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Calculatrice de polynômes

Calculer des polynômes pas à pas

La calculatrice trouvera (avec les étapes indiquées) la somme, la différence, le produit et le résultat de la division de deux polynômes (quadratique, binomial, trinomial, etc.). Il calculera également les racines des polynômes et les factorisera. Les polynômes univariés et multivariés sont acceptés.

First polynomial:

Second polynomial:

The second polynomial is needed for addition, subtraction, multiplication, division; but not for root finding, factoring.

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please contact us.

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:

(2x43x315x2+32x12)+(x24x12)=

=2x43x3+((15)+1)x2+(32+(4))x+((12)+(12))=

=2x43x314x2+28x24


Subtraction of polynomials

To subtract polynomials, combine and subtract the coefficients near the like terms:

(2x43x315x2+32x12)(x24x12)=

=2x43x3+((15)1)x2+(32(4))x+((12)(12))=

=2x43x316x2+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:

(2x43x315x2+32x12)(x24x12)=

=(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:

=2x68x524x4+

3x5+12x4+36x3+

15x4+60x3+180x2+

+32x3128x2384x+

12x2+48x+144=

Simplify further (same way as adding/subtracting polynomials):

=2x611x527x4+128x3+40x2336x+144


Division of polynomials

Perform polynomial long division (use the polynomial long division calculator to see the steps).

2x43x315x2+32x12x24x12=2x2+5x+29+208x+336x24x12


Factoring 2x43x315x2+32x12

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 2x43x315x2+32x12 by x1.

    The quotient is 2x3x216x+16, and the remainder is 4 (use the synthetic division calculator to see the steps).

  • Check 1: divide 2x43x315x2+32x12 by x+1.

    The quotient is 2x35x210x+42, and the remainder is 54 (use the synthetic division calculator to see the steps).

  • Check 2: divide 2x43x315x2+32x12 by x2.

    The quotient is 2x3+x213x+6, and the remainder is 0 (use the synthetic division calculator to see the steps).

Since the remainder is , then 2 is the root, and x2 is the factor: 2x43x315x2+32x12=(x2)(2x3+x213x+6)

(2x43x315x2+32x12)=(x2)(2x3+x213x+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+x213x+6 by x1.

    The quotient is 2x2+3x10, and the remainder is 4 (use the synthetic division calculator to see the steps).

  • Check 1: divide 2x3+x213x+6 by x+1.

    The quotient is 2x2x12, and the remainder is 18 (use the synthetic division calculator to see the steps).

  • Check 2: divide 2x3+x213x+6 by x2.

    The quotient is 2x2+5x3, and the remainder is 0 (use the synthetic division calculator to see the steps).

Since the remainder is , then 2 is the root, and x2 is the factor: 2x3+x213x+6=(x2)(2x2+5x3)

(x2)(2x3+x213x+6)=(x2)(x2)(2x2+5x3)

To factor the quadratic function 2x2+5x3, we should solve the corresponding quadratic equation 2x2+5x3=0.

Indeed, if x1 and x2 are the roots of the quadratic equation ax2+bx+c=0, then ax2+bx+c=a(xx1)(xx2).

Solve the quadratic equation 2x2+5x3=0.

The roots are x1=12, x2=3 (use the quadratic equation calculator to see the steps).

Therefore, 2x2+5x3=2(x12)(x+3).

(x2)2(2x2+5x3)=(x2)2(2(x12)(x+3))

Simplify: 2(x2)2(x12)(x+3)=(x2)2(x+3)(2x1).

2x43x315x2+32x12=(x2)2(x+3)(2x1).


Roots of the equation 2x43x315x2+32x12=0

We have already found the factorization of 2x43x315x2+32x12=(x2)2(x+3)(2x1) (see above).

Thus, we can write that 2x43x315x2+32x12=0 is equivalent to the (x2)2(x+3)(2x1)=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).

  • (x2)2=0: the root is x=2 (multiplicity: 2).
  • 2x1=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 x24x12

To factor the quadratic function x24x12, we should solve the corresponding quadratic equation x24x12=0.

Indeed, if x1 and x2 are the roots of the quadratic equation ax2+bx+c=0, then ax2+bx+c=a(xx1)(xx2).

Solve the quadratic equation x24x12=0.

The roots are x1=6, x2=2 (use the quadratic equation calculator to see the steps).

Therefore, x24x12=(x6)(x+2).

(x24x12)=(x6)(x+2)

x24x12=(x6)(x+2).


Roots of the equation x24x12=0

We have already found the factorization of x24x12=(x6)(x+2) (see above).

Thus, we can write that x24x12=0 is equivalent to the (x6)(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).

  • x6=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:

(2x43x315x2+32x12)+(x24x12)=2x43x314x2+28x24.

(2x43x315x2+32x12)(x24x12)=2x43x316x2+36x.

(2x43x315x2+32x12)(x24x12)=2x611x527x4+128x3+40x2336x+144.

2x43x315x2+32x12x24x12=2x2+5x+29+208x+336x24x12.

2x43x315x2+32x12=(x2)2(x+3)(2x1).

Roots of the equation 2x43x315x2+32x12=0:

  • 3, multiplicity 1.
  • 12, multiplicity 1.
  • 2, multiplicity 2.

x24x12=(x6)(x+2).

Roots of the equation x24x12=0:

  • 6, multiplicity 1.
  • 2, multiplicity 1.