Question

Find the product. Multiplication of the binomial.

Answer 1

We have, $(2r+3)(3r^2+4r+5)$ or more generally,

$(a+b)(c+d+e)=ac+ad+ae+bc+bd+be$

So using that, we find,

$(2r+3)(3r^2+4r+5)=6r^3+8r^2+10r+9r^2+12r+15$, then collect terms with same exponent and get $\boxed{6r^3+17r^2+22r+15}$.