Question

Factor the quadratic expression $2b^2+17b+21$ completely.

Answer 1

Factor the quadratic polynomial $\mathbf{2}{\mathit{b}}^{\mathbf{2}}\mathbf{+}\mathbf{17}\mathit{b}\mathbf{+}\mathbf{21}$:

Find the product of the quadratic term $2b^2$ and the constant term $21$, $2b^2\times 21=42b^2$ .

Through a hit and trial process, we have to find two factors of $42b^2$ that add up to give the linear term $17b$.

The two factors that satisfy these conditions are $14b$ and $3b$, because their sum is equal to the linear term $17b$ and their product is equal to $42b^2$.

Rewrite $17b$ in the expression $2b^2+17b+21$ as $14b+3b$.

Then factorize by grouping the terms:

$\begin{array}{rcl}2b^2+17b+21& =& 2b^2+14b+3b+21\\ & =& \left(2b^2+14b\right)+\left(3b+21\right)\\ & =& 2b\left(b+7\right)+3\left(b+7\right)\\ & =& \left(b+7\right)\left(2b+3\right)\end{array}$

Hence, $2b^2+17b+21$ is completely factorized to obtain $\mathbf{(}\mathit{b}\mathbf{+}\mathbf{7}\mathbf{)}\mathbf{(}\mathbf{2}\mathit{b}\mathbf{+}\mathbf{3}\mathbf{)}$.