## Algebra Polynomials

A polynomial is any multiple term function where the x terms are to the power of a positive integer.

$f(x)=a_{0}+a_{1}x+......+a_{n}x^{n}$

The value of n is the order of the polynomial and an n order polynomial can have up to n solutions, if all the coefficients are real numbers. Learning to solve polynomials algebraically will help when graphing polynomials.

### Second Order Polynomials

When n = 2 we have a second order polynomial or a quadratic:

$f(x)=ax^{2}+bx+c$

A quadratic is unique because when finding the solutions of the equation we can use the quadratic formula, which is when the equation above equals zero

$x-intercepts = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

This allows us to determine solutions of x, but can also be used to check if there are any solutions.

If $b^{2}-4ac>0$, there are two solutions found through the equation above

If $b^{2}-4ac=0$ , there is one solution and it will be $-\frac{b}{2a}$

If $b^{2}-4ac<0$ , there are no solutions, as we cannot have the square root of a negative number.

The turning point can also be easily determined as its x-coordinate is the mid-point of the two x intercepts. If there are no intercepts then it can be determined through the formula:

$x=-\frac{b}{2a}$

### Third Order polynomials

When n is = 3 we have a third order polynomial or a cubic:

$y=ax^{3}+bx^{2}+cx+d$

In order to solve cubics and higher order polynomials it is important to group all the terms on one side and equate the function to zero.

Then we need to use trial and error to factorise the equation and find the solutions. We start by guessing a value for x and then seeing if it equates the function to zero.

### Example 1:

2013 exam 2

This is a little tricky since our unknown is present as our factor and still in the equation. However, there is an easy factor we can take out which is x. Therefore we get:

$x(7x^{2}+9x-5a)$

Now our equation is a little simplier and we just need to take the factor (x+a) from:

We know:

$(7x^{2}+9x-5a)=(x+a)(bx+c)$

Therefore:

$bx\times x =7x^{2}\newline b=7$

$a\times c =-5a\newline c=-5$

And to find a, expand for the middle term:

$x\times -5 + a \times 7x =9x\newline 7a=14 \newline a=2$

### Example 2:

2014 exam 2

Here we need to perform some manipulation of a fourth order polynomial.

Step 1 Remove x

$g(x)=x(x^{3}-8)$

Step 2 use substitution to look for factors

$g(1)=1(1^{3}-8) \neq 0\newline g(2)=2 \times (2^{3}-8)=0$

Therefore 2 is a factor and (x-2) can be removed from the equation

$g(x)=x(x-2)(ax^{2}+bx+c)$

$x\times ax^{2} =x^{3}\newline a=1$

$-2\times c =-8\newline c=4$

$-2\times bx +x \times c=0\newline -2bx+4x=0 \newline b=2$

$g(x)=x(x-2)(x^{2}+2x+4)$

Step 3 separate out our square term in the last brackets.

$g(x)=x(x-2)(x^{2}+2x+1+3)\newline =x(x-2)((x+1)^{2}+3)$