How do you list all possible roots and find all factors and zeroes of 5x3+29x2+19x−5?

1 Answer
George C.
Jun 23, 2016

5x3+29x2+19x−5=(x+1)(5x−1)(x+5)

with zeros x=−1, x=15 and x=−5

Explanation:

f(x)=5x3+29x2+19x−5

Note that if you reverse the signs on the terms of odd degree then the sum of the coefficients is 0.

That is: −5+29−19−5=0


Hence f(−1)=0 and (x+1) is a factor:

5x3+29x2+19x−5=(x+1)(5x2+24x−5)


To factor the remaining quadratic use an AC method:

Look for a pair of factors of AC=5⋅5=25 which differ by B=24.

The pair 25,1 works, in that 25×1=25 and 25−1=24.

Use this pair to split the middle term and factor by grouping:

5x2+24x−5

=5x2+25x−x−5

=(5x2+25x)−(x+5)

=5x(x+5)−1(x+5)

=(5x−1)(x+5)

Hence zeros x=15 and x=−5

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