How do you find all the zeros of f(x)=x3−8x2−23x+30?

1 Answer
Burglar
Jun 20, 2016

They are x=1,x=10,x=−3.

Explanation:

To solve a cubic equation it exists the solving formula but it is very long and I do not know it.
Then I use a bit of luck to solve this equation. For example I see that x=1 is a solution because

13−8⋅12−23⋅2+30=1−8−23+30=0.

Then I know that the equation has to be in the form

(ax2+bx+c)(x−1)

To find a,b,c I just do the multiplication and compare the coefficients.

(ax2+bx+c)(x−1)

=ax3+bx2+cx−ax2−bx−c

=ax3+(b−a)x2+(c−b)x−c

Comparing this with our initial equation we see that

a=1
b−a=−8,b−1=−8,b=−7
c−b=−23,c+7=−23,c=−30
−c=30,c=−30

Our function is then

(x2−7x−30)(x−1).

Now we have to solve a second order equation (x2−7x−30) using the solution

x=−b±√b2−4ac2a

=−(−7)±√(−7)2−4⋅1⋅(−30)2

=7±√49+1202

=7±√1692

=7±132

that has the two solutions

x=7+132=10 and x=7−132=−3.

Then the function

f(x)=x3−8x2−23x+30

can be rewritten as

f(x)=(x−1)(x−10)(x+3) with the three zeros for x equal to 1,10,−3.

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