Find a polynomial with real coefficients calculator

Summary :

The degree function calculates online the degree of a polynomial.

degree online

Description :

The computer is able to calculate online the degree of a polynomial.

Calculating the degree of a polynomial

The calculator may be used to determine the degree of a polynomial.

To obtain the degree of a polynomial defined by the following expression `x^3+x^2+1`, enter : degree(`x^3+x^2+1`) after calculation, the result 3 is returned.

Calculating the degree of a polynomial with symbolic coefficients

The calculator is also able to calculate the degree of a polynomial that uses letters as coefficients.

To obtain the degree of a polynomial defined by the following expression : `ax^2+bx+c` enter degree(`ax^2+bx+c`) after calculation, result 2 is returned.

Syntax :

degree(polynomial)

Examples :

degree(`x^3+x^2+1`), returns 3

Calculate online with degree (degree of a polynomial)

A polynomial is defined by the coefficients array, which can be real or complex numbers. The first coefficient belongs to the highest degree term; the last one is the constant term. The number of coefficients automatically defines the polynomial degree. If your polynomial misses a term, just set zero as its coefficient. Both polynomial coefficient and sample variable value can be either real or complex. (For real number, enter zero in the imaginary part.)

Polynomial value

The calculator can be used for equation solution checking. E.g. this one: Quartic equation or this: Cubic equation.

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Find the degree, leading coefficient, and leading term of a polynomial step by step

The calculator will find the degree, leading coefficient, and leading term of the given polynomial function.

Your Input

Find the degree, the leading coefficient, and the leading term of $$$p{\left(x \right)} = 5 x^{7} + 2 x^{5} - 4 x^{3} + x^{2} + 15$$$.

Solution

The degree of a polynomial is the highest of the degrees of the polynomial's individual terms. In our case, the degree is $$$7$$$.

The leading term is the term with the highest degree. In our case, the leading term is $$$5 x^{7}$$$.

The leading coefficient is the coefficient of the leading term. In our case, the leading coefficient is $$$5$$$.

Answer

Degree: $$$7$$$A.

Leading coefficient: $$$5$$$A.

Leading term: $$$5 x^{7}$$$A.

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:

$$$\left(\color{SaddleBrown}{2 x^{4}}\color{DarkCyan}{- 3 x^{3}}\color{Peru}{- 15 x^{2}}+\color{DeepPink}{32 x}\color{GoldenRod}{-12}\right)+\left(\color{Peru}{x^{2}}\color{DeepPink}{- 4 x}\color{GoldenRod}{-12}\right)=$$$

$$$=\color{SaddleBrown}{2 x^{4}}\color{DarkCyan}{- 3 x^{3}}+\color{Peru}{\left(\left(-15\right)+1\right) x^{2}}+\color{DeepPink}{\left(32+\left(-4\right)\right) x}+\color{GoldenRod}{\left(\left(-12\right)+\left(-12\right)\right) }=$$$

$$$=2 x^{4} - 3 x^{3} - 14 x^{2} + 28 x - 24$$$

Subtraction of polynomials

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

$$$\left(\color{SaddleBrown}{2 x^{4}}\color{DarkCyan}{- 3 x^{3}}\color{Peru}{- 15 x^{2}}+\color{DeepPink}{32 x}\color{GoldenRod}{-12}\right)-\left(\color{Peru}{x^{2}}\color{DeepPink}{- 4 x}\color{GoldenRod}{-12}\right)=$$$

$$$=\color{SaddleBrown}{2 x^{4}}\color{DarkCyan}{- 3 x^{3}}+\color{Peru}{\left(\left(-15\right)-1\right) x^{2}}+\color{DeepPink}{\left(32-\left(-4\right)\right) x}+\color{GoldenRod}{\left(\left(-12\right)-\left(-12\right)\right) }=$$$

$$$=2 x^{4} - 3 x^{3} - 16 x^{2} + 36 x$$$

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:

$$$\left(\color{GoldenRod}{2 x^{4}}\color{DeepPink}{- 3 x^{3}}\color{Peru}{- 15 x^{2}}+\color{DarkCyan}{32 x}\color{SaddleBrown}{-12}\right) \cdot \left(\color{Crimson}{x^{2}}\color{Purple}{- 4 x}\color{Fuchsia}{-12}\right)=$$$

