Question

Factorize $6x^2+17x+5$ by using the factor theorem.

Answer 1

Step 1: Solve for the factors of the given polynomial

$$\begin{gathered} f\left(x\right)=6x^2+17x+5 \\ f\left(x\right)==6\left[x^2+\frac{17x}{6}+\frac{5}{6}\right] \end{gathered}$$

Factors of $\frac{5}{6}$are $\left(\pm 1,\pm \frac{5}{6}\right),\left(\pm \frac{1}{3},\pm \frac{5}{2}\right),.......$

Step 2: Solve for one of the roots of the polynomial

put $x=-1$

$$\begin{gathered} f(-1)=6\left[{(-1)}^2+\frac{17}{6}(-1)+\frac{5}{6}\right] \\ =6\left[1-\frac{17}{6}+\frac{5}{6}\right] \\ =6\left[-1\right] \\ \Rightarrow f(-1)\ne 0 \end{gathered}$$

$\therefore x=-1$ is not a root for the polynomial

put $x=\frac{-1}{3}$

$$\begin{gathered} f\left(\frac{-1}{3}\right)=6{\left(\frac{-1}{3}\right)}^2+17\left(\frac{-1}{3}\right)+5 \\ =\frac{2}{3}-\frac{17}{3}+5 \\ =\frac{-15}{3}+5 \\ =-5+5 \\ \Rightarrow f\left(\frac{-1}{3}\right)=0 \end{gathered}$$

$\Rightarrow x=\frac{-1}{3}$ is a root of the equation

Since the degree of the given polynomial is $2$ there will be two roots for the given polynomial for which $x=\frac{-1}{3}$ is one of the roots.

$\therefore \left(3x+1\right)$ is a factor of the given polynomial

Step 3: Solve for another factor

put $x=\frac{-5}{2}$

$$\begin{gathered} f\left(\frac{-5}{2}\right)=6{\left(\frac{-5}{2}\right)}^2+17\left(\frac{-5}{2}\right)+5 \\ =\frac{75}{2}-\frac{85}{2}+5 \\ =\frac{-10}{2}+5 \\ =-5+5 \\ \Rightarrow f\left(\frac{-5}{2}\right)=0 \end{gathered}$$

$\Rightarrow x=\frac{-5}{2}$ is a root of the equation

$\therefore \left(2x+5\right)$ is a factor of the given polynomial

Hence the factorized form of the polynomial $6x^2+17x+5$ is $(3x+1)$$(2x+5)$.