Question

what is the product?\\((-6a^{3}b + 2ab^{2})(5a^{2}-2ab^{2}-b)\\)\\(\bigcirc -30a^{6}b + 12a^{3}b^{2}+6a^{3}b + 10a^{2}b^{2}-4ab^{4}-2ab^{2}\\)\\(\bigcirc -30a^{5}b + 12a^{4}b^{3}+16a^{3}b^{2}-4a^{2}b^{4}-2ab^{3}\\)\\(\bigcirc 30a^{5}b - 12a^{4}b^{3}+4a^{3}b^{2}-4a^{2}b^{4}-2ab^{3}\\)\\(\bigcirc 30a^{6}b - 12a^{3}b^{2}-6a^{3}b + 10a^{2}b^{2}-4ab^{4}-2ab^{2}\\)

Explanation:

Step1: Apply distributive property (FOIL for polynomials)

Multiply each term in the first polynomial by each term in the second polynomial:

• Multiply \(-6a^{3}b\) with each term in \(5a^{2}-2ab^{2}-b\):
• \(-6a^{3}b\times5a^{2}=-30a^{3 + 2}b=-30a^{5}b\)
• \(-6a^{3}b\times(-2ab^{2}) = 12a^{3+1}b^{1 + 2}=12a^{4}b^{3}\)
• \(-6a^{3}b\times(-b)=6a^{3}b^{1+1}=6a^{3}b^{2}\)
• Multiply \(2ab^{2}\) with each term in \(5a^{2}-2ab^{2}-b\):
• \(2ab^{2}\times5a^{2}=10a^{1+2}b^{2}=10a^{3}b^{2}\)
• \(2ab^{2}\times(-2ab^{2})=-4a^{1 + 1}b^{2+2}=-4a^{2}b^{4}\)
• \(2ab^{2}\times(-b)=-2a b^{2+1}=-2ab^{3}\)

Step2: Combine like terms

Now, combine the terms obtained:

• For \(a^{5}b\) term: \(-30a^{5}b\) (only one term)
• For \(a^{4}b^{3}\) term: \(12a^{4}b^{3}\) (only one term)
• For \(a^{3}b^{2}\) terms: \(6a^{3}b^{2}+10a^{3}b^{2}=16a^{3}b^{2}\)
• For \(a^{2}b^{4}\) term: \(-4a^{2}b^{4}\) (only one term)
• For \(ab^{3}\) term: \(-2ab^{3}\) (only one term)

Putting it all together: \(-30a^{5}b + 12a^{4}b^{3}+16a^{3}b^{2}-4a^{2}b^{4}-2ab^{3}\)

Answer:

\(\boldsymbol{-30a^{5}b + 12a^{4}b^{3}+16a^{3}b^{2}-4a^{2}b^{4}-2ab^{3}}\) (which corresponds to the second option: \(-30a^{5}b + 12a^{4}b^{3}+16a^{3}b^{2}-4a^{2}b^{4}-2ab^{3}\))