# A-level Mathematics/OCR/C4/Algebra and Graphs

< A-level Mathematics‎ | OCR‎ | C4

## Rational Expressions

A rational expression has a polynomial in the numerator and a polynomial in the denominator: $\frac{f\left(x\right)}{g\left(x\right)}$. In some cases rational functions can be simplified, which can make differentiation, integration and just solving the equation easier. There is a simple procedure to follow if you want to simplify a fraction.

1. Fully factor the numerator and denominator.
$\frac{ax^3+bx^2+cx+d}{ax^2+bx+c}=\frac{(ex+f)(x+g)(x+h)}{(ix+j)(x+g)(x+h)}$
2. If the same factor is in both the numerator and denominator you can cancel them out.
$\frac{(ex+f)\cancel{(x + g)}\cancel{(x +h)}} {(ix+j)\cancel{(x+g)}\cancel{(x+h)}}=\frac{(ex+f)}{(ix+j)}$
3. If there multiple factors left you can recombine based on the situation.

### Example

Simplify the following expression:$\frac{5x^3+40x^2+75x}{x^2+12x+35}$.

1. Factor both the numerator and the denominator notice that a 5x can be removed in the numerator.
$\frac{5x(x^2+8x+15)}{(x+5)(x+7)}=\frac{5x(x+5)(x+3)}{(x+5)(x+7)}$
2. Cancel out the same factors
$\frac{5x(x+5)(x+3)}{(x+5)(x+7)}=\frac{5x(x+3)}{(x+7)}$
3. Recombine the factors
$\frac{5x(x+3)}{(x+7)}=\frac{5x^2+15x}{x+7}$

## Division of Polynomials

In core four you will be expected to divide polynomials of up to the fourth degree by either a linear or quadratic polynomial. This only works if the numerator is greater than the denominator. The resultant will always be in the form;$\frac{f(x)}{g(x)}=q(x) + \frac{r(x)}{g(x)}$ where q(x)is the resulting polynomial and $\frac{r(x)}{g(x)}$ is the remainder.

### Long Division

Long division of polynomials is very similar to regular long division. I will use a problem to demonstrate how long division works.

Divide $x^3 + 8x^2 - 4x + 10$ by $x^2 + 3x -1$.

The first step is to set up the equation. Make sure that it is in order from highest power of x to lowest power of x.

Then we divide the first term of the dividend by the first term of the divisor.

$\frac {x^3}{x^2} = x$

We place this resultant on top.

Then we multiply the resultant by the divisor and subtract it from the dividend.

$x \left ( x^2 + 3x -1 \right )= x^3 + 3x^2 -x$

What is left becomes the new dividend and we repeat the process again. We continue to do this until the first term has a degree less than the first term of the divisor. What is left is the remainder.

### Synthetic Division

#### Example One

Divide $2x^5 + 5x^2 - 10x^3 -30x - 171$ by x-3.

The divisor is c. So in this case it will be 3.

Then we need to arrange our divisor in order from highest to lowest degree, if a degree is missing replace it with zero.

$2x^5 + 0x^4 - 10x^3 + 5x^2 -30x - 171$

We then only look at the coefficients.

2 + 0 - 10 + 5 - 30 - 171

Now we set up our division equation.

Next we carry up the first term of the divisior.

Then we multiply the resultant by the divisor and add it to the next term.

We continue to do this until we reach the end.

Now we need to read the variables. When we read the variables we go from highest degree -1 to lowest degree. The last number is the remainder.

$2x^4 + 6x^3 + 8x^2 + 29x +57$ remainder 0.

This is the answer to the problem.

## Partial Fractions

Only proper fractions can be made into partial fractions. A proper fraction is a fraction where the numerator is of a smaller degree than the denominator. Partial Fractions can used to make differentiation, integration or work with series easier. The goal of partial fractions is to take a complex polynomial and make it several simpler polynomials. To create a partial fraction equation you need to:

