SAT Mathematics Practice Test 2

Since $2\mathit{x}\left(3\mathit{x}+5\right)+3\left(3\mathit{x}+5\right)=\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}$ for all values of $\mathit{x}$, the two sides of the equation are equal, and the value of b can be determined by simplifying the left-hand side of the equation and writing it in the same form as the right-hand side. Using the distributive property, the equation becomes $\left(6{\mathit{x}}^2+10\mathit{x}\right)+\left(9\mathit{x}+15\right)=\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}.$ Combining like terms gives $6{\mathit{x}}^2+19\mathit{x}+15=\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}$ The value of $\mathit{b}$ is the coefficient of $\mathit{x}$, which is $19$.