SOLUTION: "Find the roots of the polynomial equation." 2x^3+2x^2-19x+20 I tried factoring out a 2 but it is uneven. Can I still find the roots and solve the problem even when the x^3 coeffi

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: "Find the roots of the polynomial equation." 2x^3+2x^2-19x+20 I tried factoring out a 2 but it is uneven. Can I still find the roots and solve the problem even when the x^3 coeffi      Log On


   

Question 696202: "Find the roots of the polynomial equation." 2x^3+2x^2-19x+20
I tried factoring out a 2 but it is uneven. Can I still find the roots and solve the problem even when the x^3 coefficient is two?
I need this right now ASAP because I have an Alg. 2 final tomorrow, thanks
Found 2 solutions by solver91311, KMST:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Start with the Rational Roots Theorem. If a polynomial equation with integer coefficients has a rational root, then that root must be of the form where is a factor of the constant term and is a factor of the lead coefficient.

Hence your possible rational zeros are:



Start testing the possible zeros one at a time using Synthetic Division. Review the process for Synthetic Division here (note: there are 4 pages to review).

Once you find one that works, you will have one of your binomial factors, and the quotient of the synthetic division will give you a quadratic trinomial that you can solve with the quadratic formula. Hint: start with -4.

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The rational roots of P%28x%29=2x%5E3%2B2x%5E2-19x%2B20 (if they exist) will be of the form
p%2Fq and -p%2Fq where p is a factor of the independent term 20 and q is a factor of the leading coefficient 2.
Factors of 20 : 1, 2, 4, 5, 10, 20. Factors of 2 : 1, 2.
Possible rational roots: -20, -10, -5, -4, -2, -1, -5%2F2, -1%2F2, 1%2F2, 5%2F2, 1, 2, 4, 5, 10, and 20.
I tried them, and found that P%28-4%29=0, and that
2x%5E3%2B2x%5E2-19x%2B20-%28x%2B4%29%282x%5E2-6x%2B5%29%29
The roots would be highlight%28x=-4%29,
and the solutions to 2x%5E2-6x%2B5=0.
However, 2x%5E2-6x%2B5=0 has no real solution.
If you are studying complex numbers, you could find another two complex roots.