Expand (3-x)(2+x)^3 by using binomial expansion
To expand the expression (3-x)(2+x)^3 using binomial expansion, we will use the binomial theorem. The binomial theorem states that for any two real numbers a and b and any positive integer n, the expansion of (a+b)^n is given by:
(a+b)^n = C(n,0)*a^n*b^0 + C(n,1)*a^(n-1)*b^1 + C(n,2)*a^(n-2)*b^2 + ... + C(n,n-1)*a^1*b^(n-1) + C(n,n)*a^0*b^n
where C(n,k) is the binomial coefficient, which represents the number of ways to choose k items out of n distinct items. It can be calculated using the formula: C(n,k) = n! / (k!(n-k)!)
In our case, a = 3-x, b = 2+x, and n = 3.
Let's expand the expression step by step:
First term: C(3,0)*(3-x)^3*(2+x)^0 = 1*(3-x)^3*(2+x)^0 = (3-x)^3
Second term: C(3,1)*(3-x)^2*(2+x)^1 = 3*(3-x)^2*(2+x)^1 = 3*(9-6x+x^2)*(2+x)
Third term: C(3,2)*(3-x)^1*(2+x)^2 = 3*(3-x)^1*(2+x)^2 = 3*(3-x)*(4+4x+x^2)
Fourth term: C(3,3)*(3-x)^0*(2+x)^3 = 1*(3-x)^0*(2+x)^3 = (2+x)^3
Expanding the terms further:
(3-x)^3 = (27 - 27x + 9x^2 - x^3)
3*(9-6x+x^2)*(2+x) = (54 - 36x + 6x^2 + 27x - 18x^2 + 3x^3)
3*(3-x)*(4+4x+x^2) = (9 - 9x + 12x - 12x^2 + 3x^2 - 3x^3)
(2+x)^3 = (8 + 12x + 6x^2 + x^3)
Finally, adding all the terms together:
(27 - 27x + 9x^2 - x^3) + (54 - 36x + 6x^2 + 27x - 18x^2 + 3x^3) + (9 - 9x + 12x - 12x^2 + 3x^2 - 3x^3) + (8 + 12x + 6x^2 + x^3)
Simplifying the expression:
27 - 27x + 9x^2 - x^3 + 54 - 36x + 6x^2 + 27x - 18x^2 + 3x^3 + 9 - 9x + 12x - 12x^2 + 3x^2 - 3x^3 + 8 + 12x + 6x^2 + x^3
Combining like terms:
(27 + 54 + 9 + 8) + (-27x - 36x + 27x - 9x + 12x) + (9x^2 - 18x^2 + 6x^2 - 12x^2) + (-x^3 + 3x^3 - 3x^3 + x^3)
Simplifying further:
98 - 33x + (-9x + 9x + 12x) + (3x^2 - 18x^2 - 12x^2 + 6x^2) + 0
Final expansion result:
98 - 33x + 12x + 0x^2 - 3x^2 + 0x^3
Simplifying:
98 - 21x - 3x^2
Therefore, the expanded form of (3-x)(2+x)^3 using binomial expansion is 98 - 21x - 3x^2.