Algebra II...Rational Zero Theorem

[aives]

Dear Tutor,
We are having a great deal of difficulty in understanding the "roots and zeros", as well as the "rational zero theorem" of polynomial functions.
Can you explain these terms in low-level english, and help us to figure out how to approach these problems?

1. Identify Possible Zeros ex. f(x) = 2x3 - 11x2 + 12x +9 (numbers behind the x are exponents)

2. Find all of the zeros of f(x)= 2x4 -13x3 + 23x2 -52x + 60

Help! Thank you so much

 

[galactus]

Why not just use the ^ symbol for exponents?.

Is this it?:

\(\displaystyle 2x^{3}-11x^{2}+12x+9=0\)

\(\displaystyle 2x^{4}-13x^{3}+23x^{2}-52x+60=0\)

If so, the first one can be factored.

Rewrite:

\(\displaystyle 2x^{3}-12x^{2}+18x+x^{2}-6x+9\)

Group:

\(\displaystyle 2x(x^{2}-6x+9)+(x^{2}-6x+9)=0\)

\(\displaystyle (2x+1)(x^{2}-6x+9)\)

The quadratic also factors:

\(\displaystyle (2x+1)(x-3)^{2}\)

To use the rational root theorem, try checking possible roots. Try x=-1/2 or x=3.

When you divide by the correct root, the cubic will reduce to a quadratic or a linear. Then you know you're on the right track.

 

[stapel]

We are having a great deal of difficulty in understanding the "roots and zeros"....

"Zeroes" are values which, when plugged in for the variable, make the expression equal to zero. They are "solutions", in that they solve "(expression) = 0". "Roots" are just another term for "zeroes", though "real zeroes" mean the zeroes which, being real numbers (as opposed to imaginary numbers), show up as x-intercepts on the graph of the related function.

There are many great lessons available online:

. . . . .Google results for "roots zeroes"

. . . . .Google results for "roots zeros" (note American spelling)

. . . . .Google results for "x intercepts roots"

...as well as the "rational zero theorem" of polynomial functions. Can you explain these terms in low-level english...

Try some of these lessons:

. . . . .Google results for "rational roots test"

If none of the lessons on the first page or so of the list is helpful,then it might be useful if you replied with your understanding so far, so we can try to figure out where things are going wrong for you...?

Thank you!

Eliz.

 

[masters]

If your leading coefficient is "a[sub:51zwxu8q]0[/sub:51zwxu8q] and your constant is a[sub:51zwxu8q]n[/sub:51zwxu8q], and p/q is a rational root of y = f(x), then p is a factor of a[sub:51zwxu8q]n[/sub:51zwxu8q], and q is a factor of a[sub:51zwxu8q]0[/sub:51zwxu8q].
In your second equation, a[sub:51zwxu8q]0[/sub:51zwxu8q] = 2, and a[sub:51zwxu8q]n[/sub:51zwxu8q] = 60

p/q represents all the possible combinations of factors of 60 divided by all possible factors of 2. There's a lot of 'em.
+-1, +-2, +-3, +-4, +-5, +-6, +-10, +-15, +-30, +-60, +-1/2, +-1/3, +-1/4, etc. Get the idea?

Try 5 as a possible zero. The depressed polynomial equation would then be 2x[sup:51zwxu8q]3[/sup:51zwxu8q] - 3x[sup:51zwxu8q]2[/sup:51zwxu8q] +8x -12 = 0
Now you have fewer possible rational zeros to check. Try 3/2.

Good luck.

If you know Descartes' rule of signs, you would know that there will be 4, 2, or 0 positive real roots. There will be 0 negative real roots, so don't test any of them. You could have 4, 2, or 0 imaginary roots as well.

But, in this instance, you have 2 real rational roots and 2 imaginary ones {3/2, 5, +-2i}

 

[masters]

BTW, a graphing calculator will really cut down on the number of possible rational roots you need to try. But, of course, with a graphing calculator, who needs the rational roots theorem anyway!