First we use the distributive law, multiplying each term in the second factor by
\(2x+1\text{:}\) \((2x+1)(x^2-3x+5)=(2x+1)(x^2)+(2x+1)(-3x)+(2x+1)(5)\text{.}\)
An obtional step, but one that makes the distributive law more evident, is to us the commutative law of multiplication to reverse the order of the factors in each term:
\((2x+1)(x^2)+(2x+1)(-3x)+(2x+1)(5)=x^2(2x+1)+(-3x)(2x+1)+5(2x+1).\)
We can then use the distributive law on each term to get
\(x^2(2x+1)+(-3x)(2x+1)+5(2x+1)=x^2(2x)+x^2(1)-3x(2x)-3x(1)+5(2x)+5(1)
2x^3+x^2-6x^2-3x+10x+5 \)
Finally, we use the distributive law to combine like terms:
\(2x^3+x^2-6x^2-3x+10x+5 =2x^3+(1-6)x^2+(10-3)x+5
=2x^3-5x^2+7x+5.\)
Although the distributive law makes the process work, we typically speed the process up by multiplying each term in the first factor by each term in the second factor:
\((2x+1)(x^2-3x+5)=2x(x^2)+2x(-3x)+2x(5)+1(x^2)+1(-3x)+1(5)
=2x^3-6x^2+10x+x^2-3x+5
=2x^3-5x^2+7x+5.\)