only possible rational roots are
x = ±1, ±1/5, ± 1/25
quickly found x=-1 to work
so one factor is x+1
after reducing it to a cubic by synthetic division, it took a bit longer to find x = -1/25 to work
so (25x+1) is another factor
long algebraic divsion gave the last factor as x^2 + 5, which has no real roots.
so real roots are
x = -1 and x = -1/25
Use the rational zeros theorem to find all the real zeros of the polynomial function. Use the zeros to factor f over the real numbers.
f(x)=25x^4+26x^3+126x^2+130x+5
Find the real zeros
x=
Use the real zeros to factor f
f(x)=
3 answers
just a note:
Things worked out in this case, but x = ±5 were also candidates, since 5/1 has suitable numerator and denominator. For example,
25x^4-100x^3-124x^2-4x-5
has similar coefficients, but has real roots -1 and 5:
(x+1)(x-5)(25x^2+1)
Things worked out in this case, but x = ±5 were also candidates, since 5/1 has suitable numerator and denominator. For example,
25x^4-100x^3-124x^2-4x-5
has similar coefficients, but has real roots -1 and 5:
(x+1)(x-5)(25x^2+1)
given that f(x) = 9/x-5 and g(x) = 12/x+12 find