Find All Real Zeros of Polynomial Functions?

Question

Use Descartes's Rule of Signs and the Rational Zeros Theorem to find all the real zeros of each polynomial function.
Use the zeros to factor f over the real numbers.

(1) f(x) = 4x^4 + 15x^2 - 4

(2) f(x) = x^4 - x^3 - 6x^2 + 4x + 8

Answer 1

f(x) = 4x^4 + 15x^2 - 4 has one sign change, so by Descartes there is exactly one positive real root.
f(-x) = 4x^4 + 15x^2 - 4 has one sign change, so there is exactly one negative real root.

If the roots are rational, then they must be factors of plus or minus 4/4 by Rational Zeros Theorem : plus or minus 1/1, 1/2, 1/4, 2/1, 2/2, 2/4, 4/1, 4/2, 4/4.

1/2 and -1/2 are the real roots.

Answer 2

using the rational root theorem, attempt all the various components of the *final* term (-6) divided by all the various components of the 1st coefficient (a million, to that end, that's implied in front of the x3). So, you attempt -6/a million , -3/a million , -2/a million , -a million/a million , a million/a million , 2/a million , 3/a million , and six/a million or -6, -3, -2, -a million, a million, 2, 3, and six The a million's are least complicated to objective, So, permit's attempt plugging in +a million, first... f(a million) = a million + 2 - 5 - 6 = -8 (does not artwork) attempt -a million... f(-a million) = -a million + 2 + 5 - 6 = 0 (works!) via fact x=-a million works, then we could desire to continually have the capacity to ingredient out a (x - -a million)... or, in different words, a (x + a million). Doing long branch of polynomials, we get (x3 + 2x2 - 5x - 6) / (x + a million) = (x2 + x - 6) So, f(x) = (x + a million)(x2 + x - 6) will we ingredient that 2d polynomial? properly, if the 2d polynomial could properly be factored into some thing like (x + a)(x + b), all of us understand, from FOIL, that a*b will could desire to come out to be -6 and that a+b will could desire to be the coefficient of the 'x' term, that's +a million. What can 'a' and 'b' be such that a*b=-6 and a+b=a million ? properly, which would be a=3 and b=-2, good? So, we could desire to continually have the capacity to ingredient the 2d polynomial into (x + 3)(x - 2). And, lo, if we divide (x2 + x - 6) by (x + 3), we get (x - 2). So, it rather works. So, thoroughly factored, f(x) = (x + a million)(x + 3)(x - 2) and the zeros may be the values of x which reason *any* of those binomials to be 0... that are x = {-a million, -3, +2} For you, which would be answer D

Answer 3

(1) x^2 = z so 4z^2+15z-4 = 0

z= ((-15 +-17)/8 as z must be >= 0 z= 1/4
and x =+-1/2
(2) =(x+1)(x-2)^2(x+2) so roots are -1,2(double) and -2