How to find the zeros of a function with 2 terms

The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division repeatedly to determine all of the zeros of a polynomial function.

How To: Given a polynomial function [latex]f[/latex], use synthetic division to find its zeros

1. Use the Rational Zero Theorem to list all possible rational zeros of the function.
2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate.
3. Repeat step two using the quotient found from synthetic division. If possible, continue until the quotient is a quadratic.
4. Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula.

Example: Finding the Zeros of a Polynomial Function with Repeated Real Zeros

Find the zeros of [latex]f\left(x\right)=4{x}^{3}-3x - 1[/latex].

Show Solution

The Rational Zero Theorem tells us that if [latex]\frac{p}{q}[/latex] is a zero of [latex]f\left(x\right)[/latex], then p is a factor of –1 and q is a factor of 4.

[latex]\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 4}}\hfill \end{array}[/latex]

The factors of –1 are [latex]\pm 1[/latex] and the factors of 4 are [latex]\pm 1,\pm 2[/latex], and [latex]\pm 4[/latex]. The possible values for [latex]\frac{p}{q}[/latex] are [latex]\pm 1,\pm \frac{1}{2}[/latex], and [latex]\pm \frac{1}{4}[/latex].
These are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1.

Dividing by [latex]\left(x - 1\right)[/latex] gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as [latex]\left(x - 1\right)\left(4{x}^{2}+4x+1\right)[/latex].

The quadratic is a perfect square. [latex]f\left(x\right)[/latex] can be written as [latex]\left(x - 1\right){\left(2x+1\right)}^{2}[/latex].

We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0.

[latex]\begin{array}{l}2x+1=0\hfill \\ \text{ }x=-\frac{1}{2}\hfill \end{array}[/latex]

The zeros of the function are 1 and [latex]-\frac{1}{2}[/latex] with multiplicity 2.

Analysis of the Solution

Look at the graph of the function f. Notice that, at [latex]x=-3[/latex], the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero [latex]x=-3[/latex]. Also note the presence of the two turning points. This means that, since there is a 3rd degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of [latex]x=-3[/latex] is 1 and there are two complex solutions, which is what we found, or the multiplicity at [latex]x=-3[/latex] is three. Either way, our result is correct.

We need two equations if you need to solve for both x and b. If, however, b is a parameter and x is the independent variable, we can solve the problem as follows.

Collect the function in terms that depend on b and terms that do not, i.e. f(x,b) = 2x^4 + 2 x^3 - 10x^2 + 2x - 12 + (2b x^3 + 2b x^2 - 12bx)

This function factorizes as f(x,b) = 2*(x^2+bx+1)(x^2+x-6)

The zeros of the function require x^2+x-6 = 0 (i) or (x^2 + bx + 1) = 0 (ii)

(i) has solutions x = 2, x = -3

(ii) has solutions x = (-b ± sqrt(b^2-4))/2

Hence the four zeros of the function are x = 2, x= -3, x = (-b ± sqrt(b^2-4))/2. Note that the latter two zeros show a relation between x and b rather than specific values for x or b. The first two zeros however are independent of b and are thus solutions even for the two variable situation.

Mark M.

tutor

Ram K., by which process did you factor the fourth degree polynomial?

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01/08/17

Mark M. answered • 01/08/17

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Brielle W.

oh, i was given this by my math teacher and thats all he gave us.

 

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01/08/17

Mark M.

tutor

Did he give any other instructions/methods?

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01/08/17

Brielle W.

No, he just gave the problem and the directions said," Find all the zeros of the function".

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01/08/17

Mark M.

tutor

At the moment, synthetic substitution using the Rational Root Theorem seems to be the only possibility.

How do you find the zeros of a polynomial with two variables?

The quadratic equation of the form x² + x(a + b) + ab = 0 can be factorized as (x + a)(x + b) = 0, and we have x = -a, and x = -b as the zeros of the polynomial.

What is the formula for finding zeros of a function?

Graphically, the real zero of a function is where the graph of the function crosses the x‐axis; that is, the real zero of a function is the x‐intercept(s) of the graph of the function. Find the zeros of the function f ( x) = x ² – 8 x – 9. Find x so that f ( x) = x ² – 8 x – 9 = 0.

How do you find the zeros of a quadratic function in Algebra 2?

To find the zeros/roots of a quadratic: factor the equation, set each of the factors to 0, and solve for. In other words, given f ( x ) = a ( x − p ) ( x − q ) , find ( x − p ) = 0 and. Use the square root method for quadratic expressions in the form.