Friday Quiz
These are some homework problems that Tricia Colclaser gave her Honors PreCalculus students this week. She said they are tougher than anything they might find in their text books.
"Practicing on harder problems gives you a chance to really put all of your skills to work," the hand out says. Some of the skills that might come in handy are Descartes Rule of Signs, Boundedness, and Synthetic division.
Factor each polynomial. (Students were also asked to sketch graphs  this part is optional.)
Hint: Find all possible rational zeros and test using synthetic division.
1. f(x) = x^7  x^6  x^5  x^4 + 12x^3 + 12x^2
2. g(x) = 4x^7  16x^6 + 7x^5 + 4x^4 +92x^3  56x^2  288
3. h(x) = x^6 + 2x^5 + 2x^4 + 2x^3  x^2  4x  2
By
Michael Alison Chandler

March 4, 2009; 3:14 PM ET
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I hadn't heard of synthetic division, so I looked it up after the last post. Turns out I'd been doing essentially the same thing without knowing the name. More importantly, I also stumbled across the rational root test, which I was previously unfamiliar with.
The rational root test says that, given a polynomial with integer coefficients, the *possible* rational roots can be obtained by taking all factors of the constant term, divided by all factors of the coefficient of the highest power term. All that remains is to try them out.
So, with that in mind:
1) x^2 is a factor, leaving x^5+x^4+x^3+x^212x12. Possible roots are ±(1,2,3,4,6,12/1) = ±1,±2,±3,±4,±6,±12.
Synthetic division with 1 gives no remainder, so x+1 is a factor, leaving x^4+x^212.
If we substitute y = x^2, then this becomes y^2+y12, which is a quadratic, and factors to (y+4)(y3), with zeros at 4 and 3. So x = ±sqrt(4) = ±2i and x = ±sqrt(3) are zeros.
So, if we limit to reals, the factorization is x^2*(x+1)*(x+sqrt(3))*(xsqrt(3))*(x^2+4).
Alternately, we could write x^2*(x+1)*(x+sqrt(3))*(xsqrt(3))*(x+2i)*(x2i).
2) Ok, this one's a bear, because it's got a ton of possibilities. So I simply found the factors of 288, the factors of 4, put them into Excel, and calculated possible rational roots. The only zero I see is at x = 2. So, that gives (x2)*(4x^68x^59x^414x^3+64x^2+72x+144). I suspect there are no other real roots.
3) Well, this is more like it. x1 gives x^5+3x^4+5x^3+7x^2+6x+2. x+1 then gives x^4+2x^3+3x^2+4x+2. x+1 again gives x^3+x^2+2x+2. x+1 a third time gives x^2+2, which gives x = ±i*sqrt(2). So, (x1)*(x+1)^3*(x^2+2), or (x1)*(x+1)^3*(x+i*sqrt(2))*(xi*sqrt(2)).