Question

use the aleks graphing calculator to find all the zeros of the polynomial function. f(x)=2x³ - 2x² - 7x + 3 round to the nearest hundredth. if there is more than one answer, separate them with commas. zero(s):

Explanation:

Step1: Analyze the polynomial

We have the cubic polynomial \( f(x) = 2x^3 - 2x^2 - 7x + 3 \). To find its zeros, we can use the ALEKS graphing calculator (or we can also try rational root theorem first and then use graphing or numerical methods). The rational root theorem says possible rational roots are factors of the constant term over factors of the leading coefficient. So possible rational roots are \( \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2} \). Let's test \( x = 1 \): \( f(1)=2 - 2 - 7 + 3=-4 eq0 \). \( x = -1 \): \( f(-1)=-2 - 2 + 7 + 3 = 6 eq0 \). \( x = 3 \): \( f(3)=54 - 18 - 21 + 3 = 18 eq0 \). \( x=\frac{1}{2} \): \( f(\frac{1}{2})=2(\frac{1}{8})-2(\frac{1}{4})-7(\frac{1}{2})+3=\frac{1}{4}-\frac{1}{2}-\frac{7}{2}+3=\frac{1 - 2 - 14 + 12}{4}=\frac{-3}{4} eq0 \). \( x=\frac{3}{2} \): \( f(\frac{3}{2})=2(\frac{27}{8})-2(\frac{9}{4})-7(\frac{3}{2})+3=\frac{27}{4}-\frac{9}{2}-\frac{21}{2}+3=\frac{27 - 18 - 42 + 12}{4}=\frac{-21}{4} eq0 \). So no rational roots, we need to use graphing or numerical methods.

Step2: Use graphing calculator (conceptually)

When we graph \( y = 2x^3 - 2x^2 - 7x + 3 \), we look for x - intercepts. By analyzing the function's behavior: as \( x ightarrow\infty \), \( f(x) ightarrow\infty \); as \( x ightarrow-\infty \), \( f(x) ightarrow-\infty \). The derivative \( f^\prime(x)=6x^2 - 4x - 7 \). Setting derivative to zero: \( 6x^2 - 4x - 7 = 0 \), solutions \( x=\frac{4\pm\sqrt{16 + 168}}{12}=\frac{4\pm\sqrt{184}}{12}=\frac{4\pm2\sqrt{46}}{12}=\frac{2\pm\sqrt{46}}{6}\approx\frac{2\pm6.782}{6} \). So critical points at \( x\approx\frac{2 + 6.782}{6}\approx1.46 \) and \( x\approx\frac{2 - 6.782}{6}\approx - 0.797 \). Now, using a graphing calculator (like ALEKS), we find the x - intercepts (zeros) by looking where the graph crosses the x - axis. By numerical methods (or calculator), we find the zeros:

We can use the Newton - Raphson method or just rely on the calculator output. Let's assume we use the calculator:

First, find the intervals. Let's check \( f(-2)=2(-8)-2(4)-7(-2)+3=-16 - 8 + 14 + 3=-7 \). \( f(-1)=6 \) (as before), so there is a zero between - 2 and - 1. \( f(0)=3 \), \( f(1)=-4 \), so zero between 0 and 1. \( f(2)=16 - 8 - 14 + 3=-3 \), \( f(3)=18 \), so zero between 2 and 3.

Now, using calculator (or precise calculation):

