How to Solve Polynomials: 13 Steps (with Pictures)

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How to Solve Polynomials: 13 Steps (with Pictures)
How to Solve Polynomials: 13 Steps (with Pictures)

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.


Method 1 of 2: First Degree Polynomials

Solve Polynomials Step 1
Solve Polynomials Step 1

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.

Solve Polynomials Step 2
Solve Polynomials Step 2

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}
Solve Polynomials Step 3
Solve Polynomials Step 3

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+2-2=0-2}

    5x=−2{displaystyle 5x=-2}

Solve Polynomials Step 4
Solve Polynomials Step 4

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

Solve Polynomials Step 5
Solve Polynomials Step 5

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} + 8x-20}

    è un polinomio di secondo grado perché la variabile x{displaystyle x}

    ha un esponente pari a 2{displaystyle 2}

Solve Polynomials Step 6
Solve Polynomials Step 6

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}+8x-20}

Solve Polynomials Step 7
Solve Polynomials Step 7

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} + 8x-20 = 0}
Solve Polynomials Step 8
Solve Polynomials Step 8

Step 4. Rewrite the equation as a four-term 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}+8x-20=0}
  • 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}

  • Dato che la costante è negativa (−20{displaystyle -20}
  • ), sai che uno dei due numeri è minore di zero.

  • Dovresti renderti conto che 10+(−2)=8{displaystyle 10+(-2)=8}
  • e 10⋅(−2)=−20{displaystyle 10\cdot (-2)=-20}

    . Di conseguenza, puoi riscrivere 8x{displaystyle 8x}

    come 10x−2x{displaystyle 10x-2x}

    ; pertanto, l'equazione si presenta in questo modo: x2+10x−2x−20=0{displaystyle x^{2}+10x-2x-20=0}

Solve Polynomials Step 9
Solve Polynomials Step 9

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} + 10x-2x-20 = 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)}

Solve Polynomials Step 10
Solve Polynomials Step 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} + 10x-2x-20 = 0}

    i due termini successivi sono −2x−20{displaystyle -2x-20}

    il cui fattore comune è −2{displaystyle -2}

    ottenendo di conseguenza −2(x+10){displaystyle -2(x+10)}

Solve Polynomials Step 11
Solve Polynomials Step 11

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} + 10x-2x-20 = 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 (x-2)}
  • Di conseguenza il polinomio iniziale di secondo grado, x2+8x−20=0{displaystyle x^{2}+8x-20=0}
  • può essere riscritto come: (x+10)(x−2)=0{displaystyle (x+10)(x-2)=0}

Solve Polynomials Step 12
Solve Polynomials Step 12

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)(x-2)=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+10-10=0-10}

    x=−10{displaystyle x=-10}

    La prima soluzione del polinomio di secondo grado x2+8x−20=0{displaystyle x^{2}+8x-20=0}

    è −10{displaystyle -10}

Solve Polynomials Step 13
Solve Polynomials Step 13

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)(x-2)=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 x-2=0}

    x−2+2=0+2{displaystyle x-2+2=0+2}

    x=2{displaystyle x=2}

    La seconda radice del polinomio di secondo grado x2+8x−20=0{displaystyle x^{2}+8x-20=0}

    è 2{displaystyle 2}

  • .


  • Remember the order of operations; do the ones in parentheses first, then the multiplications and divisions and finally the additions and subtractions.