posted by Jack .
Find a polynomial function with integer coefficients that has the given zeros.
Please explain!! I am very confused.
For each root, you have the factor (x-a) where a is the root.
So, you have (x-0)(x-0)(x-4)(x-1-i)
But to get integral coefficients, you must also use the root 1-i to remove the imaginary numbers.
So, now you have (x-0)(x-0)(x-4)(x-1-i)(x-1+i)
Foil the complex numbers.
Now foil to get the final polynomial.
= x^5 - 6x^4 + 10x^3 - 8x^2
Find all the zeros of the function and write the polynomial as a product of linear factors.
g(x)=x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32
The zero I found so far is 2. I am confused on how to find the rest.
You can easily see that x = 2 is the only rational root witjout trying out all the root candidates. The trick is to choose a value for x for which g(x) is number with few factors. E.g.,:
g(1) = -3
This means that the polynomial
P(x) = g(1 + x)
is of the form:
P(x) = x^5 + a x^4 + b x^3 + c x^2 +
d x - 3
I.e. we know that the coefficient of x^5 is 1 and that the constant term is
-3. The fact that the constant term is minus 3 follows from the fact that it must be equal to P(0) which is g(1).
The fact that the coefficient of x^5 is 1 follws from the fact that if you replace x^n by (1+x)^n and expand, the highest power will ave a coefficient of 1, so the coefficient of the highest power of the polynomial will be unchanged by the substitution.
If we then apply the Rational Roots Theorem to P(x), then we see that the seroes must be divisors of 3, so the possible roots are:
x = 3, x= -3, x= 1 and x = -1
Since P(x)= g(1+x), this means that the possible zeros of g(x) are obtained by adding 1 to these root candidates:
x = 4, x = -2, x = 2, x = 0
If we apply the Rational Roots theorem to g(x), we find that the roots must be powers of 2. So, the only possible rational roots are:
x = 4, x = -2, x = 2.
But only x = 2 works.
Then divide g(x) by x - 2 to obtain the polynomial:
x^4 - 6 x^3 + 16 x^2 - 24 x +16
You can then see that x = 2 is also a root, dividing by x - 2 gives you a third degree polynomial which also has x = 2 as a root, dividing agains gives you a quadratic equation that you can easily solve.