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Summary

Algebraic division

​​In a nutshell

An algebraic fraction can be written in the form ​P(x)Q(x)\cfrac{P(x)}{Q(x)}Q(x)P(x)​, where P(x)P(x)P(x) and Q(x)Q(x)Q(x) are polynomials. If the order of P(x)P(x)P(x) is larger than or equal to the order of Q(x)Q(x)Q(x), this is an improper algebraic fraction. These types of fractions must be turned into mixed fractions in order to be expressed in terms of partial fractions. 


Note: The order of a polynomial is given by the highest power of xxx.



Algebraic division

There are two different methods used to convert an improper fraction into a mixed one:

  • Algebraic division
  • ​F(x)=Q(x)×divisor + remainderF(x) = Q(x) \times \text{divisor} \ + \ \rm remainderF(x)=Q(x)×divisor + remainder relation


Algebraic division

procedure

1.
Identify P(x), Q(x)P(x), \ Q(x)P(x), Q(x) in your division.​
2.
Write down the elements of the algebraic division.
3.
Compare the first element of P(x)P(x)P(x) and Q(x)Q(x)Q(x) to find out the multiplier (the term you multiply Q(x)Q(x)Q(x) by to get P(x)P(x)P(x)​.
4.
Multiply Q(x)Q(x)Q(x) by the multiplier and subtract that result from P(x)P(x)P(x).
5.
Repeat steps from 1-4 until you reach the final answer.


Example 1

Simplify  6x3+13x2−43x+2\cfrac{6x^3 + 13 x^2 -4}{3x+2}3x+26x3+13x2−4​​


First, write down the elements:

​3x+2) 6x3+13x2  −4‾\begin{array}{r}3x+2 {\overline{\smash{\big)}\, 6x^3+13x^2 \ \ -4}}\end{array}3x+2)6x3+13x2  −4​​​​


Compare the first element of P(x)P(x)P(x) and Q(x)Q(x)Q(x), and write down the multiplier:

​2x23x+2) 6x3+13x2  −4‾\begin{array}{r}2x^2 \hspace{19mm} \\3x+2 {\overline{\smash{\big)}\, 6x^3+13x^2 \ \ -4}}\end{array}2x23x+2)6x3+13x2  −4​​​​


Multiply Q(x)Q(x) Q(x) by the quantity found in the previous step, and subtract the result from P(x)P(x)P(x)​:​

​2x23x+2) 6x3+13x2  −4‾−6x3+4x2 0x3+9x2  −4‾\begin{array}{r}2x^2 \hspace{19mm} \\3x+2 {\overline{\smash{\big)}\, 6x^3+13x^2 \ \ -4}}\\- \hspace{25mm}\\ 6x^3 + 4x^2\hspace{9.5mm}\\\overline{\smash{}\,0x^3 +9x^2 \ \ -4\hspace{1.5mm}}\end{array}2x23x+2)6x3+13x2  −4​−6x3+4x20x3+9x2  −4​​​​


Repeat these steps until you reach the final answer.

​2x2+3x−23x+2) 6x3+13x2  −4‾−6x3+4x2 +9x2−4‾−9x2+6x  −6x−4‾−−6x−4  0‾\begin{array}{r}2x^2+3x -2 \hspace{5mm} \\3x+2 {\overline{\smash{\big)}\, 6x^3+13x^2 \ \ -4}}\\- \hspace{25mm}\\ 6x^3 + 4x^2\hspace{9.5mm}\\\overline{\smash{}\,\hspace{5mm} +9x^2 \hspace{5mm} -4\hspace{1.5mm}}\\- \hspace{22mm}\\9x^2+ 6x \hspace{4.8mm}\\\overline{\smash{}\,\hspace{10mm} \ -6x-4 }\\- \hspace{11mm}\\-6x - 4 \\\overline{\smash{}\,\hspace{10mm} \ 0 }\end{array}2x2+3x−23x+2)6x3+13x2  −4​−6x3+4x2+9x2−4​−9x2+6x −6x−4​−−6x−4 0​​​


Therefore 6x3+13x2−43x+2=2x2+3x−2+0‾\underline{\cfrac{6x^3+13x^2-4}{3x+2} =2x^2+3x-2 + 0}3x+26x3+13x2−4​=2x2+3x−2+0​​


Note: You can factorise your result, but in this case it is not necessary. 


