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 $\boxed{\left({{x^4}{y^6} + 1}\right)\left({{x^8}{y^{12}} - {x^4}{y^6} + 1}\right)}.$

Learn more:

1. Learn more about unit conversion brainly.com/question/4837736

2. Learn more about non-collinear brainly.com/question/4165000

3. Learn more aboutbinomial and trinomial brainly.com/question/1394854

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

Answer:  The required factored form of the given expression is

$(x^4y^6+1)(x^8y^{12}-x^4y^6+1).$

Step-by-step explanation:  We are given to find the factored form of the following algebraic expression:

$E=x^{12}y^{18}+1.$

We will be using the following formula:

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

Now, we have

$E\\\\=x^{12}y^{18}+1\\\\=(x^4y^6)^3+1^3\\\\=(x^4y^6+1)\{(x^4y^6)^2-x^4y^6\times1+1^2\}\\\\=(x^4y^6+1)(x^8y^{12}-x^4y^6+1).$

Thus, the required factored form of the given expression is

$(x^4y^6+1)(x^8y^{12}-x^4y^6+1).$