# A-level Mathematics/OCR/C1/Polynomials

You are already familiar with expressions such as $3x+5$ or $2x^{2}+8x+2$ and perhaps $x^{3}+2x-2$. The general word for an expression that can be written in this form is a polynomial. Excluding the constant term, every power of $x$ in a polynomial is a positive integer. Expressions that have $x$ to a non-integer or negative power are not polynomials.

## Basics of polynomials

All polynomials can be written in the form:
$a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}\dots a_{n}x^{n}$.
The numbers $a_{0},a_{1},a_{2},a_{3},a_{4}\dots$ are known as the co-efficients, and will often already be known. For example in $4x^{2}+3x+9$ the coefficient of $x^{2}$ is 4. The co-efficients may be 1 or even 0, for example $x^{3}+x$ is still a polynomial.

### Degrees of polynomials

The number $n$, i.e. the highest power in the polynomial is known as the degree, or sometimes the order, of the polynomial. For example, $x^{4}+4x^{2}+9$ is a degree 4 polynomial. Certain degrees have names, and are often written using certain letters for the coefficients when they are unknown:
* Degree 0 - Constant - $c$ or $k$
* Degree 1 - Linear - $ax+b$ or $mx+c$
* Degree 2 - Quadratic - $ax^{2}+bx+c$
* Degree 3 - Cubic - $ax^{3}+bx^{2}+cx+d$
* Degree 4 - Quartic $ax^{4}+bx^{3}+cx^{2}+dx+e$

### Notation

Most of the polynomials on this page are written in terms of $x$, for example, $4x^{2}+2x-9$, although polynomials can be written with other letters. $3z^{2}+z-2$, for example, is called a polynomial in $z$.
It is conventional to write polynomials in descending powers for clarity, although a polynomial may have the powers in any order. By writing polynomials in descending powers it is possible to see what the degree of the polynomial is simply by looking at the first term. Performing operations with two or more polynomials are also much simpler when the powers are in order, since like powers are usually grouped during calculations.
Indeed, you may lose marks in your exam if you don't simplify things like $x^{2}+5x-3x$ to $x^{2}+2x$.

## Operations with polynomials
### Addition

To add polynomials you simply add the co-efficients of each term. If that sounds confusing don't worry as you will probably already know how to do this, it is basically the same as collecting like terms. You add the co-efficients of $x$ together, the co-efficients of $x^{2}$ together, and so on. For example:
$\left(4x^{2}+2x+9\right)+\left(6x^{2}-2\right)=10x^{2}+2x+7$
Notice that $4+6=10$ and $9-2=7$. Some find it helpful to write this out like numerical addition:
| | $4x^{2}$ | $+2x$ | $+9$ |
| + | $6x^{2}$ | | $-2$ |
| | $10x^{2}$ | $+2x$ | $+7$ |
When using this method it is essential that terms with $x$ to the same power are lined up, leaving spaces if necessary as in the example above.

### Subtraction

For subtraction of polynomials, it is possible to use the same method as above, except subtracting instead of adding. It may get confusing when there are negative terms involved, so it is preferred to reverse the signs in the second row and then add the two polynomials together.
This:
| | $10x^{2}$ | $+2x$ | $+7$ |
| - | $6x^{2}$ | | $-2$ |
| | $4x^{2}$ | $+2x$ | $+9$ |
Would become this:
| | $10x^{2}$ | $+2x$ | $+7$ |
| + | $-6x^{2}$ | | $+2$ |
| | $4x^{2}$ | $+2x$ | $+9$ |
This method is desirable in the exam because subtracting negative numbers may cause confusion, and mistakes may be overlooked in the pressure of the exam.

