Calcolatrice polinomiale
Calcolo dei polinomi passo dopo passo
La calcolatrice trova (con i passaggi indicati) la somma, la differenza, il prodotto e il risultato della divisione di due polinomi (quadratici, binomiali, trinomiali, ecc.). Calcolerà anche le radici dei polinomi e li fattorizzerà. Sono accettati sia polinomi univariati che multivariati.
Solution
Your input: find the sum, difference, product of two polynomials, quotient and remainder from dividing one by another; factor them and find roots.
Addition of polynomials
To add polynomials, combine and add the coefficients near the like terms:
(2x4−3x3−15x2+32x−12)+(x2−4x−12)=
=2x4−3x3+((−15)+1)x2+(32+(−4))x+((−12)+(−12))=
=2x4−3x3−14x2+28x−24
Subtraction of polynomials
To subtract polynomials, combine and subtract the coefficients near the like terms:
(2x4−3x3−15x2+32x−12)−(x2−4x−12)=
=2x4−3x3+((−15)−1)x2+(32−(−4))x+((−12)−(−12))=
=2x4−3x3−16x2+36x
Multiplication of polynomials
To multiply polynomials, multiple each term of the first polynomial with every term of the second polynomial. Then simplify the products and add them. Finally, simplify further if possible.
So, perform the first step:
(2x4−3x3−15x2+32x−12)⋅(x2−4x−12)=
=(2x4)⋅(x2)+(2x4)⋅(−4x)+(2x4)⋅(−12)+
+(−3x3)⋅(x2)+(−3x3)⋅(−4x)+(−3x3)⋅(−12)+
+(−15x2)⋅(x2)+(−15x2)⋅(−4x)+(−15x2)⋅(−12)+
+(32x)⋅(x2)+(32x)⋅(−4x)+(32x)⋅(−12)+
+(−12)⋅(x2)+(−12)⋅(−4x)+(−12)⋅(−12)=
Simplify the products:
=2x6−8x5−24x4+
−3x5+12x4+36x3+
−15x4+60x3+180x2+
+32x3−128x2−384x+
−12x2+48x+144=
Simplify further (same way as adding/subtracting polynomials):
=2x6−11x5−27x4+128x3+40x2−336x+144
Division of polynomials
Perform polynomial long division (use the polynomial long division calculator to see the steps).
2x4−3x3−15x2+32x−12x2−4x−12=2x2+5x+29+208x+336x2−4x−12
Factoring 2x4−3x3−15x2+32x−12
Since all coefficients are integers, apply the rational zeros theorem.
The trailing coefficient (coefficient of the constant term) is −12.
Find its factors (with plus and minus): ±1,±2,±3,±4,±6,±12. These are the possible values for .
The leading coefficient (coefficient of the term with the highest degree) is 2.
Find its factors (with plus and minus): ±1,±2. These are the possible values for .
Find all possible values of ±11,±12,±21,±22,±31,±32,±41,±42,±61,±62,±121,±122.
:Simplify and remove duplicates (if any): ±1,±2,±3,±4,±6,±12,±12,±32.
If
is a root of the polynomial , then the remainder from the division of by should equal .Check 1: divide 2x4−3x3−15x2+32x−12 by x−1.
The quotient is 2x3−x2−16x+16, and the remainder is 4 (use the synthetic division calculator to see the steps).
Check −1: divide 2x4−3x3−15x2+32x−12 by x+1.
The quotient is 2x3−5x2−10x+42, and the remainder is −54 (use the synthetic division calculator to see the steps).
Check 2: divide 2x4−3x3−15x2+32x−12 by x−2.
The quotient is 2x3+x2−13x+6, and the remainder is 0 (use the synthetic division calculator to see the steps).
Since the remainder is 2 is the root, and x−2 is the factor: 2x4−3x3−15x2+32x−12=(x−2)(2x3+x2−13x+6)
, then(2x4−3x3−15x2+32x−12)=(x−2)(2x3+x2−13x+6)
Since all coefficients are integers, apply the rational zeros theorem.
The trailing coefficient (coefficient of the constant term) is 6.
Find its factors (with plus and minus): ±1,±2,±3,±6. These are the possible values for .
The leading coefficient (coefficient of the term with the highest degree) is 2.
Find its factors (with plus and minus): ±1,±2. These are the possible values for .
Find all possible values of ±11,±12,±21,±22,±31,±32,±61,±62.
