Asked by Megan Byers on Jul 06, 2024
Verified
Expand (x−3) 3( x - 3 ) ^ { 3 }(x−3) 3 and simplify.
A) x3−27x ^ { 3 } - 27x3−27
B) x3−9x2+27x−27x ^ { 3 } - 9 x ^ { 2 } + 27 x - 27x3−9x2+27x−27
C) x3+27x2−9x−27x ^ { 3 } + 27 x ^ { 2 } - 9 x - 27x3+27x2−9x−27
D) x3−27x2+9x−27x ^ { 3 } - 27 x ^ { 2 } + 9 x - 27x3−27x2+9x−27
E) x3−3x2+9x−27x ^ { 3 } - 3 x ^ { 2 } + 9 x - 27x3−3x2+9x−27
Expand
To simplify an expression or equation by multiplying out brackets and combining like terms.
- Harness the patterns of special products for the reduction of complexity in polynomial expressions.
Verified Answer
ZK
Zybrea Knight
Jul 07, 2024
Final Answer :
B
Explanation :
To expand (x−3)3( x - 3 ) ^ { 3 }(x−3)3 using the binomial theorem, we need to use the coefficients of the terms in the expansion of (x−3)3( x - 3 ) ^ { 3 }(x−3)3 . The coefficients are the same as in the expansion of (a+b)3( a + b ) ^ { 3 }(a+b)3 , where a=xa = xa=x and b=−3b = -3b=−3 . Using the binomial coefficients or Pascal's triangle, we find that the coefficients of the terms in the expansion of (x−3)3( x - 3 ) ^ { 3 }(x−3)3 are 1, -3, 3, and -1, in order. Therefore, we have
(x−3)3=1⋅x3+(−3)⋅x2⋅3+3⋅x⋅(−3)2+(−1)⋅33=x3−9x2+27x−27.( x - 3 ) ^ { 3 } = 1 \cdot x ^ { 3 } + ( -3 ) \cdot x ^ { 2 } \cdot 3 + 3 \cdot x \cdot ( -3 ) ^ { 2 } + ( -1 ) \cdot 3 ^ { 3 } = x^{3}-9x^{2}+27x-27.(x−3)3=1⋅x3+(−3)⋅x2⋅3+3⋅x⋅(−3)2+(−1)⋅33=x3−9x2+27x−27.
(x−3)3=1⋅x3+(−3)⋅x2⋅3+3⋅x⋅(−3)2+(−1)⋅33=x3−9x2+27x−27.( x - 3 ) ^ { 3 } = 1 \cdot x ^ { 3 } + ( -3 ) \cdot x ^ { 2 } \cdot 3 + 3 \cdot x \cdot ( -3 ) ^ { 2 } + ( -1 ) \cdot 3 ^ { 3 } = x^{3}-9x^{2}+27x-27.(x−3)3=1⋅x3+(−3)⋅x2⋅3+3⋅x⋅(−3)2+(−1)⋅33=x3−9x2+27x−27.
Learning Objectives
- Harness the patterns of special products for the reduction of complexity in polynomial expressions.