Calculatrice de polynômes

Your input: find the sum, difference, product of two polynomials, quotient and remainder from dividing one by another; factor them and find roots.

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

$\left(\color{Magenta}{2 x^{4}}\color{Chocolate}{- 3 x^{3}}\color{GoldenRod}{- 15 x^{2}}+\color{Brown}{32 x}\color{Violet}{-12}\right)+\left(\color{GoldenRod}{x^{2}}\color{Brown}{- 4 x}\color{Violet}{-12}\right)=$

$=\color{Magenta}{2 x^{4}}\color{Chocolate}{- 3 x^{3}}+\color{GoldenRod}{\left(\left(-15\right)+1\right) x^{2}}+\color{Brown}{\left(32+\left(-4\right)\right) x}+\color{Violet}{\left(\left(-12\right)+\left(-12\right)\right) }=$

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

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

$\left(\color{Magenta}{2 x^{4}}\color{Chocolate}{- 3 x^{3}}\color{GoldenRod}{- 15 x^{2}}+\color{Brown}{32 x}\color{Violet}{-12}\right)-\left(\color{GoldenRod}{x^{2}}\color{Brown}{- 4 x}\color{Violet}{-12}\right)=$

$=\color{Magenta}{2 x^{4}}\color{Chocolate}{- 3 x^{3}}+\color{GoldenRod}{\left(\left(-15\right)-1\right) x^{2}}+\color{Brown}{\left(32-\left(-4\right)\right) x}+\color{Violet}{\left(\left(-12\right)-\left(-12\right)\right) }=$

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

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:

$\left(\color{Violet}{2 x^{4}}\color{Brown}{- 3 x^{3}}\color{GoldenRod}{- 15 x^{2}}+\color{Chocolate}{32 x}\color{Magenta}{-12}\right) \cdot \left(\color{Peru}{x^{2}}\color{DarkBlue}{- 4 x}\color{Blue}{-12}\right)=$

$=\left(\color{Violet}{2 x^{4}}\right)\cdot \left(\color{Peru}{x^{2}}\right)+\left(\color{Violet}{2 x^{4}}\right)\cdot \left(\color{DarkBlue}{- 4 x}\right)+\left(\color{Violet}{2 x^{4}}\right)\cdot \left(\color{Blue}{-12}\right)+$

$+\left(\color{Brown}{- 3 x^{3}}\right)\cdot \left(\color{Peru}{x^{2}}\right)+\left(\color{Brown}{- 3 x^{3}}\right)\cdot \left(\color{DarkBlue}{- 4 x}\right)+\left(\color{Brown}{- 3 x^{3}}\right)\cdot \left(\color{Blue}{-12}\right)+$

$+\left(\color{GoldenRod}{- 15 x^{2}}\right)\cdot \left(\color{Peru}{x^{2}}\right)+\left(\color{GoldenRod}{- 15 x^{2}}\right)\cdot \left(\color{DarkBlue}{- 4 x}\right)+\left(\color{GoldenRod}{- 15 x^{2}}\right)\cdot \left(\color{Blue}{-12}\right)+$

$+\left(\color{Chocolate}{32 x}\right)\cdot \left(\color{Peru}{x^{2}}\right)+\left(\color{Chocolate}{32 x}\right)\cdot \left(\color{DarkBlue}{- 4 x}\right)+\left(\color{Chocolate}{32 x}\right)\cdot \left(\color{Blue}{-12}\right)+$

$+\left(\color{Magenta}{-12}\right)\cdot \left(\color{Peru}{x^{2}}\right)+\left(\color{Magenta}{-12}\right)\cdot \left(\color{DarkBlue}{- 4 x}\right)+\left(\color{Magenta}{-12}\right)\cdot \left(\color{Blue}{-12}\right)=$

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):

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

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

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

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 $\displaystyle{\large{{p}}}$.

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 $\displaystyle{\large{{q}}}$.

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

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 $\displaystyle{\large{{a}}}$ is a root of the polynomial $\displaystyle{\large{{P}{\left({x}\right)}}}$, then the remainder from the division of $\displaystyle{\large{{P}{\left({x}\right)}}}$ by $\displaystyle{\large{{x}-{a}}}$ should equal $\displaystyle{\large{{0}}}$.

• 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 $\displaystyle{\large{{0}}}$, 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 $\displaystyle{\large{{p}}}$.

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 $\displaystyle{\large{{q}}}$.

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

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

If $\displaystyle{\large{{a}}}$ is a root of the polynomial $\displaystyle{\large{{P}{\left({x}\right)}}}$, then the remainder from the division of $\displaystyle{\large{{P}{\left({x}\right)}}}$ by $\displaystyle{\large{{x}-{a}}}$ should equal $\displaystyle{\large{{0}}}$.

• 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 $\displaystyle{\large{{0}}}$, 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: $2 \left(x - 2\right)^{2} \left(x - \frac{1}{2}\right) \left(x + 3\right)=\left(x - 2\right)^{2} \left(x + 3\right) \left(2 x - 1\right)$.

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

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: $2$).

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

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

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

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 $1$.