As explained on the main polynomials page, zeroes are where the graph of a polynomial crosses the xaxis. They are also called xintercepts, roots and, less common, poles. We will use the three main terms, xintercepts, zeroes and roots, interchangeably on this page, so that you become comfortable with all of them. (The term zeroes usually refers to points on graphs while the term roots is more general and includes zeroes on the graph as well as complex numbers. We will not be picky about using these distinctions since many textbooks and instructors use them interchangebly.)
Note  This is one of the few precalculus pages where we will not assume we are talking only about real numbers. Complex numbers are also possible on this page. Click here to review complex numbers.
What Are Zeroes and Roots of a Polynomial?
The zeroes of a polynomial are where the graph of the polynomial crosses the xaxis and why they are called xintercepts. On graphs, you can also say these are points where \(y=0\), which means the same thing. These points have special meaning in many contexts.
One of the nice things about polynomials is that there are exactly the same number of roots as the highest power in the polynomial. So, if we have a polynomial \(g(x)=x^4+3x^31\), the highest power is 4 and so we know that we have 4 roots.
Special Notes
1. Notice that we did not say that the roots are all real values. From the context, it is natural to assume this but that is not the case. For example, the graph of the polynomial \(y=x^2+1\) never crosses the xaxis but it does have two roots, both complex.
2. Just because the roots exist, does not mean we can find them. In fact, it is sometimes not possible to find them exactly from the equation and we may have to go to our calculuator to approximate them.
Find The Roots Given The Polynomial
Finding the roots of a polynomial is pretty straightforward. The technique requires you to understand the concept of the zero product rule since it is used heavily when calculating the roots. Basically the technique requires us to
1. move all terms to one side of the equal sign, leaving zero on the other side; this may also entail setting the polynomial equal to zero;
2. factor the polynomial;
3. use the zero product rule and solve for the xvalues.
In the last step, we may be able to solve for the xvalues directly or we may need to approximate the values using our calculator or graphing utility. Here is a quick example.
Example
Find the roots of the polynomial \(x^24\).
\(x^24=0\) 
set the polynomial equal to zero  
\((x+2)(x2)=0\) 
factor  
\((x+2)=0 \to x=2\) or 
use the zero product rule and solve  
\(x=2; x=2\) 
final answers 
This takes some practice, so here are some practice problems.
Solve these problems giving your answers in exact form.
Find all zeros of \(p(x)=x^34x^2+x4\) given that \(+i\) is a zero.
Problem Statement 

Find all zeros of \(p(x)=x^34x^2+x4\) given that \(+i\) is a zero.
Solution 

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Find the roots of \(f(x)=8x^436x^320x^2\).
Problem Statement 

Find the roots of \(f(x)=8x^436x^320x^2\).
Solution 

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Given that two zeros are \(x=2\) and \(x=3\), find the remaining zeros for \(f(x)=x^410x^3+37x^2\) \(60x+36\).
Problem Statement 

Given that two zeros are \(x=2\) and \(x=3\), find the remaining zeros for \(f(x)=x^410x^3+37x^2\) \(60x+36\).
Solution 

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Find all the roots of \(f(x)=x^312x^2+49x78\).
Problem Statement 

Find all the roots of \(f(x)=x^312x^2+49x78\).
Solution 

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Find the roots of the polynomial \(f(x)=x^33x^213x+15\).
Problem Statement 

Find the roots of the polynomial \(f(x)=x^33x^213x+15\).
Solution 

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Find the zeros of the polynomial \(f(x)=x^34x^211x+2\).
Problem Statement 

Find the zeros of the polynomial \(f(x)=x^34x^211x+2\).
Solution 

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Find the roots of \(f(x)=x^3+16\).
Problem Statement 

Find the roots of \(f(x)=x^3+16\).
Solution 

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Find the roots of \(f(x)=x^34x^2+4x16\).
Problem Statement 

Find the roots of \(f(x)=x^34x^2+4x16\).
Solution 

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Find the roots of \(p(x)=x^3+4x^2+9x+36\).
Problem Statement 

Find the roots of \(p(x)=x^3+4x^2+9x+36\).
Solution 

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Find all the real zeros of \(f(x)=x^3+x^210x+8\).
Problem Statement 

Find all the real zeros of \(f(x)=x^3+x^210x+8\).
Solution 

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Find the roots of \(f(x)=2x^53x^4+2x^33x^2\) \(144x+216\).
Problem Statement 

Find the roots of \(f(x)=2x^53x^4+2x^33x^2\) \(144x+216\).
Solution 

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Find all the real zeros of \(f(x)=3x^48x^337x^2+\) \(2x+40\).
Problem Statement 

Find all the real zeros of \(f(x)=3x^48x^337x^2+\) \(2x+40\).
Solution 

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Find the roots of \( f(x) = x^4  3x^3  12x^2 + \) \(54x  40 \).
Problem Statement 

Find the roots of \( f(x) = x^4  3x^3  12x^2 + \) \(54x  40 \).
Solution 

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Find the real zeros of \(p(x)=2x^5+x^42x1\).
Problem Statement 