### Multiplication

To multiply polynomials you simply multiply all the terms in one by all the terms in the other and then sum the results. This method is known as the FOIL Method. It stands for First Outer Inner Last. For example:
$\left(3x^{2}+2x\right)\times \left(2x^{2}-5\right)$ can be broken down into:
Multiply the First terms together.
$3x^{2}\times 2x^{2}=6x^{4}$
Then multiply the Outer terms together.
$3x^{2}\times -5=-15x^{2}$
Then multiply the Inner terms together.
$2x\times 2x^{2}=4x^{3}$
Then multiply the Last terms together.
$2x\times -5=-10x$
Then we add the results together to give us $6x^{4}+4x^{3}-15x^{2}-10x$, written in descending powers.
As you become confident with the process, you'll be able to do the entire multiplication in one go without splitting it up, like this:
$\left(3x^{2}+2x\right)\times \left(2x^{2}-5\right)=6x^{4}+4x^{3}-15x^{2}-10x$

#### Multiplication table

Some people find it helpful to show this in a table, with one polynomial as rows and the other as columns:
| | $3x^{2}$ | $2x$ |
| $2x^{2}$ | | |
| $-5$ | | |
The headings are then multiplied to produce:
| | $3x^{2}$ | $2x$ |
| $2x^{2}$ | $6x^{4}$ | $4x^{3}$ |
| $-5$ | $-15x^{2}$ | $-10x$ |
The answer is the sum of all the cells:
$6x^{4}+4x^{3}-15x^{2}-10x$, as before. This method is particularly useful for multiplying longer polynomials where the answer might not fit on a single line and cause mistakes.

#### Long multiplication

The polynomials to be multiplied can also be laid out like regular long multiplication, like this:
| | | | $3x^{2}$ | $2x$ | |
| | $\times$ | | $2x^{2}$ | | $-5$ |
| This is the 1st row multiplied by $-5$ | | | $-15x^{2}$ | $-10x$ | |
| This is the 1st row multiplied by $2x^{2}$ | $6x^{4}$ | $4x^{3}$ | | | |
| This is the sum of the two answers | $6x^{4}$ | $4x^{3}$ | $-15x^{2}$ | $-10x$ | |

### Division

Division of polynomials can be done just like regular long division. However, you will need to be quite adept at addition, subtraction and multiplication of polynomials to divide.
For example, to divide $x^{3}-12x^{2}-42$ by $x-3$ you write it like this:
    $x-3{\overline {\vert x^{3}-12x^{2}+0x-42}}$
1. Divide the first term of the dividend by the highest term of the divisor. Place the result above the bar (x$^3$ ÷ x = x$^2$).
    ${\begin{matrix}x^{2}\\\qquad \qquad \quad x-3{\overline {\vert x^{3}-12x^{2}+0x-42}}\end{matrix}}$
2. Multiply the divisor by the result you just obtained (the first term of the eventual quotient). Write the result under the first two terms of the dividend (x$^2$ * (x-3) = x$^3$ - 3x$^2$).
    ${\begin{matrix}x^{2}\\\qquad \qquad \quad x-3{\overline {\vert x^{3}-12x^{2}+0x-42}}\\\qquad \;\;x^{3}-3x^{2}\end{matrix}}$
3. Subtract the product you just obtained from the appropriate terms of the original dividend, and write the result underneath. This can be tricky at times, because of the sign. ((x$^3$-12x$^2$) - (x$^3$-3x$^2$) = -12x$^2$ + 3x$^2$ = -9x$^2$) Then, "pull down" the next term from the dividend.
    ${\begin{matrix}x^{2}\\\qquad \qquad \quad x-3{\overline {\vert x^{3}-12x^{2}+0x-42}}\\\qquad \;\;{\underline {x^{3}-\quad 3x^{2}}}\\\qquad \qquad \qquad \quad \;-9x^{2}+0x\end{matrix}}$
4. Repeat the last three steps, except this time use the two terms that you have just written as the dividend.
    ${\begin{matrix}\;x^{2}-9x\\\qquad \quad x-3{\overline {\vert x^{3}-12x^{2}+0x-42}}\\\;\;{\underline {\;\;x^{3}-\;\;3x^{2}}}\\\qquad \qquad \quad \;-9x^{2}+0x\\\qquad \qquad \quad \;{\underline {-9x^{2}+27x}}\\\qquad \qquad \qquad \qquad \qquad -27x-42\end{matrix}}$
5. Repeat step 4. This time, there is nothing to "pull down".
    ${\begin{matrix}\qquad \quad \;\,x^{2}\;-9x\quad -27\\\qquad \quad x-3{\overline {\vert x^{3}-12x^{2}+0x-42}}\\\;\;{\underline {\;\;x^{3}-\;\;3x^{2}}}\\\qquad \qquad \quad \;-9x^{2}+0x\\\qquad \qquad \quad \;{\underline {-9x^{2}+27x}}\\\qquad \qquad \qquad \qquad \qquad -27x-42\\\qquad \qquad \qquad \qquad \qquad {\underline {-27x+81}}\\\qquad \qquad \qquad \qquad \qquad \qquad \;\;-123\end{matrix}}$
The polynomial above the bar is your answer, and the number left over (-123) is the remainder.
    ${\frac {x^{3}-12x^{2}-42}{x-3}}=x^{2}-9x-27$ remainder $-123$

