Need help!! The question is find tan A, cos A, and sin A and as a decimal correct to two places
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According to the rational root theorem, the possible rational factors, if any, of the given polynomial are the factors of 24/1=24.
Thus possible factors are
(x +/- k) where k=1,2,3,4,6,8,12,24.
Since all terms are positive, (x-k) do not exist as factors (see Descartes rule of signs) for all k.
Of the remainder, we note that the
sum of coefficients of even degree terms=1+35+24=60
sum of coefficients of odd degree terms=10+50=60
This means that (x+1) is a factor.
Divide f(x) by (x+1) =>
g(x)=f(x)/(x+1)=x^3+9x^2+26x+24 (no remainder)
We continue with x+2=0, or put x=-2 into g(x), and get
g(-2)=-8+36-52+24=0 => (x+2) is another rational factor.
Again, divide g(x) by (x+2) to get
h(x)=x^2+7x+12 (remainder=0)
We can readily factor h(x) by inspection as h(x)=(x+3)(x+4)
[from 3+4=7,3*4=12]
Therefore
f(x)=x^4+10*x^3+35*x^2+50*x+24=(x+1)(x+2)(x+3)(x+4)
Thus possible factors are
(x +/- k) where k=1,2,3,4,6,8,12,24.
Since all terms are positive, (x-k) do not exist as factors (see Descartes rule of signs) for all k.
Of the remainder, we note that the
sum of coefficients of even degree terms=1+35+24=60
sum of coefficients of odd degree terms=10+50=60
This means that (x+1) is a factor.
Divide f(x) by (x+1) =>
g(x)=f(x)/(x+1)=x^3+9x^2+26x+24 (no remainder)
We continue with x+2=0, or put x=-2 into g(x), and get
g(-2)=-8+36-52+24=0 => (x+2) is another rational factor.
Again, divide g(x) by (x+2) to get
h(x)=x^2+7x+12 (remainder=0)
We can readily factor h(x) by inspection as h(x)=(x+3)(x+4)
[from 3+4=7,3*4=12]
Therefore
f(x)=x^4+10*x^3+35*x^2+50*x+24=(x+1)(x+2)(x+3)(x+4)