Find the polynomials roots to each of the following problems: #1) x^2+3x+1 #2) x^2+4x+3=0 #3) -2x^2+4x-5 #3 is not an equation. Dod you omit "= 0" at the end? #2 can be factored into (x+1)(x+3) = 0, so the roots are x=-1 and -3.
Factor this polynomial: F(x)=x^3-x^2-4x+4 Try to find the rational roots. If p/q is a root (p and q having no factors in common), then p must divide 4 and q must divide 1 (the coefficient of x^3). The rational roots can thuis be
Use the rational root theorem to list all possible rational roots for the equation. X^3+2x-9=0. Use the rational root theorem to list all possible rational roots for the equation. 3X^3+9x-6=0. A polynomial function P(x) with
If 5x4-14x³+18x²+40x+16=(x²-4x+8)(ax²+bc+c) find a,b and c and hence find the four solutions of the equation 5x4-14x³+18x²+40x+16 Given that x³-1=(x-&)(ax²+bx+c) find the values of a,b and c and hence find the three roots
If the equation 3x^4+2x^3-6x^2-6x+p=0 has two equal roots, find the possible values of p. x-1 is a factor so after replacing x=1 I got p=7. But I don't know any other possible values of p. I got p=1 but I solved it through
(1)given that -2x^2+6x+7=0, if alfa and B are the roots of the equation, form new equation whose roots are(a)alfa/B and B/alfa (b)alfa/B^2and B/alfa^2(b) if alfa and B are the roots of the equation 4x^+8x-1=0. find the values of
"Show that x^6 - 7x^3 - 8 = 0 has a quadratic form. Then find the two real roots and the four imaginary roots of this equation." I used synthetic division to get the real roots 2 and -1, but I can't figure out how to get the