## Curves of polynomials

It is possible to look at a polynomial and get an idea to the general shape of its curve. The most important factors are the degree of the polynomial, and the sign of the co-efficient of the highest power. For a more accurate sketch of the graph, you will need to know where the curve has turning points and where it intersects the $x$ and $y$ axis. A sketch with this information will often be sufficiently detailed, and is usually much quicker and easier than plotting the curve point by point.

### Turning points

A turning point is where a curve that is decreasing changes to increasing, or vice versa. On a graph, the curve simply changes direction, hence the name turning point. In $y=x^{2}$, the turning point is at (0,0), the trough of the "bucket". The turning point in $y=x^{2}$ is a minimum (plural minima), and a maximum (plural maxima) is the same bucket shape, but upside down. For a polynomial of degree $n$, its curve will have at most $n-1$ turning points.

### Behaviour with extreme values of $x$

What happens to a polynomial at large positive and negative values for $x$?
If we say $1000$ is our large number, we can put it into $6x^{4}+4x^{3}-15x^{2}-10x$, we can take a look at what happens:
First for $x=1000$: $6000000000000+4000000000-15000000-10000=6003970000000$
Now for $x=-1000$: $6000000000000-4000000000-15000000+10000=5995985010000$
The most change in the final answer is made by the $x^{4}$ term. This is the dominant term. For a polynomial of degree $n$, $x^{n}$ is the dominant term as $x\rightarrow \pm \infty$.
Once you have investigated the behaviour of polynomials with extreme values of $x$, you may notice that if $n$ is even, it will share the same general shape as $x^{2}$, and if $n$ is odd it will share the same general shape as $x^{3}$.
If the coefficient of $x^{n}$ is positive, curves with an even value of $n$ will have a bucket shape, and curves with an odd value of $n$ will generally be negative when $x$ is negative, and positive when $x$ is positive (like $y=x$).
If the coefficient of $x^{n}$ is negative, curves with an even value of $n$ will have an upside down bucket shape, and curves with an odd value of $n$ will generally be positive when $x$ is negative, and negative when $x$ is positive (similar to $y=-x$).

### Intersections with axes

To find where the curve crosses the $y$ axis, you can simply look at the constant term. $y=x^{3}+2x^{2}-9x-18$, for example, would cross the $y$ axis at $-18$. If there is no constant term, the curve will go through the origin, (0,0), and would therefore cross the $y$ axis at $0$.
In factorised form, you can find the points at which the curve crosses the $x$ axis. $y=(x+3)(x+2)(x-3)$ would cross the $x$ axis at $-3$, $-2$, and $3$, because by making one of the brackets equal to $0$, it makes the whole polynomial equal to $0$, and $y$ equal to $0$. $y=0$ is of course the line of the $x$ axis. Each instance of the curve crossing the $x$ axis is a root of the equation. A polynomial of degree $n$ will have at most $n$ roots.
To solve an equation means to find every root. Polynomials of degree 2 can be solved using the quadratic formula. For higher degree polynomials you may need to factorise the equation like above, plot the graph point by point and look at where the curve crosses the $x$ axis, or use a numerical method. You can check your answers by substituting them into the original equation and seeing if the result is $0$.