Find the real zeros of \(p(x)=2x^5+x^42x1\).
Solution 

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Find The Polynomial Given The Roots
From the discussion above, finding the roots is pretty straightforward as long as you can factor the polynomial. Another thing you will be asked to do is find the polynomial given the roots. This, also, is pretty straightforward if you remember just a couple of things.
1. In precalculus you will almost certainly always be working with polynomials with real coefficients. This is important since the next point follows from and is dependent on this fact.
2. For complex roots of polynomials with real coefficients, the complex roots always appear in complex conjugate pairs. For example, if one of your roots is \(2+3i\), the other root is guaranteed to be \(23i\).
3. Another situation to watch for is something called multiplicity. The root can appear more than once and, therefore, have a multiplicity greater than one. For example, the roots of \(x^26x+9=(x3)^2\) are both \(x=3\), so we say that the roots are 3 with multiplicity 2 (since it appears twice). Multiplicity just tells us how many times the root exists.
To solve a problem where you are given the roots and you need to come up with the polynomial, you just write the factors and multiply out. Also, most instructors will want the resulting polynomial to have integer coefficients. For example, if you multiply out and get a polynomial like \((1/2)x^2+3\), multiply by 2 to get \(x^3+6\). Notice this polynomial has the same roots as the first one even though it is not the same polynomial. Of course, it depends on the problem statement, so check with your instructor to see what they require.
Before jumping into the practice problems, let's watch a quick video explaining these techniques in more detail and how the roots look on a graph. This video also includes lots of examples.
video by MIP4U 

Okay, time for you to try your hand at these practice problems.
Solve these problems giving your answers in exact form.
Find the polynomial \(f(x)\) of degree 3 with zeros \(1, ~2, ~4\) where \(f(1)=8\).
Problem Statement 

Find the polynomial \(f(x)\) of degree 3 with zeros \(1, ~2, ~4\) where \(f(1)=8\).
Solution 

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Find a polynomial function with real coefficients having zeros \(4\) and \(2+3i\).
Problem Statement 

Find a polynomial function with real coefficients having zeros \(4\) and \(2+3i\).
Solution 

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Find a polynomial function with real coefficients having zeros \(6\) and \(i\).
Problem Statement 

Find a polynomial function with real coefficients having zeros \(6\) and \(i\).
Solution 

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Find a polynomial with real coefficients having roots \(2/3, 3/7, 1\).
Problem Statement 

Find a polynomial with real coefficients having roots \(2/3, 3/7, 1\).
Solution 

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Find a polynomial with real coefficients having roots \(4, 1, 3\).
Problem Statement 

Find a polynomial with real coefficients having roots \(4, 1, 3\).
Solution 

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Find a polynomial with real coefficients of degree 3 with roots \(x=2\) (multiplicity 2) and \(x=3\) through the point \((2,80)\).
Problem Statement 

Find a polynomial with real coefficients of degree 3 with roots \(x=2\) (multiplicity 2) and \(x=3\) through the point \((2,80)\).
Solution 

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Find the real polynomial of degree 4 having roots \(x=4\) (multiplicity 2), \(x=3\) and \(x=0\) passing through the point \((2,36)\).
Problem Statement 

Find the real polynomial of degree 4 having roots \(x=4\) (multiplicity 2), \(x=3\) and \(x=0\) passing through the point \((2,36)\).
Solution 

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Find the polynomial \(f(x)\) of degree 3 where two roots are \(x=1\) and \(x=2\), the leading coefficient is \(1\) and \(f(3)=48\).
Problem Statement 

Find the polynomial \(f(x)\) of degree 3 where two roots are \(x=1\) and \(x=2\), the leading coefficient is \(1\) and \(f(3)=48\).
Solution 

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Write an expression for a polynomial \(f(x)\) of degree 3 with zeros \(x=2\) and \(x=2\), leading coefficient of 1 and \(f(4)=30\).
Problem Statement 

Write an expression for a polynomial \(f(x)\) of degree 3 with zeros \(x=2\) and \(x=2\), leading coefficient of 1 and \(f(4)=30\).
Solution 

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Find a polynomial with real coefficients having roots \(x=2\) (multiplicity 2) and \(x=12i\).
Problem Statement 

Find a polynomial with real coefficients having roots \(x=2\) (multiplicity 2) and \(x=12i\).
Solution 

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Find a polynomial with real coefficients and zeros \(3, 1, 4+2i\).
Problem Statement 

Find a polynomial with real coefficients and zeros \(3, 1, 4+2i\).
Solution 

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Find a polynomial with real coefficients having roots \( 1/4, 2/3, 2i\).
Problem Statement 

Find a polynomial with real coefficients having roots \( 1/4, 2/3, 2i\).
Solution 

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Rational Roots Test
The Rational Roots Test is quick way to get a list of all possible rational roots to a polynomial. It is a very good place to start when trying to find roots of higher order polynomials (3 or higher). Once you have a rational root or two, you can reduce the order of the polynomial using synthetic division to find irrational and complex roots.
However, that said, we never used this technique in calculus, so as far as we are concerned, you do not need to know this for calculus. But, as usual, check with your instructor to see what they require.
If you need to know this, we recommend this video explaining this technique and showing an example.
video by PatrickJMT 

Here is a playlist of the videos on this page.
Really UNDERSTAND Precalculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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