Find all the zeros of f(x). f(x) = x^4 - 4x^3 - 7x^2 + 28x First factor out an x, then get the p/q list. List answers from smallest to largest. If there is a double root, list it twice....
Find all the zeros of f(x). f(x) = x^4 - 4x^3 - 7x^2 + 28x
First factor out an x, then get the p/q list.
List answers from smallest to largest. If there is a double root, list it twice. Keep fractions in fractional form.
Since `f(x) = x^4 -4x^3 -7x^2 + 28x` is a fourth degree polynomial, there can be up to 4 real zeros, but some of them might be the same.
As suggested, start by factoring out x:
`f(x) = x(x^3 -4x^2 -7x + 28)`
So the first zero is `x = 0` . Now we are looking for zeros of the cubic polynomial `x^3-4x^2-7x+28` .
"p/q" list refers to all possible numbers such the numerator, p, is a factor of the constant term (here, 28) and the denominator, q, is a factor of the leading coefficient (here, 1). Since the leading coefficient is 1 the only possible value for q is 1, so all zeros of this polynomial are integers.
All possible factors of 28 are `p=+- 1, +-2, +-4, +-7, +-28` .
Check which one of these numbers is a zero of the above polynomial.
1: `1^3 -4*1^2 -7*1 - 28 != 0` , so 1 is not a zero
-1: `(-1)^3 -4*(-1)^2 -7*(-1) - 28 != 0`
so -1 is not a zero
Similarly, `+-2` are not zeros either.
However, pluggin in x = 4 yields a zero. We can also notice that the polynomial can be factored by grouping:
`x^3-4x^2-7x+28=x^2(x-4)-7(x-4) = (x-4)(x^2 - 7)`
So x = 4 is a zero and two remaining zeros can be found by setting
`x^2 - 7 = 0`
This results in `x= +-sqrt(7)` .
Therefore, the zeros of f(x), listed the results in increasing order, are
`-sqrt(7), 0, sqrt(7), 4` .