## Quadratic expressions

A quadratic is a polynomial of degree 2, in the form $f(x)=ax^{2}+bx+c$.

### Graph

A quadratic graph is one that can be written in the form $y=ax^{2}+bx+c$. The graph of $y=2x^{2}+8x+2$ is shown on the right, and as you can see it has the same characteristic "bucket" shape that all quadratics have, called a parabola. The vertex is the maximum or minimum point. The line of symmetry is the line that divides the graph into two mirror halves. It can be found by the formula ${\frac {-b}{2a}}$.
However, these properties can be more easily deduced from its completed square form ( $a(x+d)^{2}+e$ ). In this form we know that, - d is the axis of symmetry, and that e is maximum or minimum point. If a is greater 0 then the vertex (-d,e) will be a minimum point. If a is less than 0 the vertex (-d,e) will be a maximum point.

### Completing the square

Completing the square is the process of changing a quadratic from the form $ax^{2}+bx+c$ to the equivalent form $a(x+d)^{2}+e$, where a, d and e are constants. For example, the quadratic $2x^{2}+8x+2$ would become $2(x+2)^{2}-6$.
Changing a quadratic to completed square form makes it easy to find several things, such as the roots of the quadratic and the vertex of the quadratic, without even requiring a sketch.
Here are the steps for completing the square. Don't worry, it's easier than it looks.
| Step | Action | Example | General case |
| 1. | Ensure the quadratic is in the conventional form: $ax^{2}+bx+c$. | $2x^{2}+8x+2$ | $ax^{2}+bx+c$ |
| 2. | Unless $a=1$, "pull/factor out a", that is, divide the entire quadratic by $a$ and put it outside a bracket. Note: If the quadratic is part of an equation you can divide each side by $a$, for example $2x^{2}+8x+2=0$ simply becomes $x^{2}+4x+1=0$. | $2\left(x^{2}+4x+1\right)$ | $a\left(x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}\right)$ |
| 3. | Replace the $x^{2}+kx$ part with $\left(x+{\frac {k}{2}}\right)^{2}$. It is important to realise that $\left(x+{\frac {k}{2}}\right)^{2}=x^{2}+kx+{\frac {k^{2}}{4}}$ which is close to what it was there before but not equal. This will be corrected in the next step. To prevent writing something that isn't actually equal it is a good idea to do both these steps at once in your working once you have got used to the method. | $2\left((x+2)^{2}+1\right)$ | $a\left(\left(x+{\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}\right)$ |
| 4. | Correct the error introduced in the previous step by inserting the subtraction of a suitable number. This suitable number can be found in two ways: By expanding the term inserted in the previous step and comparing it to the original; or By remembering that the error is always ${\frac {k^{2}}{4}}$. This step is known as "completing the square" and gives the method its name. | $(x+2)^{2}=x^{2}+4x+4$ which is 3 greater than $x^{2}+4x+1$, so $-3$ is inserted: $2\left((x+2)^{2}-3\right)$; or ${\frac {4^{2}}{4}}=4$ so the error is 4: $2\left((x+2)^{2}-4+1\right)$$=2\left((x+2)^{2}-3\right)$ | $a\left(\left(x+{\frac {b}{2a}}\right)^{2}-{\frac {b^{2}}{4a^{2}}}+{\frac {c}{a}}\right)$ |
| 5. | If step 2 was necessary then simplify the result a bit by expanding the outer bracket. Then you are finished. | $2(x+2)^{2}-6$ | $a\left(x+{\frac {b}{2a}}\right)^{2}-{\frac {b^{2}}{4a}}+c$ |
| 6. | Check that what you have obtained expands back to what you started with. Note: You may feel confident enough to skip this step. | $2\left(x^{2}+4x+4\right)-6$ $=2x^{2}+8x+8-6$ $=2x^{2}+8x+2$ | $a\left(x^{2}+{\frac {b}{a}}x+{\frac {b^{2}}{4a^{2}}}\right)-{\frac {b^{2}}{4a}}+c$ $=ax^{2}+bx+{\frac {b^{2}}{4a}}-{\frac {b^{2}}{4a}}+c$ $=ax^{2}+bx+c$ |
So the completed square form of $y=2x^{2}+8x+2$ is $2(x+2)^{2}-6$. The $-6$ tells us that the lowest point of the curve is at $y=-6$ and the $x+2$ tells us that the line of symmetry is at $x+2=0$ or $x=-2$. Therefore the vertex is at $(-2,-6)$, and if you look at the graph, you can see that is the case.

