus sign remains for the final step.

Find A then C by substitution

x=−13:19−23+5=109A ⟹ 109A=409 ⟹ A=4,x=0:5=4+C ⟹ C=1.\begin{aligned}x=-\frac13:&\quad\frac19-\frac23+5=\frac{10}9A\ \Longrightarrow\ \frac{10}9A=\frac{40}9\ \Longrightarrow\ A=4,\\x=0:&\quad5=4+C\ \Longrightarrow\ C=1.\end{aligned}

The printed values are x=−1/3 and x=0.

Use x=1 for the last coefficient

1+2+5=4(2)+(B+1)(4) ⟹ B=−1\begin{aligned}&1+2+5\\ & =4(2)+(B+1)(4)\ \Longrightarrow\ B\\ & =-1\end{aligned}

This gives B=−1, so the linear numerator is 1−x.

How did this part go?

Restore the outer subtraction

Answer

1−43x+1−1−xx2+11-\dfrac4{3x+1}-\dfrac{1-x}{x^2+1}.

Subtract both proper-fraction terms

1−(43x+1+1−xx2+1)=1−43x+1−1−xx2+1\begin{aligned}&1-\left(\frac4{3x+1}+\frac{1-x}{x^2+1}\right)\\ & =1-\frac4{3x+1}-\frac{1-x}{x^2+1}\end{aligned}

The negative remainder changes the sign of the entire decomposed expression. For real x, x²+1 is positive; only x=−1/3 is excluded. The domain check is independent of the printed working.

How did this part go?

Common mistakes

Dividing an improper fraction produces a polynomial part plus a proper remainder. Do not discard the quotient 1 or apply the remainder minus sign to only one term.