Question

drag the operator to the correct location on the image. which operation results in a binomial? - + · (3y⁶ + 4) (9y¹² - 12y⁶ + 16)

Explanation:

Step1: Recall Polynomial Operations

We know that the formula for the sum of cubes is \(a^3 + b^3=(a + b)(a^2 - ab + b^2)\) and the formula for the difference of cubes is \(a^3 - b^3=(a - b)(a^2+ab + b^2)\). Also, multiplication of a binomial and a trinomial (here \((3y^{6}+4)\) is a binomial and \((9y^{12}-12y^{6}+16)\) is a trinomial) using the distributive property (or the sum/difference of cubes formula) will be considered. Let's check each operation:

• Addition: \((3y^{6}+4)+(9y^{12}-12y^{6}+16)=9y^{12}-9y^{6}+20\), which is a trinomial (degree 12, three terms).
• Subtraction: \((3y^{6}+4)-(9y^{12}-12y^{6}+16)= - 9y^{12}+15y^{6}-12\), which is a trinomial (degree 12, three terms).
• Multiplication: Let \(a = 3y^{6}\) and \(b = 4\). Then \(a^2=(3y^{6})^2 = 9y^{12}\), \(ab=(3y^{6})\times4 = 12y^{6}\), \(b^2 = 16\). The trinomial \(9y^{12}-12y^{6}+16\) can be written as \(a^2 - ab + b^2\) (since \(ab = 12y^{6}\), so \(-ab=- 12y^{6}\)). Then by the sum of cubes formula \(a^3 + b^3=(a + b)(a^2 - ab + b^2)\), so \((3y^{6}+4)\times(9y^{12}-12y^{6}+16)=(3y^{6})^3+4^3=27y^{18}+64\), which is a binomial (two terms: \(27y^{18}\) and \(64\)).

Step2: Conclusion

So the operation that results in a binomial is multiplication (the \(\cdot\) operator).

Answer:

The operator is \(\boldsymbol{\cdot}\) (multiplication), so we drag the \(\cdot\) operator to the location between \((3y^{6}+4)\) and \((9y^{12}-12y^{6}+16)\).