# How do you use a graph, synthetic division, and factoring to find all the roots of an equation x^3 + 5x^2 + 3x -9 = 0?

Aug 31, 2015

Graphing shows us roots $x = 1$, $x = - 3$ (repeated).

Alternatively, synthetic division and factoring gives us the same roots.

#### Explanation:

Let $f \left(x\right) = {x}^{3} + 5 {x}^{2} + 3 x - 9$

The word 'graph' put me off answering this sooner. To plot a graph I usually try to identify turning points, roots, etc first - not the other way round. But if you do plot a graph by picking a few $x$ values, then you would pretty quickly find that $x = 1$ is a root and you would probably notice that $x = - 3$ is a repeated root.

graph{x^3+5x^2+3x-9 [-10.51, 9.49, -9.56, 0.44]}

Alternatively, first note that the sum of the coefficients is $0$, implying that $x = 1$ is a root.

So $\left(x - 1\right)$ is a factor. Divide by this using synthetic division:

So ${x}^{3} + 5 {x}^{2} + 3 x - 9 = \left(x - 1\right) \left({x}^{2} + 6 x + 9\right)$

${x}^{2} + 6 x + 9$ is a perfect square trinomial, recognisable by being of the form ${A}^{2} + 2 A B + {B}^{2} = {\left(A + B\right)}^{2}$ :

${x}^{2} + 6 x + 9 = \left({x}^{2} + 2 \cdot x \cdot 3 + {3}^{2}\right) = {\left(x + 3\right)}^{2}$

Another little trick for recognising this perfect square trinomial is that $169 = {13}^{2}$ is a perfect square. So the pattern of coefficients $1$, $6$, $9$ corresponds to a square of a binomial with coefficients $1$, $3$.

So $f \left(x\right) = \left(x - 1\right) {\left(x + 3\right)}^{2}$ and $f \left(x\right) = 0$ has roots $1 , - 3 , - 3$