Polynomials can be divided like numerical constants, either by factoring or by long division. The method you use depends on how complex the dividend and divisor of the polynomial are.
Steps
Method 1 of 3: Part 1 of 3: Choose the appropriate approach
Step 1. Observe the complexity of the divider
The level of complexity of the divisor (the polynomial you are dividing by) versus the dividend (the polynomial you are dividing into) determines the best approach to use.
- If the divisor is a monomial (single-term polynomial), or a variable with a coefficient or a constant (a number not followed by a variable), you can probably factor the dividend and cancel one of the resulting factors and dividends. See Part 2 for instructions and examples.
- If the divisor is a binomial (2-term polynomial), you may be able to break down the dividend and cancel out one of the resulting factors and divisors.
- If the divisor is a trinomial (3-term polynomial), you may be able to factor both the dividend and the divisor, cancel the common factor, and then either further break down the dividend or use long division.
- If the divisor is a polynomial with more than 3 factors, you will probably need to use long division. See Part 3 for instructions and examples.
Step 2. Look at the complexity of the dividend
If the polynomial divisor of the equation does not suggest you try to break down the dividend, look at the dividend itself.
- If the dividend has 3 or less than 3 terms, you can probably break it down and cross out the divisor.
- If the dividend has more than 3 terms, you will probably need to divide the divisor by it using long division.
Method 2 of 3: Part 2 of 3: Break down the dividend
Step 1. Check if all the terms of the dividend contain a factor in common with the divisors
If that's the case, you can break it down and probably get rid of the divider.
- If you are dividing the binomial 3x - 9 by 3, you can decompose the 3 from both terms of the binomial, making it 3 (x - 3). You can later cancel out the divisor 3, giving you a quotient of x - 3.
- If you are dividing by 6x the binomial 24x3 - 18x2, you can decompose 6x from both terms of the binomial, making it 6x (4x2 - 3). You can then cancel the divisor, leaving a quotient of 4x2 - 3.
Step 2. Look for particular sequences in the dividend that indicate the possibility of breaking it down
Certain polynomials show terms that tell you they can be factored. If one of those factors matches the divisor, you can cancel it out, leaving the remaining factor as the quotient. Here are some sequences to look for:
- Perfect difference of squares. This is the combination of form '' a 2x2 - b '', in which the values of '' a 2'' And '' b 2’’ Are perfect squares. This binomial breaks down into two binomials (ax + b) (ax - b), where a and b are the square roots of the coefficient and the constant of the previous binomial.
- Perfect square trinomial. This trinomial has the form a2x2 + 2abx + b 2. It breaks down into (ax + b) (ax + b), which can also be written as (ax + b)2. If the sign in front of the second term is a minus, the binomial decompositions will be expressed as follows: (ax - b) (ax - b).
- Sum or difference of cubes. This binomial has the form a3x3 + b3 or a3x3 - b3, in which the values of '' a 3'' And '' b 3’’ Are perfect cubes. This binomial breaks down into a binomial and a trinomial. A sum of cubes is decomposed into (ax + b) (a2x2 - abx + b2). A difference of cubes is decomposed into (ax - b) (a2x2 + abx + b2).
Step 3. Use trial and error to break down the dividend
If you don't see a special sequence in the dividend that tells you how to break it down, you can try different possible combinations for the breakdown. You can do this by looking first at the constant and finding various decompositions for it, then at the coefficient of the central term.
- For example, if the dividend were x2 - 3x - 10, you would look at the factors of 10 and use the 3 to help you determine which pair of factors is correct.
- The number 10 can be factored into 1 and 10 or 2 and 5. Since the sign in front of 10 is negative, one of the binomial factors must have a negative number in front of its constant.
- The number 3 is the difference between 2 and 5, so these must be the constants of the decomposed binomials. Since the sign in front of the 3 is negative, the pairing with the 5 must be the negative one. The binomial decompositions will therefore be (x - 5) (x + 2). If the divisor is one of these two decompositions, that can be eliminated, and the other is the quotient.
Method 3 of 3: Part 3 of 3: Using long polynomial division
Step 1. Prepare the division
Write long polynomial division the same way you would divide numbers. The dividend goes below the long dividing line, while the divider goes to the left.
If you are dividing x2 + 11 x + 10 for x +1, x2 + 11 x + 10 goes below the line, while x + 1 goes to the left.
Step 2. Divide the first term of the divisor into the first term of the dividend
The result of this division goes to the top of the division line.
For our example, dividing x2, the first term of the dividend, for x, the first term of the divisor yields x. You will write an x at the top of the dividing line, above x2.
Step 3. Multiply the x in the quotient position by the divisor
Write the result of the multiplication under the leftmost terms of the dividend.
Continuing with our example, multiplying x + 1 by x gives x2 + x. You will write this under the first two terms of the dividend.
Step 4. Subtract from the dividend
To do this, first invert the signs of the product of multiplication. After subtracting, bring in the remaining terms of the dividend.
The inversion of the signs of x2 + x creates - x2 - x. Subtracting this from the first two terms of the dividend we get 10x. After bringing down the remaining terms of the dividend, we have 10x + 10 as a provisional quotient on which to continue the splitting process.
Step 5. Repeat the previous three steps on the provisional quotient
Divide the first term of the divisor back into the provisional quotient, write the result at the top of the dividing line after the first term of the quotient, multiply the result by the divisor, and then calculate what to subtract from the provisional quotient.
- Since x is 10 times in 10x, you will write “+ 10” after the x in the position of the quotient on the division bar.
- Multiplying x +1 by 10 yields 10x + 10. Write this under the provisional quotient and reverse the signs for the subtraction, making it -10x - 10.
- When you do the subtraction, you have a remainder of 0. Now, dividing x2 + 11 x + 10 times x +1 you get a quotient of x + 10. (You could have done the same by factoring, but this example was chosen to keep the division relatively simple).
Advice
- If, during a long division on a polynomial, you have a remainder not equal to 0, you can make that remainder part of the quotient by writing it as a fraction that has the remainder as its numerator and the divisor as its denominator. If, in our example, the dividend was x2 + 11 x + 12 instead of x2 + 11 x + 10, dividing by x +1 would leave a remainder of 2. The complete quotient would then be written as: x + 10 + 2x + 1 { displaystyle x + 10 + { frac {2} {x +1}}}
- se il dividendo ha un vuoto nei gradi dei propri termini, tipo 3x3+9x2+18, puoi inserire il termine mancante con un coefficiente di 0, in questo caso 0x, per rendere più facile il posizionamento degli altri termini nella divisione. fare questo non cambia il valore del dividendo.
- sii consapevole che alcuni libri di algebra tendono a giustificare l’impaginazione di quoziente e dividendo nelle divisioni polinomiali, o a presentare i termini in modo che elementi con lo stesso grado in entrambi i polinomi risultino allineati l’un l’altro. potresti trovare più semplice, tuttavia, quando fai le divisioni a mano, giustificare sulla sinistra quoziente e dividendo come descritto nei passaggi precedenti.
avvertenze
- mantieni le colonne allineate mentre dividi polinomi lunghi per evitare di sottrarre i termini sbagliati.
- quando scrivi il quoziente di una divisione polinomiale che include un elemento frazionale, usa sempre un segno più tra l’intero numero (o l’intera variabile) e l’elemento frazionale.