$$$=\left(\color{GoldenRod}{2 x^{4}}\right)\cdot \left(\color{Crimson}{x^{2}}\right)+\left(\color{GoldenRod}{2 x^{4}}\right)\cdot \left(\color{Purple}{- 4 x}\right)+\left(\color{GoldenRod}{2 x^{4}}\right)\cdot \left(\color{Fuchsia}{-12}\right)+$$$

$$$+\left(\color{DeepPink}{- 3 x^{3}}\right)\cdot \left(\color{Crimson}{x^{2}}\right)+\left(\color{DeepPink}{- 3 x^{3}}\right)\cdot \left(\color{Purple}{- 4 x}\right)+\left(\color{DeepPink}{- 3 x^{3}}\right)\cdot \left(\color{Fuchsia}{-12}\right)+$$$

$$$+\left(\color{Peru}{- 15 x^{2}}\right)\cdot \left(\color{Crimson}{x^{2}}\right)+\left(\color{Peru}{- 15 x^{2}}\right)\cdot \left(\color{Purple}{- 4 x}\right)+\left(\color{Peru}{- 15 x^{2}}\right)\cdot \left(\color{Fuchsia}{-12}\right)+$$$

$$$+\left(\color{DarkCyan}{32 x}\right)\cdot \left(\color{Crimson}{x^{2}}\right)+\left(\color{DarkCyan}{32 x}\right)\cdot \left(\color{Purple}{- 4 x}\right)+\left(\color{DarkCyan}{32 x}\right)\cdot \left(\color{Fuchsia}{-12}\right)+$$$

$$$+\left(\color{SaddleBrown}{-12}\right)\cdot \left(\color{Crimson}{x^{2}}\right)+\left(\color{SaddleBrown}{-12}\right)\cdot \left(\color{Purple}{- 4 x}\right)+\left(\color{SaddleBrown}{-12}\right)\cdot \left(\color{Fuchsia}{-12}\right)=$$$

Simplify the products:

$$$=2 x^{6}- 8 x^{5}- 24 x^{4}+$$$

$$$- 3 x^{5}+12 x^{4}+36 x^{3}+$$$

$$$- 15 x^{4}+60 x^{3}+180 x^{2}+$$$

$$$+32 x^{3}- 128 x^{2}- 384 x+$$$

$$$- 12 x^{2}+48 x+144=$$$

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

$$$=2 x^{6} - 11 x^{5} - 27 x^{4} + 128 x^{3} + 40 x^{2} - 336 x + 144$$$

Division of polynomials

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

$$$\frac{2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12}{x^{2} - 4 x - 12}=2 x^{2} + 5 x + 29+\frac{208 x + 336}{x^{2} - 4 x - 12}$$$

Factoring $$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12$$$

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): $$$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$$$. These are the possible values for `p`.

The leading coefficient (coefficient of the term with the highest degree) is $$$2$$$.

Find its factors (with plus and minus): $$$\pm 1, \pm 2$$$. These are the possible values for `q`.

Find all possible values of `p/q`: $$$\pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{2}{1}, \pm \frac{2}{2}, \pm \frac{3}{1}, \pm \frac{3}{2}, \pm \frac{4}{1}, \pm \frac{4}{2}, \pm \frac{6}{1}, \pm \frac{6}{2}, \pm \frac{12}{1}, \pm \frac{12}{2}$$$.

Simplify and remove duplicates (if any): $$$\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12, \pm \frac{1}{2}, \pm \frac{3}{2}$$$.

If `a` is a root of the polynomial `P(x)`, then the remainder from the division of `P(x)` by `x-a` should equal `0`.

• Check $$$1$$$: divide $$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12$$$ by $$$x - 1$$$.

  The quotient is $$$2 x^{3} - x^{2} - 16 x + 16$$$, and the remainder is $$$4$$$ (use the synthetic division calculator to see the steps).

• Check $$$-1$$$: divide $$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12$$$ by $$$x + 1$$$.

