A polynomial is a mathematical expression made up of terms that are added or subtracted from each other; a term can consist of constants, coefficients and variables. When solving a polynomial you usually have to find the value for which x = 0. Those of lower degree have one or two solutions, based on whether they are linear or second degree. This kind of polynomials is easily solved by exploiting elementary algebraic concepts and with methods of factorization. To learn how to solve the higher grade ones you can read this article.
Steps
Method 1 of 2: First Degree Polynomials
Step 1. Determine if it is a linear polynomial
This term indicates an expression of the first degree. This means that there are no variables with exponent (or better with an exponent greater than 1); since it is a first degree expression, it has only one solution, or root.

For example, 5x + 2 { displaystyle 5x + 2}
è un polinomio di primo grado perché la variabile x{displaystyle x}
ha esponente uguale a 1.
Step 2. Set the equation equal to zero
This step is necessary to solve all polynomials.
 For example, 5x + 2 = 0 { displaystyle 5x + 2 = 0}
Step 3. Isolate the term of the variable
Add or subtract the constant on both sides of the equation; the constant is the parameter without variable.

For example, to isolate x { displaystyle x}
nell'equazione 5x+2=0{displaystyle 5x+2=0}
dovresti sottrarre 2{displaystyle 2}
da entrambi i lati:
5x+2=0{displaystyle 5x+2=0}
5x+2−2=0−2{displaystyle 5x+22=02}
5x=−2{displaystyle 5x=2}
Step 4. Solve for the variable
Typically, you must divide both terms of the equation by the coefficient of the variable; in this way you get the root, or solution, of the polynomial.

For example, to find the value of x { displaystyle x}
in 5x=−2{displaystyle 5x=2}
dovresti dividere entrambi i termini per 5{displaystyle 5}
:
5x=−2{displaystyle 5x=2}
5x5=−25{displaystyle {frac {5x}{5}}={frac {2}{5}}}
x=−25{displaystyle x={frac {2}{5}}}
di conseguenza, la soluzione di 5x+2{displaystyle 5x+2}
è x=−25{displaystyle x={frac {2}{5}}}
Metodo 2 di 2: Polinomi di Secondo Grado
Step 1. Determine if you have a second degree polynomial
Sometimes, it is also referred to as a quadratic polynomial; this means that there is no variable with an exponent greater than 2. Since it is a second degree polynomial, it has two roots, or solutions.

For example, x2 + 8x − 20 { displaystyle x ^ {2} + 8x20}
è un polinomio di secondo grado perché la variabile x{displaystyle x}
ha un esponente pari a 2{displaystyle 2}
Step 2. Make sure it's written in rank order
This means that the term with exponent equal to 2 { displaystyle 2}
deve essere il primo a sinistra, seguito da quello di primo grado e infine dalla costante.
 Per esempio, dovresti riscrivere 8x+x2−20{displaystyle 8x+x^{2}20}
in questa forma: x2+8x−20{displaystyle x^{2}+8x20}
Step 3. Set up an equation of zero
This step is necessary to solve all polynomials.
 For example: x2 + 8x − 20 = 0 { displaystyle x ^ {2} + 8x20 = 0}
Step 4. Rewrite the equation as a fourterm expression
To proceed you need to separate the first degree term (the term x { displaystyle x}
); devi trovare due numeri la cui somma sia pari al coefficiente della variabile e il prodotto pari alla costante.
 Per esempio, se consideri il polinomio di secondo grado x2+8x−20=0{displaystyle x^{2}+8x20=0}
 Dato che la costante è negativa (−20{displaystyle 20}
 Dovresti renderti conto che 10+(−2)=8{displaystyle 10+(2)=8}
devi trovare i due numeri (a{displaystyle a}
e b{displaystyle b}
) per cui a+b=8{displaystyle a+b=8}
e a⋅b=−20{displaystyle a\cdot b=20}
), sai che uno dei due numeri è minore di zero.
e 10⋅(−2)=−20{displaystyle 10\cdot (2)=20}
. Di conseguenza, puoi riscrivere 8x{displaystyle 8x}
come 10x−2x{displaystyle 10x2x}
; pertanto, l'equazione si presenta in questo modo: x2+10x−2x−20=0{displaystyle x^{2}+10x2x20=0}
Step 5. Collect in factors
Extract the common one for the first two terms of the polynomial.

In the example considered so far, x2 + 10x − 2x − 20 = 0 { displaystyle x ^ {2} + 10x2x20 = 0}
i primi due termini sono x2+10x{displaystyle x^{2}+10x}
. Il fattore comune è x{displaystyle x}
quindi puoi riscriverli come x(x+10){displaystyle x(x+10)}
Step 6. Factor in the second group
Extract the common factor of the other two terms of the polynomial.

For example, looking at x2 + 10x − 2x − 20 = 0 { displaystyle x ^ {2} + 10x2x20 = 0}
i due termini successivi sono −2x−20{displaystyle 2x20}
il cui fattore comune è −2{displaystyle 2}
ottenendo di conseguenza −2(x+10){displaystyle 2(x+10)}
Step 7. Rewrite the polynomial as two binomials
A binomial is an expression made up of two terms. One is already present and is represented by the expressions collected in brackets and which are equal to each other; the other is the combination of common factors which lie outside the parentheses, but which can be combined for the distributive property.

For example, after collecting the common factors, the equation x2 + 10x − 2x − 20 = 0 { displaystyle x ^ {2} + 10x2x20 = 0}
diventa x(x+10)−2(x+10)=0{displaystyle x(x+10)2(x+10)=0}
 Il primo binomio è (x+10){displaystyle (x+10)}
 Il secondo è (x−2){displaystyle (x2)}
 Di conseguenza il polinomio iniziale di secondo grado, x2+8x−20=0{displaystyle x^{2}+8x20=0}
può essere riscritto come: (x+10)(x−2)=0{displaystyle (x+10)(x2)=0}
Step 8. Find the first solution, or root
To do this, you have to solve the first binomial for x { displaystyle x}
 Nell'esempio considerato, per calcolare la prima radice di (x+10)(x−2)=0{displaystyle (x+10)(x2)=0}
devi imporre il primo binomio uguale a 0{displaystyle 0}
e trovare x{displaystyle x}
. Quindi:
x+10=0{displaystyle x+10=0}
x+10−10=0−10{displaystyle x+1010=010}
x=−10{displaystyle x=10}
La prima soluzione del polinomio di secondo grado x2+8x−20=0{displaystyle x^{2}+8x20=0}
è −10{displaystyle 10}
Step 9. Find the second root, or solution
In that case, you have to solve the second binomial for x { displaystyle x}
 Per esempio, per calcolare la seconda radice di (x+10)(x−2)=0{displaystyle (x+10)(x2)=0}
 .
devi imporre il secondo binomio pari a 0{displaystyle 0}
e svolgere le operazioni per trovare x{displaystyle x}
. Quindi:
x−2=0{displaystyle x2=0}
x−2+2=0+2{displaystyle x2+2=0+2}
x=2{displaystyle x=2}
La seconda radice del polinomio di secondo grado x2+8x−20=0{displaystyle x^{2}+8x20=0}
è 2{displaystyle 2}
Advice
 Remember the order of operations; do the ones in parentheses first, then the multiplications and divisions and finally the additions and subtractions.