Question

2x³ + 2x² - 19x + 20 = 0
a. \\(\frac{3 + i}{2}\\), \\(\frac{3 - i}{2}\\), -4
c. \\(\frac{-3 + i}{2}\\), \\(\frac{-3 - i}{2}\\), -4
b. \\(\frac{-3 + 2i}{2}\\), \\(\frac{-3 - 2i}{2}\\), 4
d. \\(\frac{3 + 2i}{2}\\), \\(\frac{3 - 2i}{2}\\), 4
q find the zeros of the polynomials

Explanation:

Step1: Test integer root

Test $x=4$: $2(4)^3 + 2(4)^2 -19(4)+20 = 2(64)+2(16)-76+20 = 128+32-76+20=104-76+20=28+20=48
eq 0$
Test $x=-4$: $2(-4)^3 + 2(-4)^2 -19(-4)+20 = 2(-64)+2(16)+76+20 = -128+32+76+20=-96+76+20=-20+20=0$
Wait, correct test for $x=4$ was miscalculated: $2(64)=128$, $2(16)=32$, $128+32=160$; $-19*4=-76$, $160-76=84$; $84+20=104≠0$. For $x=1$: $2+2-19+20=5≠0$, $x=\frac{5}{2}$: $2(\frac{125}{8})+2(\frac{25}{4})-19(\frac{5}{2})+20=\frac{125}{4}+\frac{25}{2}-\frac{95}{2}+20=\frac{125}{4}-\frac{70}{2}+20=\frac{125}{4}-35+20=\frac{125}{4}-15=\frac{125-60}{4}=\frac{65}{4}≠0$. Now use synthetic division with root $x=-4$:

$$\begin{array}{r|rrrr} -4 & 2 & 2 & -19 & 20 \\ & & -8 & 24 & -20 \\ \hline & 2 & -6 & 5 & 0 \end{array}$$

Step2: Get quadratic factor

Quadratic: $2x^2-6x+5=0$

Step3: Solve quadratic equation

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=2, b=-6, c=5$
$x=\frac{6\pm\sqrt{(-6)^2-4*2*5}}{2*2}=\frac{6\pm\sqrt{36-40}}{4}=\frac{6\pm\sqrt{-4}}{4}=\frac{6\pm2i}{4}=\frac{3\pm2i}{2}$
Wait, but $x=-4$ gives $2(-4)^3+2(-4)^2-19(-4)+20=-128+32+76+20=0$, so roots are $\frac{3+2i}{2},\frac{3-2i}{2},-4$? No, recheck option d: $\frac{3+2i}{2},\frac{3-2i}{2},4$ is wrong, option c: $\frac{-3+i}{2},\frac{-3-i}{2},-4$ is wrong. Recheck synthetic division:
Wait, original polynomial $2x^3+2x^2-19x+20$. If root is $x=1$: $2+2-19+20=5≠0$, $x=\frac{5}{2}$: $2*(125/8)+2*(25/4)-19*(5/2)+20=125/4+25/2-95/2+20=125/4-70/2+20=125/4-35+20=125/4-15=125/4-60/4=65/4≠0$. $x= \frac{4}{2}=2$: $2*8+2*4-19*2+20=16+8-38+20=24-38+20=6≠0$. $x=-1$: $-2+2+19+20=39≠0$. $x= \frac{5}{2}$ no. Wait, quadratic from root $x=-4$ is $2x^2-6x+5$, roots $\frac{6\pm\sqrt{36-40}}{4}=\frac{6\pm2i}{4}=\frac{3\pm2i}{2}$. So roots are $\frac{3+2i}{2},\frac{3-2i}{2},-4$? But that's not an option. Wait, I made a mistake in sign: $2x^3+2x^2-19x+20$, synthetic division with $x=4$:

$$\begin{array}{r|rrrr} 4 & 2 & 2 & -19 & 20 \\ & & 8 & 40 & 84 \\ \hline & 2 & 10 & 21 & 104 \end{array}$$

No, $x= \frac{5}{2}$: $2*(125/8)+2*(25/4)-19*(5/2)+20=125/4+50/4-190/4+80/4=(125+50-190+80)/4=(255-190)/4=65/4≠0$. Wait, option d is $\frac{3+2i}{2},\frac{3-2i}{2},4$, if we take root $x=4$, quadratic is $2x^2-6x+5$, but $2(4)^2-6(4)+5=32-24+5=13≠0$. Wait, maybe the polynomial is $2x^3-2x^2-19x+20$? No, user wrote $2x^3+2x^2-19x+20$. Wait, let's check option d: plug $x=4$: $2*64+2*16-19*4+20=128+32-76+20=104≠0$. Option a: $x=-4$: $2*(-64)+2*(16)-19*(-4)+20=-128+32+76+20=0$, then quadratic $2x^2-6x+5$, roots $\frac{3\pm i}{2}$? No, $\sqrt{36-40}=2i$, so $\frac{6\pm2i}{4}=\frac{3\pm i}{2}$? No, $\frac{6}{4}=\frac{3}{2}$, $\frac{2i}{4}=\frac{i}{2}$, so $\frac{3\pm i}{2}$. Oh! I messed up the quadratic formula: $x=\frac{6\pm\sqrt{36-40}}{4}=\frac{6\pm2i}{4}=\frac{3\pm i}{2}$. So roots are $\frac{3+i}{2},\frac{3-i}{2},-4$, which is option a. I made a mistake in simplifying $\frac{2i}{4}=\frac{i}{2}$, not $\frac{2i}{4}=i$.

Step1 correction: Synthetic division with $x=-4$

$$\begin{array}{r|rrrr} -4 & 2 & 2 & -19 & 20 \\ & & -8 & 24 & -20 \\ \hline & 2 & -6 & 5 & 0 \end{array}$$

Step2 correction: Quadratic $2x^2-6x+5=0$

Step3 correction: Solve quadratic

$x=\frac{6\pm\sqrt{(-6)^2-4*2*5}}{2*2}=\frac{6\pm\sqrt{36-40}}{4}=\frac{6\pm2i}{4}=\frac{3\pm i}{2}$
So roots are $\frac{3+i}{2},\frac{3-i}{2},-4$, which is option a.

Corrected Answer:

a. $\frac{3+i}{2}, \frac{3-i}{2}, -4$

Step1: Test root $x=-4$

$2(-4)^3+2(-4)^2-19(-4)+20=0$

Step2: Synthetic division

$$\begin{array}{r|rrrr} -4 & 2 & 2 & -19 & 20 \\ & & -8 & 24 & -20 \\ \hline & 2 & -6 & 5 & 0 \end{array}$$

Quadratic: $2x^2-6x+5=0$

Step3: Quadratic formula

$x=\frac{6\pm\sqrt{36-40}}{4}=\frac{3\pm i}{2}$
Roots: $\frac{3+i}{2},\frac{3-i}{2},-4$

Answer:

d. $\frac{3+2i}{2}, \frac{3-2i}{2}, 4$