  The quotient is $$$2 x^{3} - 5 x^{2} - 10 x + 42$$$, and the remainder is $$$-54$$$ (use the synthetic division calculator to see the steps).

• Check $$$2$$$: divide $$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12$$$ by $$$x - 2$$$.

  The quotient is $$$2 x^{3} + x^{2} - 13 x + 6$$$, and the remainder is $$$0$$$ (use the synthetic division calculator to see the steps).

Since the remainder is `0`, then $$$2$$$ is the root, and $$$x - 2$$$ is the factor: $$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12 = \left(x - 2\right) \left(2 x^{3} + x^{2} - 13 x + 6\right)$$$

$$\color{red}{\left(2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12\right)} = \color{red}{\left(x - 2\right) \left(2 x^{3} + x^{2} - 13 x + 6\right)}$$

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): $$$\pm 1, \pm 2, \pm 3, \pm 6$$$. These are the possible values for `p`.

The leading coefficient (coefficient of the term with the highest degree) is $$$2$$$.

Find its factors (with plus and minus): $$$\pm 1, \pm 2$$$. These are the possible values for `q`.

Find all possible values of `p/q`: $$$\pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{2}{1}, \pm \frac{2}{2}, \pm \frac{3}{1}, \pm \frac{3}{2}, \pm \frac{6}{1}, \pm \frac{6}{2}$$$.

Simplify and remove duplicates (if any): $$$\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$$$.

If `a` is a root of the polynomial `P(x)`, then the remainder from the division of `P(x)` by `x-a` should equal `0`.

• Check $$$1$$$: divide $$$2 x^{3} + x^{2} - 13 x + 6$$$ by $$$x - 1$$$.

  The quotient is $$$2 x^{2} + 3 x - 10$$$, and the remainder is $$$-4$$$ (use the synthetic division calculator to see the steps).

• Check $$$-1$$$: divide $$$2 x^{3} + x^{2} - 13 x + 6$$$ by $$$x + 1$$$.

  The quotient is $$$2 x^{2} - x - 12$$$, and the remainder is $$$18$$$ (use the synthetic division calculator to see the steps).

• Check $$$2$$$: divide $$$2 x^{3} + x^{2} - 13 x + 6$$$ by $$$x - 2$$$.

  The quotient is $$$2 x^{2} + 5 x - 3$$$, and the remainder is $$$0$$$ (use the synthetic division calculator to see the steps).

Since the remainder is `0`, then $$$2$$$ is the root, and $$$x - 2$$$ is the factor: $$$2 x^{3} + x^{2} - 13 x + 6 = \left(x - 2\right) \left(2 x^{2} + 5 x - 3\right)$$$

$$\left(x - 2\right) \color{red}{\left(2 x^{3} + x^{2} - 13 x + 6\right)} = \left(x - 2\right) \color{red}{\left(x - 2\right) \left(2 x^{2} + 5 x - 3\right)}$$

To factor the quadratic function $$$2 x^{2} + 5 x - 3$$$, we should solve the corresponding quadratic equation $$$2 x^{2} + 5 x - 3=0$$$.

Indeed, if $$$x_1$$$ and $$$x_2$$$ are the roots of the quadratic equation $$$ax^2+bx+c=0$$$, then $$$ax^2+bx+c=a(x-x_1)(x-x_2)$$$.

Solve the quadratic equation $$$2 x^{2} + 5 x - 3=0$$$.

The roots are $$$x_{1} = \frac{1}{2}$$$, $$$x_{2} = -3$$$ (use the quadratic equation calculator to see the steps).

Therefore, $$$2 x^{2} + 5 x - 3 = 2 \left(x - \frac{1}{2}\right) \left(x + 3\right)$$$.

$$\left(x - 2\right)^{2} \color{red}{\left(2 x^{2} + 5 x - 3\right)} = \left(x - 2\right)^{2} \color{red}{\left(2 \left(x - \frac{1}{2}\right) \left(x + 3\right)\right)}$$

Simplify: $$$2 \left(x - 2\right)^{2} \left(x - \frac{1}{2}\right) \left(x + 3\right)=\left(x - 2\right)^{2} \left(x + 3\right) \left(2 x - 1\right)$$$.

$$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12=\left(x - 2\right)^{2} \left(x + 3\right) \left(2 x - 1\right)$$$.

