Question

find all zeros of $f(x) = x^3 + 3x^2 + x - 1$. enter the zeros separated by commas. enter exact value, not decimal approximations.

Explanation:

Step1: Analyze the function

The function is \( f(x) = x^3 + 3x^2 + x - 1 \). We can use the Rational Root Theorem first to check for possible rational roots. The possible rational roots are factors of the constant term (1) divided by factors of the leading coefficient (1), so possible roots are \( \pm1 \).

Let's test \( x = 1 \): \( f(1) = 1 + 3 + 1 - 1 = 4 eq 0 \).

Test \( x = -1 \): \( f(-1) = -1 + 3 - 1 - 1 = 0 \). Wait, no, \( f(-1)=(-1)^3 + 3(-1)^2 + (-1) - 1=-1 + 3 - 1 - 1 = 0 \)? Wait, \( -1 + 3 = 2 \), \( 2 - 1 = 1 \), \( 1 - 1 = 0 \). Oh, \( x = -1 \) is a root? Wait, no, let's recalculate: \( (-1)^3=-1 \), \( 3(-1)^2 = 3(1)=3 \), \( (-1) \) is -1, and -1. So \( -1 + 3 - 1 - 1 = 0 \). So \( x = -1 \) is a root? Wait, but when we factor \( f(x) \) as \( (x + 1)(x^2 + 2x - 1) \), let's check: \( (x + 1)(x^2 + 2x - 1)=x^3 + 2x^2 - x + x^2 + 2x - 1=x^3 + 3x^2 + x - 1 \), which matches. So now we can solve the quadratic \( x^2 + 2x - 1 = 0 \) using the quadratic formula.

Step2: Solve the quadratic equation

For a quadratic equation \( ax^2 + bx + c = 0 \), the solutions are \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 2 \), \( c = -1 \).

So \( x = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \).

Now we have the roots: \( x = -1 \), \( x = -1 + \sqrt{2} \approx -1 + 1.4142 = 0.4142 \), and \( x = -1 - \sqrt{2} \approx -1 - 1.4142 = -2.4142 \). Wait, but earlier when we tested \( x = -1 \), \( f(-1) = 0 \), so that's a root. Wait, but let's check the original function at \( x = -1 \): \( (-1)^3 + 3(-1)^2 + (-1) - 1 = -1 + 3 - 1 - 1 = 0 \), correct. So the roots are \( -1 - \sqrt{2} \), \( -1 \), and \( -1 + \sqrt{2} \)? Wait, no, wait the quadratic was \( x^2 + 2x - 1 \), so solving that gives \( x = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2} \), and we also have \( x = -1 \) from the linear factor. Wait, no, wait the factorization: we had \( f(x)=(x + 1)(x^2 + 2x - 1) \), so the roots are \( x = -1 \), and the roots of \( x^2 + 2x - 1 = 0 \), which are \( -1 \pm \sqrt{2} \). Let's verify with the original function:

For \( x = -1 - \sqrt{2} \approx -2.4142 \): \( f(-2.4142) \approx (-2.4142)^3 + 3(-2.4142)^2 + (-2.4142) - 1 \). Let's calculate \( (-2.4142)^3 \approx -14.2426 \), \( 3(-2.4142)^2 \approx 3(5.8284) \approx 17.4852 \), so total is \( -14.2426 + 17.4852 - 2.4142 - 1 \approx 0 \), correct.

For \( x = -1 \): \( f(-1) = 0 \), correct.

For \( x = -1 + \sqrt{2} \approx 0.4142 \): \( f(0.4142) \approx (0.4142)^3 + 3(0.4142)^2 + 0.4142 - 1 \approx 0.0718 + 3(0.1716) + 0.4142 - 1 \approx 0.0718 + 0.5148 + 0.4142 - 1 \approx 1.0008 - 1 \approx 0.0008 \), which is approximately 0 (due to rounding), correct.

So the zeros are \( -1 - \sqrt{2} \approx -2.414 \), \( -1 \), and \( -1 + \sqrt{2} \approx 0.414 \). Wait, but let's check the initial factorization again. Wait, when we divided \( f(x) \) by \( (x + 1) \), we get \( x^2 + 2x - 1 \), so the roots are \( x = -1 \), and \( x = \frac{-2 \pm \sqrt{4 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2} \). So the three roots are \( -1 - \sqrt{2} \), \( -1 \), and \( -1 + \sqrt{2} \). Converting to decimal approximations: \( -1 - \sqrt{2} \approx -2.414 \), \( -1 \), \( -1 + \sqrt{2} \approx 0.414 \).

Wait, but let's check the function's derivative to see the number of real roots. The derivative \( f'(x) = 3x^2 + 6x + 1 \). The discriminant of the derivative is \( 36 - 12 = 24 \), so the derivative has two real roots, meaning the function has a local maximum and minimum. Let's find the critical points: \( x = \frac{-6 \pm \sqrt{24}}{6} = \frac{-6 \pm 2\sqrt{6}}{6} = \frac{-3 \pm \sqrt{6}}{3} \approx \frac{-3 \pm 2.4495}{3} \), so approximately \( x \approx -1.8165 \) and \( x \approx -0.1835 \). Now, let's evaluate \( f(x) \) at these critical points:

At \( x \approx -1.8165 \): \( f(-1.8165) \approx (-1.8165)^3 + 3(-1.8165)^2 + (-1.8165) - 1 \approx -5.97 + 3(3.299) - 1.8165 - 1 \approx -5.97 + 9.897 - 2.8165 \approx 1.1105 \) (local maximum).

At \( x \approx -0.1835 \): \( f(-0.1835) \approx (-0.1835)^3 + 3(-0.1835)^2 + (-0.1835) - 1 \approx -0.0061 + 3(0.0337) - 0.1835 - 1 \approx -0.0061 + 0.1011 - 1.1835 \approx -1.0885 \) (local minimum).

Since the local maximum is positive (≈1.11) and the local minimum is negative (≈-1.09), the function crosses the x-axis three times: once left of the local maximum (x < -1.8165), once between the local maximum and minimum (-1.8165 < x < -0.1835), and once right of the local minimum (x > -0.1835). We found \( x = -1 - \sqrt{2} \approx -2.414 \) (left of -1.8165), \( x = -1 \) (between -1.8165 and -0.1835, since -1.8165 < -1 < -0.1835), and \( x = -1 + \sqrt{2} \approx 0.414 \) (right of -0.1835). So these are the three real roots.

Answer:

\( -2.414, -1, 0.414 \) (or the exact forms \( -1 - \sqrt{2}, -1, -1 + \sqrt{2} \))