1. Factor the denominator and the numerator.
$\frac{ax^2+bx+c}{ax^3+bx^2+cx+d}\equiv\frac{(ix+j)(kx+l)}{(ax+b)(cx+d)(ex+f)}$
2. Create the partial fraction by making each factor the denominator, with a unique variable for the numerator.
$\frac{(ix+j)(kx+l)}{(ax+b)(cx+d)(ex+f)}\equiv\frac{A}{(ax+b)}+\frac{B}{(cx+d)}+\frac{C}{(ex+f)}$
3. Multiply every fraction by the denominator of original fraction.
$\frac{(ix+j)(kx+l)(ax+b)(cx+d)(ex+f)}{(ax+b)(cx+d)(ex+f)}\equiv$
$\frac{A(ax+b)(cx+d)(ex+f)}{(ax+b)}+\frac{B(ax+b)(cx+d)(ex+f)}{(cx+d)}+\frac{C(ax+b)(cx+d)(ex+f)}{(ex+f)}$
4. Cancel all out all factors that are the same.
$(ix+j)(kx+l)\equiv A(cx+d)(ex+f)+B(ax+b)(ex+f)+C(ax+b)(cx+d)$
5. Multiply all factors together.
$ikx^2+ilx+jlx+jl \equiv$
$(Acex^2+Acfx+Adex+Adf)+(Baex^2+Bafx+Bbex+Bbf)+(Cacx^2+Cadx+Cbdx+Cbd)\,$
6. Group all terms with x to the same index together.
$ikx^2+(il+kj)x+jl\equiv$
$(Ace+Bae+Cac)x^2+\left(Acf+Ade+Baf+Bbe+Cad+Cbd\right)x+\left(Adf+Bbf+Cbd\right)$
7. The co-efficient of x raised to the same index is the same on both sides of the identity. Using this fact solve for the variables in the numerator, using simultaneous equations. This is known as equating coefficients.
$ik\equiv\left(Ace+Bae+Cac\right)$
$\left(il+kj\right)\equiv\left(Acf+Ade+Baf+Bbe+Cad+Cbd\right)$
$jl\equiv\left(Adf+Bbf+Cbd\right)$
8. In step 2 replace the variables A, B, C with the value of the variable obtained from the previous step.

### Example One

Rewrite as a partial fraction the following expression: $\frac{x^2+13x+40}{2x^3+27x^2+111x+140}$.

1. Using the knowledge from Further Pure 1: Roots of Polynomial Equations. We can factor the denominator as such.
$\frac{x^2+13x+40}{2x^3+27x^2+111x+140} \equiv \frac{(x+5)(x+8)}{(2x+5)(x+7)(x+4)}$
2. Now we start making the partial fraction.
$\frac{x^2+13x+40}{(2x+5)(x+7)(x+4)}\equiv\frac{A}{(2x+5)}+\frac{B}{(x+7)}+\frac{C}{(x+4)}$
3. Next we multiply everything by the denominator of the original expression.
$\frac{x^2+13x+40(2x+5)(x+7)(x+4)}{(2x+5)(x+7)(x+4)}\equiv$
$\frac{A(2x+5)(x+7)(x+4)}{(2x+5)}+\frac{B(2x+5)(x+7)(x+4)}{(x+7}+\frac{C(2x+5)(x+7)(x+4)}{(x+4)}$
4. Now we cancel out terms.
$(x+7)(x+8)\equiv A(x+7)(x+4)+B(2x+5)(x+4)+C(2x+5)(x+7)$
5. Then multiply all the factors together.
$x^2+13x+40=Ax^2+11Ax+28A+2Bx^2+13Bx+20B+2Cx^2+19Cx+35C\,$
6. We then group the terms together.
$x^2+13x+40=(A+2B+2C)x^2+(11A+13B+19C)x+(28A+20B+35C)\,$
7. Now we equate coefficients. Then we need to solve the simultaneous equations.
$1\equiv(A+2B+2C)\equiv\,$
$13\equiv(11A+13B+19C)\,$
$40\equiv(28A+20B+35C)\,$
1. First we equate A
$1\equiv(A+2B+2C)\equiv A=1-2B-2C\,$
Then
$13\equiv11(1-2B-2C)+13B+19C\equiv11-22B-22C+13B+19C\equiv11-9B-3C$
$9B\equiv3C+2$
$B\equiv-\frac{3C+2}{9}$
2. We can further equate A to
$A\equiv1-2(-\frac{3C+2}{9})-2C$
$A\equiv\frac{13-12C}{9}$
3. Now we can solve for C
$40\equiv(28(\frac{13-12C}{9})+20(-\frac{3C+2}{9})+35C)\,$
$40\equiv((\frac{364-336C}{9})+(-\frac{60C+40}{9})+35C)\,$
$40=36-9C\,$
$C=\frac{-4}{9}$
4. Now we can solve for the rest
$A\equiv\frac{13-12\frac{-4}{9}}{9}=\frac{55}{27}$
$B\equiv-\frac{3C+2}{9}=\frac{-2}{27}$
$C\equiv\frac{-4}{9}$
8. Now we can write out our partial fraction:
$\frac{x^2+13x+40}{(2x+5)(x+7)(x+4)}\equiv\frac{55}{27(2x+5)}-\frac{2}{27(x+7)}-\frac{4}{9(x+4)}$