:Simplify and remove duplicates (if any): ±1,±2,±3,±6,±12,±32.
If
is a root of the polynomial , then the remainder from the division of by should equal .Check 1: divide 2x3+x2−13x+6 by x−1.
The quotient is 2x2+3x−10, and the remainder is −4 (use the synthetic division calculator to see the steps).
Check −1: divide 2x3+x2−13x+6 by x+1.
The quotient is 2x2−x−12, and the remainder is 18 (use the synthetic division calculator to see the steps).
Check 2: divide 2x3+x2−13x+6 by x−2.
The quotient is 2x2+5x−3, and the remainder is 0 (use the synthetic division calculator to see the steps).
Since the remainder is 2 is the root, and x−2 is the factor: 2x3+x2−13x+6=(x−2)(2x2+5x−3)
, then(x−2)(2x3+x2−13x+6)=(x−2)(x−2)(2x2+5x−3)
To factor the quadratic function 2x2+5x−3, we should solve the corresponding quadratic equation 2x2+5x−3=0.
Indeed, if x1 and x2 are the roots of the quadratic equation ax2+bx+c=0, then ax2+bx+c=a(x−x1)(x−x2).
Solve the quadratic equation 2x2+5x−3=0.
The roots are x1=12, x2=−3 (use the quadratic equation calculator to see the steps).
Therefore, 2x2+5x−3=2(x−12)(x+3).
(x−2)2(2x2+5x−3)=(x−2)2(2(x−12)(x+3))
Simplify: 2(x−2)2(x−12)(x+3)=(x−2)2(x+3)(2x−1).
2x4−3x3−15x2+32x−12=(x−2)2(x+3)(2x−1).
Roots of the equation 2x4−3x3−15x2+32x−12=0
We have already found the factorization of 2x4−3x3−15x2+32x−12=(x−2)2(x+3)(2x−1) (see above).
Thus, we can write that 2x4−3x3−15x2+32x−12=0 is equivalent to the (x−2)2(x+3)(2x−1)=0.
It is known that the product is zero when at least one factor is zero, so we just need to set the factors equal to zero and solve the corresponding equations (some equations have already been solved, some can't be solved by hand).
- (x−2)2=0: the root is x=2 (multiplicity: 2).
- 2x−1=0: the root is x=12.
- x+3=0: the root is x=−3.
Therefore, the roots of the initial equation are: x1=−3; x2=12; x3=2 (multiplicity: 2).
Factoring x2−4x−12
To factor the quadratic function x2−4x−12, we should solve the corresponding quadratic equation x2−4x−12=0.
Indeed, if x1 and x2 are the roots of the quadratic equation ax2+bx+c=0, then ax2+bx+c=a(x−x1)(x−x2).
Solve the quadratic equation x2−4x−12=0.
The roots are x1=6, x2=−2 (use the quadratic equation calculator to see the steps).
Therefore, x2−4x−12=(x−6)(x+2).
(x2−4x−12)=(x−6)(x+2)
x2−4x−12=(x−6)(x+2).
Roots of the equation x2−4x−12=0
We have already found the factorization of x2−4x−12=(x−6)(x+2) (see above).
Thus, we can write that x2−4x−12=0 is equivalent to the (x−6)(x+2)=0.
It is known that the product is zero when at least one factor is zero, so we just need to set the factors equal to zero and solve the corresponding equations (some equations have already been solved, some can't be solved by hand).
- x−6=0: the root is x=6.
- x+2=0: the root is x=−2.
Therefore, the roots of the initial equation are: x1=6; x2=−2.
Answer:
(2x4−3x3−15x2+32x−12)+(x2−4x−12)=2x4−3x3−14x2+28x−24.
(2x4−3x3−15x2+32x−12)−(x2−4x−12)=2x4−3x3−16x2+36x.
(2x4−3x3−15x2+32x−12)⋅(x2−4x−12)=2x6−11x5−27x4+128x3+40x2−336x+144.
2x4−3x3−15x2+32x−12x2−4x−12=2x2+5x+29+208x+336x2−4x−12.
2x4−3x3−15x2+32x−12=(x−2)2(x+3)(2x−1).
Roots of the equation 2x4−3x3−15x2+32x−12=0:
- −3, multiplicity 1.
- 12, multiplicity 1.
- 2, multiplicity 2.
x2−4x−12=(x−6)(x+2).
Roots of the equation x2−4x−12=0:
- 6, multiplicity 1.
- −2, multiplicity 1.