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#?

1 Answer
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(x) = x^3+5x^2+3x-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 #(x-1)# is a factor. Divide by this using synthetic division:

enter image source here

So #x^3+5x^2+3x-9 = (x-1)(x^2+6x+9)#

#x^2+6x+9# is a perfect square trinomial, recognisable by being of the form #A^2+2AB+B^2 = (A+B)^2# :

#x^2+6x+9 = (x^2+2*x*3+3^2) = (x+3)^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(x) = (x-1)(x+3)^2# and #f(x) = 0# has roots #1, -3, -3#