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Calcolatrice di Fattorizzazione polinomiale

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1

Here, we show you a step-by-step solved example of polynomial factorization. This solution was automatically generated by our smart calculator:

$4x^4-29x^2+25$
2

We can factor the polynomial $4x^4-29x^2+25$ using the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $25$

$1, 5, 25$
3

Next, list all divisors of the leading coefficient $a_n$, which equals $4$

$1, 2, 4$
4

The possible roots $\pm\frac{p}{q}$ of the polynomial $4x^4-29x^2+25$ will then be

$\pm\frac{1}{2},\:\pm\frac{1}{4},\:\pm1,\:\pm\frac{5}{2},\:\pm\frac{5}{4},\:\pm5,\:\pm\frac{25}{2},\:\pm\frac{25}{4},\:\pm25$
5

Trying all possible roots, we found that $1$ is a root of the polynomial. When we evaluate it in the polynomial, it gives us $0$ as a result

$4\cdot 1^4-29\cdot 1^2+25=0$
6

Now, divide the polynomial by the root we found $\left(x-1\right)$ using synthetic division (Ruffini's rule). First, write the coefficients of the terms of the numerator in descending order. Then, take the first coefficient $4$ and multiply by the factor $1$. Add the result to the second coefficient and then multiply this by $1$ and so on

$\left|\begin{array}{c}4 & 0 & -29 & 0 & 25 \\ & 4 & 4 & -25 & -25 \\ 4 & 4 & -25 & -25 & 0\end{array}\right|1$
7

In the last row of the division appear the new coefficients, with remainder equals zero. Now, rewrite the polynomial (a degree less) with the new coefficients, and multiplied by the factor $\left(x-1\right)$

$\left(4x^{3}+4x^{2}-25x-25\right)\left(x-1\right)$
8

We can factor the polynomial $\left(4x^{3}+4x^{2}-25x-25\right)$ using the rational root theorem, which guarantees that for a polynomial of the form $a_nx^n+a_{n-1}x^{n-1}+\dots+a_0$ there is a rational root of the form $\pm\frac{p}{q}$, where $p$ belongs to the divisors of the constant term $a_0$, and $q$ belongs to the divisors of the leading coefficient $a_n$. List all divisors $p$ of the constant term $a_0$, which equals $-25$

$1, 5, 25$
9

Next, list all divisors of the leading coefficient $a_n$, which equals $4$

$1, 2, 4$
10

The possible roots $\pm\frac{p}{q}$ of the polynomial $\left(4x^{3}+4x^{2}-25x-25\right)$ will then be

$\pm\frac{1}{2},\:\pm\frac{1}{4},\:\pm1,\:\pm\frac{5}{2},\:\pm\frac{5}{4},\:\pm5,\:\pm\frac{25}{2},\:\pm\frac{25}{4},\:\pm25$
11

Trying all possible roots, we found that $-1$ is a root of the polynomial. When we evaluate it in the polynomial, it gives us $0$ as a result

$4\cdot {\left(-1\right)}^{3}+4\cdot {\left(-1\right)}^{2}-25\cdot -1-25=0$
12

Now, divide the polynomial by the root we found $\left(x+1\right)$ using synthetic division (Ruffini's rule). First, write the coefficients of the terms of the numerator in descending order. Then, take the first coefficient $4$ and multiply by the factor $-1$. Add the result to the second coefficient and then multiply this by $-1$ and so on

$\left|\begin{array}{c}4 & 4 & -25 & -25 \\ & -4 & 0 & 25 \\ 4 & 0 & -25 & 0\end{array}\right|-1$
13

In the last row of the division appear the new coefficients, with remainder equals zero. Now, rewrite the polynomial (a degree less) with the new coefficients, and multiplied by the factor $\left(x+1\right)$

$\left(4x^{2}-25\right)\left(x+1\right)\left(x-1\right)$

The power of a product is equal to the product of it's factors raised to the same power

$\left(2x+\sqrt{25}\right)\left(x+1\right)\left(x-1\right)\left(\sqrt{4x^{2}}-\sqrt{25}\right)$

Calculate the power $\sqrt{25}$

$\left(2x+5\right)\left(x+1\right)\left(x-1\right)\left(\sqrt{4x^{2}}-\sqrt{25}\right)$

The power of a product is equal to the product of it's factors raised to the same power

$\left(2x+5\right)\left(x+1\right)\left(x-1\right)\left(2x-\sqrt{25}\right)$

Calculate the power $\sqrt{25}$

$\left(2x+5\right)\left(x+1\right)\left(x-1\right)\left(2x- 5\right)$

Multiply $-1$ times $5$

$\left(2x+5\right)\left(x+1\right)\left(x-1\right)\left(2x-5\right)$
14

Factor the difference of squares $\left(4x^{2}-25\right)$ as the product of two conjugated binomials

$\left(2x+5\right)\left(x+1\right)\left(x-1\right)\left(2x-5\right)$

Risposta finale al problema

$\left(2x+5\right)\left(x+1\right)\left(x-1\right)\left(2x-5\right)$

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