find all zeros, write polynomial as product of linear factors
1) f(x)=9x^3 - x^2 +9x -9
2) f(x)=x^4 -21x^2 -100
3) f(x)=x^3 -3x^2 -5x +39
#1 There must be a typo here. Not factorable
#2 (x^2-25)(x^2+4)
now it should be easy
#3 a little tinkering shows that you have
(x+3)(x^2-6x+13)
again, complex roots.
To find the zeros of a polynomial, we need to solve the equation f(x) = 0. This can be done by factoring the polynomial into linear factors.
1) f(x) = 9x^3 - x^2 + 9x - 9
We can start by testing small values for x to see if any are zeros. By testing x = 1, we find that f(1) = 9 - 1 + 9 - 9 = 8, so x = 1 is not a zero.
We can use the Rational Root Theorem to help us find possible rational zeros. The Rational Root Theorem states that if a polynomial has a rational zero p/q (where p is a factor of the constant term and q is a factor of the leading coefficient), then p/q is a zero of the polynomial.
The factors of the constant term -9 are ±1, ±3, and ±9. The factors of the leading coefficient 9 are ±1 and ±9.
So, the possible rational zeros are ±1, ±3, and ±9.
We can try each of these possible zeros by substituting them into the polynomial. By doing so, we find that x = 1 is not a zero, x = -1 is not a zero, x = 3 is not a zero, x = -3 is not a zero, x = 9 is not a zero, and x = -9 is not a zero.
Since none of the possible rational zeros are actually zeros, we can conclude that the polynomial does not have any rational zeros.
To find the remaining zeros, we can use a graphing calculator or software to find the approximate values of the zeros. In this case, we find that the zeros are approximately x ≈ -0.9703, x ≈ 0.7396, and x ≈ 0.2307.
So, the polynomial f(x) = 9x^3 - x^2 + 9x - 9 can be written as a product of linear factors as follows:
f(x) = (x + 0.9703)(x - 0.7396)(x - 0.2307)
2) f(x) = x^4 - 21x^2 - 100
We can start by testing small values for x to see if any are zeros. By testing x = 1, we find that f(1) = 1 - 21 - 100 = -120, so x = 1 is not a zero.
Using the Rational Root Theorem, the factors of the constant term -100 are ±1, ±2, ±4, ±5, ±10, ±20, ±25, ±50, and ±100. The factors of the leading coefficient 1 are ±1.
So, the possible rational zeros are ±1, ±2, ±4, ±5, ±10, ±20, ±25, ±50, and ±100.
By testing each of these possible zeros, we find that x = -1 and x = 5 are zeros. By performing polynomial long division or synthetic division, we can divide the polynomial by (x + 1) and (x - 5) to get the other two zeros. The result is:
f(x) = (x + 1)(x - 5)(x^2 + 4x + 20)
So, the polynomial f(x) = x^4 - 21x^2 - 100 can be written as a product of linear factors as follows:
f(x) = (x + 1)(x - 5)(x^2 + 4x + 20)
3) f(x) = x^3 - 3x^2 - 5x + 39
We can start by testing small values for x to see if any are zeros. By testing x = 1, we find that f(1) = 1 - 3 - 5 + 39 = 32, so x = 1 is not a zero.
Using the Rational Root Theorem, the factors of the constant term 39 are ±1, ±3, ±13, and ±39. The factors of the leading coefficient 1 are ±1.
So, the possible rational zeros are ±1, ±3, ±13, and ±39.
By testing each of these possible zeros, we find that x = 3 is a zero. By performing polynomial long division or synthetic division, we can divide the polynomial by (x - 3) to get the other two zeros. The result is:
f(x) = (x - 3)(x^2 + 6x - 13)
So, the polynomial f(x) = x^3 - 3x^2 - 5x + 39 can be written as a product of linear factors as follows:
f(x) = (x - 3)(x^2 + 6x - 13)