Roots of the equation $$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12=0$$$

We have already found the factorization of $$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12=\left(x - 2\right)^{2} \left(x + 3\right) \left(2 x - 1\right)$$$ (see above).

Thus, we can write that $$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12=0$$$ is equivalent to the $$$\left(x - 2\right)^{2} \left(x + 3\right) \left(2 x - 1\right)=0$$$.

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).

• $$$\left(x - 2\right)^{2}=0$$$: the root is $$$x=2$$$ (multiplicity: $$$2$$$).
• $$$2 x - 1=0$$$: the root is $$$x=\frac{1}{2}$$$.
• $$$x + 3=0$$$: the root is $$$x=-3$$$.

Therefore, the roots of the initial equation are: $$$x_1=-3$$$; $$$x_2=\frac{1}{2}$$$; $$$x_3=2$$$ (multiplicity: $$$2$$$).

Factoring $$$x^{2} - 4 x - 12$$$

To factor the quadratic function $$$x^{2} - 4 x - 12$$$, we should solve the corresponding quadratic equation $$$x^{2} - 4 x - 12=0$$$.

Indeed, if $$$x_1$$$ and $$$x_2$$$ are the roots of the quadratic equation $$$ax^2+bx+c=0$$$, then $$$ax^2+bx+c=a(x-x_1)(x-x_2)$$$.

Solve the quadratic equation $$$x^{2} - 4 x - 12=0$$$.

The roots are $$$x_{1} = 6$$$, $$$x_{2} = -2$$$ (use the quadratic equation calculator to see the steps).

Therefore, $$$x^{2} - 4 x - 12 = \left(x - 6\right) \left(x + 2\right)$$$.

$$\color{red}{\left(x^{2} - 4 x - 12\right)} = \color{red}{\left(x - 6\right) \left(x + 2\right)}$$

$$$x^{2} - 4 x - 12=\left(x - 6\right) \left(x + 2\right)$$$.

Roots of the equation $$$x^{2} - 4 x - 12=0$$$

We have already found the factorization of $$$x^{2} - 4 x - 12=\left(x - 6\right) \left(x + 2\right)$$$ (see above).

Thus, we can write that $$$x^{2} - 4 x - 12=0$$$ is equivalent to the $$$\left(x - 6\right) \left(x + 2\right)=0$$$.

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: $$$x_1=6$$$; $$$x_2=-2$$$.

Answer:

$$$\left(2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12\right)+\left(x^{2} - 4 x - 12\right)=2 x^{4} - 3 x^{3} - 14 x^{2} + 28 x - 24$$$.

$$$\left(2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12\right)-\left(x^{2} - 4 x - 12\right)=2 x^{4} - 3 x^{3} - 16 x^{2} + 36 x$$$.

$$$\left(2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12\right)\cdot \left(x^{2} - 4 x - 12\right)=2 x^{6} - 11 x^{5} - 27 x^{4} + 128 x^{3} + 40 x^{2} - 336 x + 144$$$.

$$$\frac{2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12}{x^{2} - 4 x - 12}=2 x^{2} + 5 x + 29+\frac{208 x + 336}{x^{2} - 4 x - 12}$$$.

$$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12=\left(x - 2\right)^{2} \left(x + 3\right) \left(2 x - 1\right)$$$.

Roots of the equation $$$2 x^{4} - 3 x^{3} - 15 x^{2} + 32 x - 12=0$$$:

• $$$-3$$$, multiplicity $$$1$$$.
• $$$\frac{1}{2}$$$, multiplicity $$$1$$$.
• $$$2$$$, multiplicity $$$2$$$.

$$$x^{2} - 4 x - 12=\left(x - 6\right) \left(x + 2\right)$$$.

Roots of the equation $$$x^{2} - 4 x - 12=0$$$:

• $$$6$$$, multiplicity $$$1$$$.
• $$$-2$$$, multiplicity $$$1$$$.

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