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
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 + ...
To expand and simplify the first three terms of each binomial power, you need to apply the distributive property. Distribute each term from the first binomial to each term in the second binomial, and then combine like terms if possible. Let's go through each example:
a. (2z^3 - 3y^2)*5
To expand it, multiply each term in the first binomial by each term in the second binomial:
2z^3 * 5 = 10z^3
- 3y^2 * 5 = - 15y^2
So, the expanded form is: 10z^3 - 15y^2.
b. (3b^2 - 2/b)*14
To expand it, multiply each term in the first binomial by each term in the second binomial:
3b^2 * 14 = 42b^2
- 2/b * 14 = - 28/b
So, the expanded form is: 42b^2 - 28/b.
c. (5x^3 + 3y^2)*8
To expand it, multiply each term in the first binomial by each term in the second binomial:
5x^3 * 8 = 40x^3
3y^2 * 8 = 24y^2
So, the expanded form is: 40x^3 + 24y^2.
d. (√a + √5)*10
To expand it, multiply each term in the first binomial by each term in the second binomial:
√a * 10 = 10√a
√5 * 10 = 10√5
So, the expanded form is: 10√a + 10√5.
Now you have the expanded forms of the first three terms for each binomial power.