​F(x)=Q(x)×divisor + remainder\bf F(x) = Q(x) \times \textbf{divisor} \ + \ \rm\bf remainder F(x)=Q(x)×divisor + remainder relation

This method is really useful when dealing with expressions with the form: 

​ F(x)=Q(x)×D(x) + R(x)\ F(x) = Q(x) \times D(x) \ + \ R(x) F(x)=Q(x)×D(x) + R(x)​​


Where D(x)D(x)D(x) is the divisor and R(x)R(x)R(x) the remainder. 


Procedure

1.
Multiply out the brackets if necessary.
2.
Group together the cubic, quadratic, linear and constant terms.
3.
Equate these terms on both sides of the expression.
4.
Solve for the values, using substitution.


Example 2

Given that 6x3+13x2−4=(3x+2)×(Ax2+Bx+C)+R{6x^3+13x^2-4}=({3x+2})\times (Ax^2+Bx+C) + R6x3+13x2−4=(3x+2)×(Ax2+Bx+C)+R, find the values A, B, CA, \ B, \ CA, B, C and RRR​​


Multiply out the brackets.

6x3+13x2−4=3Ax3+3Bx2+3Cx+2Ax2+2Bx+2C+R{6x^3+13x^2-4}=3Ax^3 + 3Bx^2 + 3Cx + 2Ax^2 + 2Bx + 2C + R6x3+13x2−4=3Ax3+3Bx2+3Cx+2Ax2+2Bx+2C+R​​


Group the cubic, quadratic, linear and constant terms together: 

​6x3+13x2−4=3Ax3+x2(3B+2A)+x(3C+2B)+(2C+R){6x^3+13x^2-4}=3Ax^3 + x^2 (3B + 2A) + x (3C + 2B) + (2C + R)6x3+13x2−4=3Ax3+x2(3B+2A)+x(3C+2B)+(2C+R)​​

​

Equate the cubic, quadratic, linear and constant terms: 

​{6x3=3Ax313x2=x2(3B+2A)0x=x(3C+2B)−4=2C+R\begin{cases}\begin{aligned}6x^3 &= 3Ax^3\\13x^2 &= x^2 (3B+2A) \\0x & = x(3C + 2 B)\\-4 &= 2C + R\end{aligned}\end{cases}⎩⎨⎧​6x313x20x−4​=3Ax3=x2(3B+2A)=x(3C+2B)=2C+R​​​​


From the first equation you have that A=2‾\underline{A = 2}A=2​. Now onto the other equations:

​{13x2=x2(3B+4) ⟹ B=3‾0x=x(3C+2B) ⟹ C=−2‾−4=2C+R ⟹ R=0‾\begin{cases}\begin{aligned}13x^2 &= x^2 (3B+4) \implies \underline{B = 3}\\0x & = x(3C + 2 B)\implies \underline{C=-2}\\-4 &= 2C + R \implies \underline{R=0}\end{aligned}\end{cases}⎩⎨⎧​13x20x−4​=x2(3B+4)⟹B=3​=x(3C+2B)⟹C=−2​=2C+R⟹R=0​​​​​

​

In conclusion:

​6x3+13x2−4=(3x+2)×(2x2+3x−2)‾\underline{6x^3 +13x^2 -4 =(3x+2)\times (2x^2 +3x -2) }6x3+13x2−4=(3x+2)×(2x2+3x−2)​



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FAQs - Frequently Asked Questions

If the order of Q(x) (the denominator) is larger than the order of P(x) (the numerator), is it an improper fraction?

No. It is a proper fraction if order Q(x)>order P(x). If order P(x)>order Q(x), then it is an improper fraction.

What are the two methods used to convert an improper fraction into a mixed one?

There are two methods: algebraic division and the relation F(x) = Q(x) * divisor + remainder

What is an improper fraction?

An improper fraction is a fraction in which the order of P(x) (the numerator) is larger than or equal to the order of Q(x) (the denominator).

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