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ANSWER
[tex]6y^6-15y^5-9y^4[/tex]EXPLANATION
We want to find the product of:
[tex]3y^4(2y\text{ + 1)(y - 3)}[/tex]To do this, let us first expand the two brackets. We do that by using the components in the first bracket to multiply the second:
[tex]\begin{gathered} 3y^4\lbrack(2y\cdot\text{ y) - (2y }\cdot\text{ 3) + (1 }\cdot\text{ y) + (1 }\cdot\text{ -3)\rbrack} \\ =\text{ }3y^4(2y^2\text{ - 6y + y - 3)} \end{gathered}[/tex]Now, expand this bracket by multiplying the term outside the bracket with those in it:
[tex]\begin{gathered} (3y^4\cdot2y^2)\text{ - (}3y^4\cdot\text{ 6y) + (}3y^4\cdot\text{ y) - }(3y^4\cdot\text{ 3)} \\ =\text{ }6y^{4\text{ + 2}}-18y^{4\text{ + 1}}+3y^{4\text{ + 1}}-9y^{4\text{ }} \\ =6y^6-18y^5+3y^5-9y^4 \\ =6y^6-15y^5-9y^4 \end{gathered}[/tex]That is the answer.
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