Question

section 3.5
score: 3/6 answered: 3/6
question 4
0/1 pt
find all zeros of f(x) = x³ + 3x² + x - 1. enter the zeros separated by commas. enter exact value, not decimal approximations.
question help: video post to forum
save progress

Explanation:

Step1: Analyze the function

We have the cubic function \( f(x) = x^3 + 3x^2 + x - 1 \). First, we can try to find rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term (\(\pm1\)) divided by factors of the leading coefficient (\(\pm1\)), so possible roots are \(\pm1\).

Let's test \( x = 1 \): \( f(1)=1 + 3+1 - 1=4 eq0 \).

Test \( x=-1 \): \( f(-1)=-1 + 3 - 1 - 1=0 \). Wait, no, \( (-1)^3+3(-1)^2+(-1)-1=-1 + 3-1 - 1=0 \)? Wait, \(-1 + 3=2\), \(2-1 = 1\), \(1-1 = 0\). Oh, \( x=-1 \) is a root? Wait, no, let's recalculate: \( (-1)^3=-1 \), \(3(-1)^2 = 3(1)=3\), \(x=-1\), so \( f(-1)=-1 + 3-1 - 1=0 \). Wait, but when we test \( x = 0.2679\) (we'll see later), but first, maybe I made a mistake. Wait, no, let's check the derivative to see the number of real roots. The derivative \( f'(x)=3x^2 + 6x + 1 \). The discriminant of the derivative is \( \Delta=36 - 12=24>0 \), so the function has a local maximum and minimum. Let's find the critical points: \( x=\frac{-6\pm\sqrt{24}}{6}=\frac{-6\pm2\sqrt{6}}{6}=\frac{-3\pm\sqrt{6}}{3}=-1\pm\frac{\sqrt{6}}{3}\approx -1\pm0.8165 \). So critical points at \( x\approx -1.8165 \) and \( x\approx -0.1835 \). Now, evaluate \( f(x) \) at these critical points:

\( f(-1.8165)\approx (-1.8165)^3+3(-1.8165)^2+(-1.8165)-1\approx -6.03 + 3(3.299)+(-1.8165)-1\approx -6.03+9.897 - 1.8165 - 1\approx 1.0505 \)

\( f(-0.1835)\approx (-0.1835)^3+3(-0.1835)^2+(-0.1835)-1\approx -0.006 + 3(0.0337)+(-0.1835)-1\approx -0.006+0.1011 - 0.1835 - 1\approx -1.0884 \)

Since \( f(-2)=(-8)+12-2 - 1=1 \), \( f(-1.8165)\approx1.05 \), \( f(-0.1835)\approx -1.088 \), \( f(0)=-1 \), \( f(1)=4 \). So the function crosses the x - axis three times? Wait, no, at \( x < -1.8165 \), as \( x\to-\infty \), \( f(x)\to-\infty \), \( f(-2)=1 \), so between \( -\infty \) and \( -1.8165 \), it goes from \( -\infty \) to \( 1 \), but the local max at \( x=-1.8165 \) is \( \approx1.05 \), then decreases to local min at \( x=-0.1835 \approx -1.088 \), then increases to \( \infty \) as \( x\to\infty \). So \( f(-2)=1 \), \( f(-1.8165)\approx1.05 \), \( f(-1)=0 \)? Wait, earlier calculation of \( f(-1) \): \( (-1)^3+3(-1)^2+(-1)-1=-1 + 3-1 - 1=0 \). Oh! So \( x = -1 \) is a root. Then we can factor \( f(x)=(x + 1)(x^2 + 2x - 1) \). Now, solve \( x^2 + 2x - 1=0 \) using the quadratic formula: \( x=\frac{-2\pm\sqrt{4 + 4}}{2}=\frac{-2\pm2\sqrt{2}}{2}=-1\pm\sqrt{2} \). Wait, but earlier when we calculated \( f(-1) \), we got 0, so the roots are \( x=-1 \), \( x=-1+\sqrt{2}\approx -1 + 1.4142=0.4142 \), \( x=-1-\sqrt{2}\approx -1 - 1.4142=-2.4142 \). Wait, but when we calculated \( f(-2)=1 \), \( f(-2.4142)\): Let's check \( x=-1-\sqrt{2}\approx -2.4142 \), \( f(-2.4142)=(-2.4142)^3+3(-2.4142)^2+(-2.4142)-1\approx -14.2 + 3(5.83)+(-2.4142)-1\approx -14.2+17.49 - 2.4142 - 1\approx -0.1242 \approx0 \)? Wait, no, my earlier factoring must be wrong. Wait, \( (x + 1)(x^2+2x - 1)=x^3+2x^2 - x+x^2+2x - 1=x^3+3x^2+x - 1 \), which is correct. So the roots are \( x=-1 \), \( x=-1\pm\sqrt{2} \). Wait, but when we calculated \( f(-1 - \sqrt{2})\approx (-2.4142)^3+3(-2.4142)^2+(-2.4142)-1 \). Let's compute \( (-2.4142)^3\approx -14.207 \), \( 3(-2.4142)^2=3\times5.83=17.49 \), so \( -14.207+17.49 - 2.4142 - 1= -14.207-2.4142 - 1+17.49=-17.6212 + 17.49=-0.1312\approx0 \) (due to rounding). And \( f(-1 + \sqrt{2})\approx(0.4142)^3+3(0.4142)^2+0.4142 - 1\approx0.071 + 3(0.1716)+0.4142 - 1\approx0.071+0.5148+0.4142 - 1=0 \). And \( f(-1)=0 \). Wait, but earlier when we calculated the critical points, we had a local max at \( x\approx -1.8165 \) and local min at \( x\approx -0.1835 \). But \( x=-1 - \sqrt{2}\approx -2.4142 \), which is less than \( -1.8165 \), and \( x=-1+\sqrt{2}\approx0.4142 \), which is greater than \( -0.1835 \). So the three roots are \( x=-1-\sqrt{2}\approx -2.4142 \), \( x=-1 \), \( x=-1+\sqrt{2}\approx0.4142 \). Wait, but when we tested \( x=-1 \), we got \( f(-1)=0 \), so that's a root.

Step2: Solve for roots

We factored \( f(x)=(x + 1)(x^2 + 2x - 1) \). The roots of \( x^2+2x - 1=0 \) are given by the quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), \( c=-1 \). So \( x=\frac{-2\pm\sqrt{4 + 4}}{2}=\frac{-2\pm2\sqrt{2}}{2}=-1\pm\sqrt{2} \). And we also have the root \( x=-1 \) from the linear factor \( (x + 1) \).

Answer:

\(-1-\sqrt{2}, -1, -1+\sqrt{2}\) (or approximately \(-2.4142, -1, 0.4142\))