Asked by Rose
Can someone show me the steps for how to solve:
1.) f(x) = 4x^4 -3x^3 +2x^2 +x-3
2.) f(x) = x^3 - 2x^2 -5x +6
3.) f(x) = -3x^3 -2x^2 + x -1
Thank you so much :)
1.) f(x) = 4x^4 -3x^3 +2x^2 +x-3
2.) f(x) = x^3 - 2x^2 -5x +6
3.) f(x) = -3x^3 -2x^2 + x -1
Thank you so much :)
Answers
Answered by
Damon
You only typed half the problem. Perhaps you are supposed to graph, not solve?
Answered by
Rose
sorry, i forgot to say the instructions said "find all possible rational roots"
Answered by
Damon
Result:
The roots of the polynomial 4x4 - 3x3 + 2x2 + x - 3 are:
X1 = 0.301 + 1.002*i
X2 = 0.301 - 1.002*i
X3 = 0.905
X4 = -0.757
The roots of the polynomial 4x4 - 3x3 + 2x2 + x - 3 are:
X1 = 0.301 + 1.002*i
X2 = 0.301 - 1.002*i
X3 = 0.905
X4 = -0.757
Answered by
Damon
That is from:
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-roots-calculator.php
Answered by
Damon
As you can see you really have to graph that polynomial to find the roots. If you have a graphing calculator, try that
Notice that it only crosses the x axis twice and then has a complex pair for the third and fourth roots
Notice that it only crosses the x axis twice and then has a complex pair for the third and fourth roots
Answered by
Damon
For the second one you can see that one is a root by inspection.
Then you can divide by (x-1) to get a quadratic which you can solve.
Then you can divide by (x-1) to get a quadratic which you can solve.
Answered by
Damon
By the way if you tried all possible rational roots results on the first problem using factors of the first and last term 3 and 4 ( 3, 2, 4) you would have found that it has no rational factors by the rational root test.
Answered by
Rose
thank you so much for your help ! :)