For the zero between - 2 and - 1: Let's use Newton - Raphson. Let \( x_0=-1.5 \), \( f(-1.5)=2(-3.375)-2(2.25)-7(-1.5)+3=-6.75 - 4.5 + 10.5 + 3 = 2.25 \). \( f^\prime(-1.5)=6(2.25)-4(-1.5)-7 = 13.5 + 6 - 7 = 12.5 \). Next iteration: \( x_1=x_0-\frac{f(x_0)}{f^\prime(x_0)}=-1.5-\frac{2.25}{12.5}=-1.5 - 0.18=-1.68 \). \( f(-1.68)=2(-4.741)-2(2.822)-7(-1.68)+3=-9.482 - 5.644 + 11.76 + 3 = -0.366 \). \( f^\prime(-1.68)=6(2.822)-4(-1.68)-7 = 16.932 + 6.72 - 7 = 16.652 \). \( x_2=-1.68-\frac{-0.366}{16.652}\approx - 1.68 + 0.022\approx - 1.658 \). \( f(-1.658)\approx2(-4.59)-2(2.74)-7(-1.658)+3\approx - 9.18 - 5.48 + 11.606 + 3\approx0.046 \). \( x_3=-1.658-\frac{0.046}{f^\prime(-1.658)} \). \( f^\prime(-1.658)=6(2.74)-4(-1.658)-7\approx16.44 + 6.632 - 7 = 16.072 \). \( x_3\approx - 1.658 - \frac{0.046}{16.072}\approx - 1.658 - 0.0029\approx - 1.6609 \). So approximately - 1.66 (rounded to hundredth).

For the zero between 0 and 1: Let \( x_0 = 0.5 \), \( f(0.5)=-0.75 \) (as before), \( f(1)=-4 \), wait no, \( f(0)=3 \), \( f(1)=-4 \), so between 0 and 1. Wait, \( f(0)=3 \), \( f(1)=-4 \), so let's take \( x_0 = 0.5 \), \( f(0.5)=-0.75 \), \( f(0)=3 \). Newton - Raphson: \( x_1=0.5-\frac{-0.75}{f^\prime(0.5)} \). \( f^\prime(0.5)=6(0.25)-4(0.5)-7 = 1.5 - 2 - 7=-7.5 \). \( x_1=0.5-\frac{-0.75}{-7.5}=0.5 - 0.1 = 0.4 \). \( f(0.4)=2(0.064)-2(0.16)-7(0.4)+3=0.128 - 0.32 - 2.8 + 3 = 0.008 \). Oh, \( f(0.4)\approx0.008 \approx0 \). So \( x = 0.4 \) is a zero? Wait, \( f(0.4)=2(0.064)-2(0.16)-7(0.4)+3=0.128 - 0.32 - 2.8 + 3=0.008\approx0 \). So \( x = 0.4 \) (exact? Wait, \( 2(0.4)^3-2(0.4)^2-7(0.4)+3=2(0.064)-2(0.16)-2.8 + 3=0.128 - 0.32 - 2.8 + 3=0.008\approx0 \), so \( x = 0.4 \) (or 0.40 when rounded to hundredth).

For the zero between 2 and 3: Let \( x_0 = 2 \), \( f(2)=-3 \), \( x_0 = 3 \), \( f(3)=18 \). Newton - Raphson: \( x_1=2-\frac{-3}{f^\prime(2)} \). \( f^\prime(2)=6(4)-4(2)-7=24 - 8 - 7 = 9 \). \( x_1=2+\frac{3}{9}=2.333 \). \( f(2.333)=2(12.703)-2(5.444)-7(2.333)+3=25.406 - 10.888 - 16.331 + 3=1.187 \). \( f^\prime(2.333)=6(5.444)-4(2.333)-7=32.664 - 9.332 - 7 = 16.332 \). \( x_2=2.333-\frac{1.187}{16.332}\approx2.333 - 0.0727\approx2.2603 \). \( f(2.2603)=2(11.54)-2(5.108)-7(2.2603)+3=23.08 - 10.216 - 15.822 + 3=0.042 \). \( x_3=2.2603-\frac{0.042}{f^\prime(2.2603)} \). \( f^\prime(2.2603)=6(5.108)-4(2.2603)-7=30.648 - 9.041 - 7 = 14.607 \). \( x_3\approx2.2603 - 0.0029\approx2.2574 \). So approximately 2.26 (rounded to hundredth).

So the zeros are approximately - 1.66, 0.40, 2.26.

Answer:

-1.66, 0.40, 2.26