Q. One root of \(f(x)=2x^3+9x^2+7x-6\) is \(-3\). How to find the factors of the polynomial.

Problem: Given that \(x=-3\) is a root of the polynomial \(f(x)=2x^{3}+9x^{2}+7x-6\), find all the factors of the polynomial.

Step 1 – Identify the first linear factor

By the Factor Theorem, since \(x=-3\) is a root, \(x-(-3)=x+3\) is a factor. Thus \(x+3\) is a factor.

Step 2 – Use synthetic division to find the depressed polynomial

Use the root \(-3\) and the coefficients \(2,\,9,\,7,\,-6\). The synthetic-division table is:

\[
\begin{array}{r|rrrr}
-3 & 2 & 9 & 7 & -6\\\hline
& & -6 & -9 & 6\\
& 2 & 3 & -2 & 0
\end{array}
\]

The remainder is \(0\), confirming \(x+3\) is a factor. The quotient coefficients \(2,\,3,\,-2\) give the quadratic
\[
Q(x)=2x^{2}+3x-2.
\]

Step 3 – Factor the quadratic result

Factor \(2x^{2}+3x-2\). Look for two numbers that multiply to \(2\cdot(-2)=-4\) and sum to \(3\); these are \(4\) and \(-1\). Rewrite and group:

\[
2x^{2}+3x-2 = 2x^{2}+4x – x – 2 = 2x(x+2)-1(x+2) = (2x-1)(x+2).
\]

Answer

Combining the factors,
\[
2x^{3}+9x^{2}+7x-6 = (x+3)(2x-1)(x+2).
\]