Respuesta :
Given:
[tex](v^2+10v+11)(v^2+3v-4)[/tex]To find- the simplification.
Explanation-
We know that the distribution property of multiplication over addition says
[tex]a(b+c)=ab+ac[/tex]Use this property to simplify, and we get
[tex]\begin{gathered} =(v^2+10v+11)(v^2+3v-4) \\ =v^2(v^2+3v-4)+10v(v^2+3v-4)+11(v^2+3v-4) \end{gathered}[/tex]Multiply by opening the bracket, and we get
[tex]=(v^4+3v^3-4v^2)+(10v^3+30v^2-40v)+(11v^2+33v-44)[/tex]Now, open the bracket and combine the like terms.
[tex]\begin{gathered} =v^4+3v^3-4v^2+10v^3+30v^2-40v+11v^2+33v-44 \\ =v^4+(3v^3+10v^3)+(11v^2-4v^2+30v^2)-40v+33v-44 \end{gathered}[/tex]On further solving, we get
[tex]=v^4+13v^3+37v^2-7v-44[/tex]Thus, from the distributive property of multiplication over addition, we get v⁴+13v³+37v²-7v-44.
The answer is v⁴ + 13v³ + 37v² - 7v - 44.
