find the real zeros in increasing order of the polynominial P(x)=x^4-3x^3-5x^2+13x+6?

Question

can someone please show me how to get the real zeros, there should be four i think, i am not just not sure how to get there??

Answer 1

Gauss theorem says that if m/n is a rational root of a polynomial, with m and n coprimes, m must divide the independent term and n must divide the principal coefficient.

Therefore, the only possible rational roots of P are +-1,+-2 , +-3 and +-6.

P(-2)= P(3) =0, we've found two roots.

By dividing the polinomial, we have

P(x) = (x+2)*(x-3)*(x^2-2*x-1)

The remaining roots are easily found:

-1+-sqrt(1+1)

Then

P(x) = (x+2)*(x-(1-sqrt(2))*(x-(1+sqrt(2))*(x-3),

ordered in ascending order.

Roots = {-2, 1-sqrt(2), 1+sqrt(2), 3}

Answer 2

The Rational Root Theorem will tell you what the possible rational roots are. (I assume you're already familiar with this theorem and its usage.) In this case, the possible rational roots are ±1, ±2, ±3, and ±6. You can "cheat" like I do and graph the polynomial first so you know which of these to check. The rational roots are 3 and -2. So (x-3) and (x+2) are both factors of P(x).

Divide P(x) by (x-3), and divide the quotient by (x+2); that will give you a quadratic. You can use the quadratic formula to find the other two roots (which are real but not rational).

Answer 3

We can factorize step by step as follows:

x^4-3x^3-5x^2+13x+6=
(x+2)(x^3-5x^2+5x+3)=
(x+2)(x-3)(x^2-2x-1)

and the roots of the quadratic polynomial are 1+sqrt(2) and 1-sqrt(2). since sqrt(2) is approximately equal to 1.4, the roots are:

-2, 1-sqrt(2), 1+sqrt(2), and 3.

Answer 4

x=-2 and x=3 solutions by trial and error.
(x+2)(x-3) are factors.
There are 4 roots and 4 solutions as this is a 4th degree polynomial.
Divide p(x) by (x-2)(x-3)=x^2-5x+6
You'll get a 2nd degree equation which you can factor.
Two of the answers are -2,3