Asked by yasmine harper on Jun 13, 2024

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Subtract 3y2−3y3 from 9y5+9y33 y ^ { 2 } - 3 y ^ { 3 } \text { from } 9 y ^ { 5 } + 9 y ^ { 3 }3y23y3 from 9y5+9y3 .

A) 6y3+126 y ^ { 3 } + 126y3+12
B) −9y5−12y3+3y2- 9 y ^ { 5 } - 12 y ^ { 3 } + 3 y ^ { 2 }9y512y3+3y2
C) −6y3−12- 6 y ^ { 3 } - 126y312
D) 9y5+12y3−3y29 y ^ { 5 } + 12 y ^ { 3 } - 3 y ^ { 2 }9y5+12y33y2
E) 6y5+12y36 y ^ { 5 } + 12 y ^ { 3 }6y5+12y3

Cubic Polynomial

A polynomial of degree three, characterized by an equation of the form ax^3 + bx^2 + cx + d, where a, b, c, and d are constants and a ≠ 0.

Quintic Polynomial

A polynomial of degree five, represented as \(ax^5 + bx^4 + cx^3 + dx^2 + ex + f\) where \(a\), \(b\), \(c\), \(d\), \(e\), and \(f\) are constants and \(a \neq 0\).

  • Attain understanding in the basic mathematical operations, specifically addition, subtraction, multiplication, and division, as they apply to polynomials.
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MR
Martin RodriguezJun 19, 2024
Final Answer :
D
Explanation :
We have to subtract 3y2−3y3{3 y ^ { 2 } - 3 y ^ { 3 }}3y23y3 from 9y5+9y3{9 y ^ { 5 } + 9 y ^ { 3 }}9y5+9y3 .
So, the difference can be found by adding the negative of the second polynomial to the first polynomial.
(9y5+9y3)−(3y2−3y3)=9y5+9y3−(3y2−3y3)=9y5+9y3−3y2+3y3=9y5+12y3−3y2.\begin{aligned} \left( 9 y ^ { 5 } + 9 y ^ { 3 } \right) - \left( 3 y ^ { 2 } - 3 y ^ { 3 } \right) &= 9 y ^ { 5 } + 9 y ^ { 3 } - \left( 3 y ^ { 2 } - 3 y ^ { 3 } \right) \\ &= 9 y ^ { 5 } + 9 y ^ { 3 } - 3 y ^ { 2 } + 3 y ^ { 3 } \\ &= \boxed{9 y ^ { 5 } + 12 y ^ { 3 } - 3 y ^ { 2 }}. \end{aligned}(9y5+9y3)(3y23y3)=9y5+9y3(3y23y3)=9y5+9y33y2+3y3=9y5+12y33y2. Therefore, the correct answer is (D).