Asked by cathy
expand a simplify the first three terms of each binomial power
a. (2z*3 - 3y*2)*5
b. (3b*2 - 2/b)*14
c. (5x*3 + 3y*2)*8
d. (sqrt.a + sqrt.5)*10
sqrt= square root
a. (2z*3 - 3y*2)*5
b. (3b*2 - 2/b)*14
c. (5x*3 + 3y*2)*8
d. (sqrt.a + sqrt.5)*10
sqrt= square root
Answers
Answered by
Reiny
I will assume that your * means exponent
In most typing notations we use the ^ to show exponents
so your first would be
(2z^3 - 3y^2)^5
you will have to know the expansion for
(a+b)^n = a^n + C(n,1) (a)^(n-1) b + C(n,2) a^(n-2) b^2 + ...
I will do the 2nd and last, you do the others
#2
(3b^2 - 2/b)^14
= (3b^2)^14 + C(14,1) (3b^2)^13 (-2/b) + C(14,2) (3b^2)^12 (-2/b)^2 + ....
= 4782969b^28 + 14(1594323 b^26 (-2/b) + 91(531441 b^24 (4/b^2) + ..
= 4782969 b^28 - 44641044 b^25 + 193444524 b^22 + ...
check my arithmetic
d) (√a + √5)^10
= (√a)^10 + 10(√a)^9 √b + 45(√a)^8 (√b)^2 + ...
= a^5 + 10a^4 √a√b + 45a^4 b + ...
In most typing notations we use the ^ to show exponents
so your first would be
(2z^3 - 3y^2)^5
you will have to know the expansion for
(a+b)^n = a^n + C(n,1) (a)^(n-1) b + C(n,2) a^(n-2) b^2 + ...
I will do the 2nd and last, you do the others
#2
(3b^2 - 2/b)^14
= (3b^2)^14 + C(14,1) (3b^2)^13 (-2/b) + C(14,2) (3b^2)^12 (-2/b)^2 + ....
= 4782969b^28 + 14(1594323 b^26 (-2/b) + 91(531441 b^24 (4/b^2) + ..
= 4782969 b^28 - 44641044 b^25 + 193444524 b^22 + ...
check my arithmetic
d) (√a + √5)^10
= (√a)^10 + 10(√a)^9 √b + 45(√a)^8 (√b)^2 + ...
= a^5 + 10a^4 √a√b + 45a^4 b + ...
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