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Math  /  Algebra

QuestionFind all zeros of f(x)=x3−3x2−3x+1f(x)=x^{3}-3 x^{2}-3 x+1. Enter the zeros separated by commas. Enter exact value, not decimal approximations.

Studdy Solution

STEP 1

1. We are given the polynomial function f(x)=x3−3x2−3x+1 f(x) = x^3 - 3x^2 - 3x + 1 .
2. We need to find all zeros of the function, which are the values of x x for which f(x)=0 f(x) = 0 .

STEP 2

1. Use the Rational Root Theorem to identify possible rational roots.
2. Test possible rational roots using synthetic division or direct substitution.
3. Factor the polynomial using any found roots.
4. Solve the resulting polynomial equation(s) to find all zeros.

STEP 3

Apply the Rational Root Theorem. The possible rational roots are the factors of the constant term divided by the factors of the leading coefficient. Here, the constant term is 1 1 and the leading coefficient is 1 1 .
Possible rational roots are: ±1 \pm 1 .

STEP 4

Test the possible rational roots by substituting them into the polynomial:
First, test x=1 x = 1 :
f(1)=13−3(1)2−3(1)+1=1−3−3+1=−4 f(1) = 1^3 - 3(1)^2 - 3(1) + 1 = 1 - 3 - 3 + 1 = -4
Since f(1)≠0 f(1) \neq 0 , x=1 x = 1 is not a root.
Next, test x=−1 x = -1 :
f(−1)=(−1)3−3(−1)2−3(−1)+1=−1−3+3+1=0 f(-1) = (-1)^3 - 3(-1)^2 - 3(-1) + 1 = -1 - 3 + 3 + 1 = 0
Since f(−1)=0 f(-1) = 0 , x=−1 x = -1 is a root.

STEP 5

Since x=−1 x = -1 is a root, factor the polynomial f(x) f(x) by dividing it by x+1 x + 1 .
Use synthetic division:
−11−3−31−14−11−410\begin{array}{r|rrrr} -1 & 1 & -3 & -3 & 1 \\ & & -1 & 4 & -1 \\ \hline & 1 & -4 & 1 & 0 \\ \end{array}
The quotient is x2−4x+1 x^2 - 4x + 1 .
Thus, f(x)=(x+1)(x2−4x+1) f(x) = (x + 1)(x^2 - 4x + 1) .

STEP 6

Solve the quadratic equation x2−4x+1=0 x^2 - 4x + 1 = 0 using the quadratic formula:
The quadratic formula is given by:
x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
For x2−4x+1=0 x^2 - 4x + 1 = 0 , a=1 a = 1 , b=−4 b = -4 , c=1 c = 1 .
x=−(−4)±(−4)2−4(1)(1)2(1) x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} x=4±16−42 x = \frac{4 \pm \sqrt{16 - 4}}{2} x=4±122 x = \frac{4 \pm \sqrt{12}}{2} x=4±232 x = \frac{4 \pm 2\sqrt{3}}{2} x=2±3 x = 2 \pm \sqrt{3}
The zeros of the polynomial are:
−1,2+3,2−3 \boxed{-1, 2 + \sqrt{3}, 2 - \sqrt{3}}

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