Solve f(v)=(v+3)^2(v-1)^3

$$f ( v ) = ( v + 3 ) ^ { 2 } ( v - 1 ) ^ { 3 }$$

Evaluate

$$\left(v+3\right)^{2}\left(v-1\right)^{3}$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(v+3\right)^{2}.

$$\left(v^{2}+6v+9\right)\left(v-1\right)^{3}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(v-1\right)^{3}.

$$\left(v^{2}+6v+9\right)\left(v^{3}-3v^{2}+3v-1\right)$$

Use the distributive property to multiply v^{2}+6v+9 by v^{3}-3v^{2}+3v-1 and combine like terms.

$$v^{5}+3v^{4}-6v^{3}-10v^{2}+21v-9$$

Expand

$$v^{5}+3v^{4}-6v^{3}-10v^{2}+21v-9$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(v+3\right)^{2}.

$$\left(v^{2}+6v+9\right)\left(v-1\right)^{3}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(v-1\right)^{3}.

$$\left(v^{2}+6v+9\right)\left(v^{3}-3v^{2}+3v-1\right)$$

Use the distributive property to multiply v^{2}+6v+9 by v^{3}-3v^{2}+3v-1 and combine like terms.

$$v^{5}+3v^{4}-6v^{3}-10v^{2}+21v-9$$