# Zeros

## Key Questions

Graph the function on a graphing calculator to see what the x-coordinates are where the function intersects the x-axis.

#### Explanation:

The zeros of a function are found by determining what x-values will cause the y-value to be equal to zero. One way to find the zeros is to graph the function on a graphing calculator to see what the x-coordinates are where the function intersects the x-axis.

• A zero of a function is an interception between the function itself and the X-axis.
The possibilities are:

• no zero (e.g. $y = {x}^{2} + 1$) graph{x^2 +1 [-10, 10, -5, 5]}
• one zero (e.g. $y = x$) graph{x [-10, 10, -5, 5]}
• two or more zeros (e.g. $y = {x}^{2} - 1$) graph{x^2-1 [-10, 10, -5, 5]}
• infinite zeros (e.g. $y = \sin x$) graph{sinx [-10, 10, -5, 5]}

To find the eventual zeros of a function it is necessary to solve the equation system between the equation of the function and the equation of the X-axis ($y = 0$).

It depends...

#### Explanation:

Here are some cases...

Polynomial with coefficients with zero sum

If the sum of the coefficients of a polynomial is zero then $1$ is a zero. If the sum of the coefficients with signs inverted on the terms of odd degree is zero then $- 1$ is a zero.

Any polynomial with rational roots

Any rational zeros of a polynomial with integer coefficients of the form ${a}_{n} {x}^{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{0}$ are expressible in the form $\frac{p}{q}$ where $p , q$ are integers, $p$ a divisor of ${a}_{0}$ and $q$ a divisor of ${a}_{n}$.

Polynomials with degree <= 4

$a x + b = 0 \implies x = - \frac{b}{a}$

$a {x}^{2} + b x + c = 0 \implies x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

There are formulas for the general solution to a cubic, but depending on what form you want the solution in and whether the cubic has $1$ or $3$ Real roots, you may find some methods preferable to others.

In the case of one Real root and two Complex ones, my preferred method is Cardano's method. The symmetry of this method gives neater result formulations than Vieta's substitution.

In the case of three Real roots, it may be preferable to use the trigonometric substitution that squeezes a cubic into the identity $\cos 3 \theta = 4 {\cos}^{3} \theta - 3 \cos \theta$, thereby finding zeros in terms of $\cos$ and $\arccos$.

There are general formulas for the solution of quartic equations, but it's generally easier to work with the individual cases.

In the worst cases, you can transform $a {x}^{4} + b {x}^{3} + c {x}^{2} + \mathrm{dx} + e$ into a monic quartic by dividing by $a$, get into the form ${t}^{4} + p {t}^{2} + q t + r$ using the substitution $t = x + \frac{b}{4 a}$, then look at factorisations of the form:

${t}^{4} + p {t}^{2} + q t + r = \left({t}^{2} + A t + B\right) \left({t}^{2} - A t + C\right)$

multiplying out and equating coefficients to get 3 simultaneous equations in $A$, $B$ and $C$. Then use ${\left(B + C\right)}^{2} = {\left(C - B\right)}^{2} + 4 B C$ to derive a cubic equation in ${A}^{2}$. By now you hopefully know how to solve cubics, so you can find $A$, hence $B$ and $C$, etc.

• The zeros of a function represent the x value(s) that result in the y value being 0.

The zeros of a function represent the x-intercept(s) when the function is graphed.

The zeros of a function represent the root(s) of a function.

The zeros of a function represent the solution(s) of a function.