Question

rewrite without parentheses.
$-7x^{2}y^{4}(9y^{5}-6x + 8)$
simplify your answer as much as possible.

Explanation:

Step1: Distribute the monomial

We use the distributive property (also known as the distributive law of multiplication over addition/subtraction), which states that \(a(b + c)=ab+ac\) and \(a(b - c)=ab - ac\). Here, \(a=-7x^{2}y^{4}\), \(b = 9y^{5}\), \(c=-6x\), and \(d = 8\) (so we have \(a(b + c + d)=ab+ac+ad\) in this case with three terms inside the parentheses).
First, multiply \(-7x^{2}y^{4}\) by \(9y^{5}\):
\[
-7x^{2}y^{4}\times9y^{5}=(-7\times9)x^{2}(y^{4}\times y^{5})
\]
Using the rule of exponents \(y^{m}\times y^{n}=y^{m + n}\), we get:
\[
(-7\times9)x^{2}y^{4 + 5}=-63x^{2}y^{9}
\]
Next, multiply \(-7x^{2}y^{4}\) by \(-6x\):
\[
-7x^{2}y^{4}\times(-6x)=(-7\times(-6))(x^{2}\times x)y^{4}
\]
Using the rule of exponents \(x^{m}\times x^{n}=x^{m + n}\), we have:
\[
42x^{2+1}y^{4}=42x^{3}y^{4}
\]
Then, multiply \(-7x^{2}y^{4}\) by \(8\):
\[
-7x^{2}y^{4}\times8=(-7\times8)x^{2}y^{4}=-56x^{2}y^{4}
\]

Step2: Combine the results

Now, we combine the three products we got from the distribution:
\[
-63x^{2}y^{9}+42x^{3}y^{4}-56x^{2}y^{4}
\]
Since there are no like terms (terms with the same variables raised to the same powers) among \(-63x^{2}y^{9}\), \(42x^{3}y^{4}\), and \(-56x^{2}y^{4}\), this is the simplified form.

Answer:

\(-63x^{2}y^{9}+42x^{3}y^{4}-56x^{2}y^{4}\)