### The quadratic formula

The quadratic formula is derived from the general case of completing the square:
$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$
It can be used to find the roots of a quadratic by putting numbers directly into it. For example, for $y=2x^{2}+8x+2$:
$x={\frac {-8\pm {\sqrt {64-16}}}{4}}=-2\pm {\sqrt {3}}$
so $x=-2+{\sqrt {3}}$ and $x=-2-{\sqrt {3}}$. The factors of the equation are $x+2-{\sqrt {3}}$ and $x+2+{\sqrt {3}}$.
Here is the way that the quadratic formula is derived from the completing the square formula.
| Step | Action | Example | General case |
| 1. | To solve an equation in the form $ax^{2}+bx+c=0$ first complete the square using the method above. | $2(x+2)^{2}-6=0$ | $a\left(x+{\frac {b}{2a}}\right)^{2}-{\frac {b^{2}}{4a}}+c=0$ |
| 2. | Isolate the $(x+k)^{2}$ term. | $2(x+2)^{2}=6$ $(x+2)^{2}=3$ | $a\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}}{4a}}-c$ $\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}$ |
| 3. | Square root each side, including a $\pm$ as the bit inside the bracket might be negative or positive. | $x+2=\pm {\sqrt {3}}$ | $x+{\frac {b}{2a}}=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}$ Then some simplification: $x+{\frac {b}{2a}}=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {4ac}{4a^{2}}}}}$ $x+{\frac {b}{2a}}=\pm {\sqrt {\frac {b^{2}-4ac}{4a^{2}}}}$ $x+{\frac {b}{2a}}=\pm {\frac {\sqrt {b^{2}-4ac}}{\sqrt {4a^{2}}}}$ $x+{\frac {b}{2a}}=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}$ |
| 4. | Isolate $x$. | $x=\pm {\sqrt {3}}-2$ $x={\sqrt {3}}-2$ or $-{\sqrt {3}}-2$ $x\approx -0.268$ or $-3.73$ | $x=\pm {\frac {\sqrt {b^{2}-4ac}}{2a}}-{\frac {b}{2a}}$ $x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$ |

#### The discriminant

Notice that the quadratic equation contains $b^{2}-4ac$ inside a square root sign. This part is called the discriminant and can be considered on its own to determine the number of roots of the equation.
* If $b^{2}-4ac<0$ then you will be unable to find the square roots as you don't know how to square root a negative number. The type of numbers you have encountered so far are known as real numbers and so it is said that the quadratic has no real roots.
* If $b^{2}-4ac=0$ then changing the $\pm$ sign in front of the square root won't make any difference, because it is zero either way. You will therefore get the same root twice, so it is said the quadratic has one repeated root.
* If $b^{2}-4ac>0$ then the $\pm$ will mean you get two answers, and so you can say the quadratic has two distinct (i.e. different) roots.
This is part of the C1 (Core Mathematics 1) module of the A-level Mathematics text.