Question

What is the factored form of x^12y^18+1

Answer 1

The factorized form of the equation ${x^{12}}{y^{18}} + 1$ is $\boxed{\left( {{x^4}{y^6} + 1} \right)\left( {{x^8}{y^{12}} - {x^4}{y^6} + 1} \right)}.$

Further Explanation:

The rules of exponents are as follows,

1.$\boxed{\left( {{x^m}} \right) \times \left( {{x^n}} \right) = {x^{m + n}}}$

2. $\boxed{\frac{{{x^m}}}{{{x^n}}} = {x^{m - n}}}$

3. $\boxed{{{\left( {{x^a}} \right)}^b} = {x^{a \times b}}}$

4. $\boxed{{x^{\frac{m}{n}}} = \sqrt[n]{{{x^m}}}}$

Given:

The polynomial function is ${x^{12}}{y^{18}} + 1.$

Calculation:

The cubic formula can be expressed as follows,

$\boxed{{a^3} + {b^3}=\left( {a + b} \right)\left( {{a^2} - ab + {b^2}}\right)}$

The given polynomial function is ${x^{12}}{y^{18}} + 1.$

$\begin{aligned}P\left( x \right) &= {x^{12}}{y^{18}} + 1\\&= {\left( {{x^4}{y^6}} \right)^3} + {\left( 1 \right)^3}\\\end{aligned}$

Use the identity ${a^3} + {b^3}=\left( {a + b}\right)\left( {{a^2} - ab + {b^2}}\right)$ in above expression.

$\begin{aligned}P\left( x \right)&= \left( {{x^4}{y^6} + 1}\right)\left[ {{{\left( {{x^4}{y^6}} \right)}^2} - {x^4}{y^6} \times 1 + {1^2}} \right]\\&= \left( {{x^4}{y^6} + 1} \right)\left( {{x^8}{y^{12}} - {x^4}{y^6} + 1} \right) \\\end{aligned}$

The factorized form of the equation ${x^{12}}{y^{18}} + 1$ is $\boxed{\left({{x^4}{y^6} + 1}\right)\left({{x^8}{y^{12}} - {x^4}{y^6} + 1}\right)}.$

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Answer details:

Grade: High School

Subject: Mathematics

Chapter: Exponents and Powers

Keywords: Solution, factorized form, $x^12y^18+1$, exponents, power, equation, power rule, exponent rule.

Answer 2

The factorized form of the expression is the simplest form of the expression which can be obtained by using the correct identity and by using the identity the factorized form of the given expression can be written as :

$f(x)=(x^4y^6+1)(x^8y^{12}-x^4y^6+1]$

Step-by-step explanation:

Given information:

The expression $x^{12}y^{18}+1$

To get the factored form of the expression

We have:

$a^3+b^3=(a+b)(a^2-ab+b^2)$

on converting the given expression to the above format

$f(x)=x^{12}y^{18}+1$

using the identity

$f(x)=(x^4y^6)^3+1^3\\f(x)=(x^4y^6+1)[(x^4y^6)^2-x^4y^6\times 1+1^2]\\f(x)=(x^4y^6+1)(x^8y^{12}-x^4y^6+1]$

Hence the factored form of the expression is :

$f(x)=(x^4y^6+1)(x^8y^{12}-x^4y^6+1]$

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