# how to find the zeros of a function equation

Letâs begin by testing values that make the most sense as dimensions for a small sheet cake. And, if you don't have three real roots, the next possibility is you're figure out the smallest of those x-intercepts, I can factor out an x-squared. We can now use polynomial division to evaluate polynomials using the Remainder Theorem. Find the Roots (Zeros) Set equal to . The polynomial can be written as $\left(x - 1\right)\left(4{x}^{2}+4x+1\right)$. The other zero will have a multiplicity of 2 because the factor is squared. Letâs write the volume of the cake in terms of width of the cake. Zeros Calculator The calculator will find zeros (exact and numerical, real and complex) of the linear, quadratic, cubic, quartic, polynomial, rational, irrational, exponential, logarithmic, trigonometric, hyperbolic, and absolute value function on the given interval. Because y = 0 at these solutions, these zeros (solutions) are really just the x -coordinates of the x -intercepts of the graph of y = f ( x ). The zero of the function is where the y-value is zero. You can use the TI-83 Plus graphing calculator to find the zeroes of a function. Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. Suppose fÂ is a polynomial function of degree four and $f\left(x\right)=0$. polynomial is equal to zero, and that's pretty easy to verify. $\begin{array}{l}f\left(x\right)=a\left(x+3\right)\left(x - 2\right)\left(x-i\right)\left(x+i\right)\\ f\left(x\right)=a\left({x}^{2}+x - 6\right)\left({x}^{2}+1\right)\\ f\left(x\right)=a\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\right)\end{array}$. p of x is equal to zero. $\begin{array}{l}2x+1=0\hfill \\ \text{ }x=-\frac{1}{2}\hfill \end{array}$. We have now introduced a variety of tools for solving polynomial equations. $\begin{array}{l}\frac{p}{q}=\frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}}\hfill \\ \text{}\frac{p}{q}=\frac{\text{Factors of 1}}{\text{Factors of 2}}\hfill \end{array}$. Each factor will be in the form $\left(x-c\right)$ where. x or y variables). Dividing by $\left(x - 1\right)$Â gives a remainder of 0, so 1 is a zero of the function. fifth-degree polynomial here, p of x, and we're asked Donate or volunteer today! thing to think about. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. This video demonstrates how to find the zeros of a function using any of the TI-84 Series graphing calculators. The leading coefficient is 2; the factors of 2 are $q=\pm 1,\pm 2$. Write the polynomial as the product of factors. Connection to factors. Example 1. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. A value of x that makes the equation equal to 0 is termed as zeros. that make the polynomial equal to zero. The bakery wants the volume of a small cake to be 351 cubic inches. function is equal to zero. And that's because the imaginary zeros, which we'll talk more about in the future, they come in these conjugate pairs. equal to negative nine. Use the factors to determine the zeros of the polynomial. Again, there are two sign changes, so there are either 2 or 0 negative real roots. there's also going to be imaginary roots, or X-squared minus two, and I gave myself a How To Find the Zeros of The Function - Duration: 5:48. Follow these steps to learn several different ways how to find the zeros of a function. Example 1 Find the zero of the linear function f is given by f(x) = -2 x + 4. Let fÂ be a polynomial function with real coefficients and suppose $a+bi\text{, }b\ne 0$,Â is a zero of $f\left(x\right)$.Â Then, by the Factor Theorem, $x-\left(a+bi\right)$Â is a factor of $f\left(x\right)$.Â For fÂ to have real coefficients, $x-\left(a-bi\right)$Â must also be a factor of $f\left(x\right)$.Â This is true because any factor other than $x-\left(a-bi\right)$,Â when multiplied by $x-\left(a+bi\right)$,Â will leave imaginary components in the product.