# Solve the factorization of x^{12}+x^{7}+x^{5}+1 Question
Polynomial factorization Solve the factorization of $$x^{12}+x^{7}+x^{5}+1$$ 2020-11-03
Formula used:
The factors of a polynomial can be find by taking a common factor and this method is called factor by grouping,
$$ab+ac+bd+cd=a(b+c)+d(b+c)$$
$$=(a+d)(b+c)$$
Or,
$$ab-ac+bd-cd=a(b-c)+d(b-c)$$
$$=(a+d)(b-c)$$
Calculation:
Consider the polynomial $$x^{12} + x^{7} +x^{5} +1$$.
This is a four term polynomial, factorization of this polynomial can be find by factor by grouping as,
$$x^{12}+x^{7}+x^{5}+1 = (x^{12}+x^{7})(x^{5} +1)$$
$$=x^{7}(x^{5}+1)+1(x^{5}+1)$$
As, $$(x^{5} + 1)$$ is the common factor of the polynomial,
The polynomial can be factorized as,
$$x^{7}(x^{5}+1)+1(x^{5}+1)=(x^{5}+1)(x^{7}+1)$$
Therefore, the factorization of the polynomial $$x^{12} + x^{7} + x^{5} + 1$$ is $$(x^{5} +1) (x^{7} +1)$$.

### Relevant Questions Use the prime factorizations $$\displaystyle{a}={2}^{{4}}\times{3}^{{4}}\times{5}^{{2}}\times{7}^{{3}}{\quad\text{and}\quad}{b}={2}^{{2}}\times{3}\times{5}^{{3}}\times{11}$$ to find the prime factorization of the following.
(a) LCM(a, b)
(b) GCF(a, b) Need to calculate:The factorization of $$x^{3}+4x^{2}+3x+12$$. Solve the qudaratic equation by factorization
$$\displaystyle{x}^{{2}}+{x}-{\left({a}+{1}\right)}{\left({a}+{2}\right)}={0}$$ Does the equation $$\displaystyle{x}^{{2}}\equiv{x}\cdot{x}\equiv{2}{x}\cdot{4}{x}\text{mod}{7}$$ show that factorization of polynomials mod 7 is not unique? Why or why not? Solve. The factorization of the polynomial is $$(3x+2)(2x^{2}+1)$$.
Given Information:
The provided polynomial is $$6x^{3}+4x^{2}+3x+2$$. Complete Factorization Factor the polynomial completely, and find all its zeros.State the multiplicity of each zero.
$$P(x) = x^{5}+7x^{3}$$ Find the LU-Factorization of the matrix A below
$$\displaystyle{A}={\left[\begin{array}{ccc} {2}&{1}&-{1}\\-{2}&{0}&{3}\\{2}&{1}&-{4}\\{4}&{1}&-{4}\\{6}&{5}&-{2}\end{array}\right]}$$ $$\displaystyle{A}={\left[\begin{array}{ccc} -{4}&{0}&{4}\\{12}&{2}&-{9}\\{12}&{8}&{9}\end{array}\right]}$$ $$\displaystyle{4}{x}^{{2}}-{2}{\left({a}^{{2}}+{b}^{{2}}\right)}{x}+{a}^{{2}}{b}^{{2}}={0}$$ $$\displaystyle{x}^{{2}}-{5}{x}-{24}={0}$$