Question

Multiply the monomials :

Answer 1

here we want to multiply different monomials, the particular important thing about this problem is learning how to simplify these multiplications.

Here we need to remember the property:

$a^n*a^m = a^{n + m}$

Now we want to directly multiply the monomials:

a) a³ , -6 a² b and 2 b³

Here we just have:

$a^3*(-6*a^2*b)*(2*b^3)$

Let's reorder that, so we can separate the variables and the coefficients, so then we can use the above property to solve this:

$a^3*(-6*a^2*b)*(2*b^3) = (-6*2)*(a^3*a^2)*(b*b^3) = -12*a^5*b^4$

b) 16 x⁶ , -10 xy² and 3/5 x² y²

Like before, we want to simplify:

$(16*x^6)*(-10*x*y^2)*(\frac{3}{5}*x^2*y^2)$

Again, reordering that we get:

$(16*\frac{3}{5}*-10)*(x^6*x*x^2)*(y^2*y^2)$

Now solving each parenthesis:

$(16*\frac{3}{5}*-10)*(x^6*x*x^2)*(y^2*y^2) = -96*x^{6 +1 + 2}*y^{2 + 2}$

Simplifying the last expression we get:

$-96*x^{6 +1 + 2}*y^{2 + 2} = -96*x^9*y^4$

c)  -4 p² q² and 3/8 pq²

Here we want to simplify:

$(-4*p^2*q^2)*(\frac{3}{8}*p*q^2)$

Using the same approach as before, we will get:

$(-4*p^2*q^2)*(\frac{3}{8}*p*q^2) = (-4*\frac{3}{8})*(p^2*p)*(q^2*q^2) = \frac{-3}{2}*p^3*q^4$

If you want to learn more, you can read:

brainly.com/question/17825040