SOLUTION: How do i find the zeros of this equation {{{ 6x^3+5x^2-9x+2 }}}
I have tried to factor out an x and than factor it as i would with
a normal quadratic but i reached this answer
Algebra.Com
Question 979438: How do i find the zeros of this equation
I have tried to factor out an x and than factor it as i would with
a normal quadratic but i reached this answer and
and i needed to multiply the x back in but i
did not know how to or what i should do after that. also because this is a
3rd degree polynomial than shouldn't it have 3 zeros. if possible please
show me step by step how to do this , thank you so much
Found 2 solutions by KMST, Fombitz:
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
If has a rational zero,
you should be able to use it to factor the polynomial,
and find the exact value of all of its real zeros.
If does not have a rational zero,
you could only find approximate values by guided trial and error.
If has a rational zero , it must be that
is a factor of the constant term , and
is a factor of the leading coefficient .
Factors of are , and .
Factors of are , , , and .
The possible values for are
-2, -1, -2/3, -1/3, -1/6, 1/6, 1/3, 2/3, 1, and 2.
You are expected to use synthetic division
(long division works too, but uses more ink)
to find that is a zero.
(The other nine possible values are not zeros).
Synthetic division tells you
not only that is zero for ,
but also that
<--->
The other two zeros of are the zeros of .
You can find if they are real zeros, and their value by solving
.
You could do that by completing the square, or by applying
the quadratic formula ,
which gives the solutions to
In this case, with , the solutions to are
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
First step, graph the function.
.
.
.
.
.
Look like is a root.
So then,
is a factor of the cubic polynomial.
Divide it out.
Now you can use the quadratic formula to solve for the other two roots.
Verify that your solutions match the solutions in the graph.
If there wasn't an obvious root, like we had at , the next option was to use the cubic formula similar to the quadratic formula but much more complicated.
You can Google it to see.
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