11\nfill in the blank 5 points\nsolve the equation $2x^3 + 7x^2 + 2x - 3 = 0$\nvideo tutorial on solving…

11\nfill in the blank 5 points\nsolve the equation $2x^3 + 7x^2 + 2x - 3 = 0$\nvideo tutorial on solving polynomial equations:\nthe solutions are x = type your answer..., x = type your answer..., and x = type your answer...\nprevious

11\nfill in the blank 5 points\nsolve the equation $2x^3 + 7x^2 + 2x - 3 = 0$\nvideo tutorial on solving polynomial equations:\nthe solutions are x = type your answer..., x = type your answer..., and x = type your answer...\nprevious

Answer

Explanation:

Step1: Test rational root candidates

Using Rational Root Theorem, possible roots are $\pm1, \pm3, \pm\frac{1}{2}, \pm\frac{3}{2}$. Test $x=\frac{1}{2}$: $2\left(\frac{1}{2}\right)^3 +7\left(\frac{1}{2}\right)^2 +2\left(\frac{1}{2}\right)-3 = 2\cdot\frac{1}{8}+7\cdot\frac{1}{4}+1-3 = \frac{1}{4}+\frac{7}{4}-2 = 2-2=0$. So $x=\frac{1}{2}$ is a root, factor is $(2x-1)$.

Step2: Polynomial long division

Divide $2x^3+7x^2+2x-3$ by $(2x-1)$: $$ \begin{align*} \frac{2x^3+7x^2+2x-3}{2x-1}&=x^2+4x+3 \end{align*} $$

Step3: Factor quadratic

Factor $x^2+4x+3$: $x^2+4x+3=(x+1)(x+3)$

Step4: Find remaining roots

Set factors to 0: $x+1=0 \implies x=-1$; $x+3=0 \implies x=-3$

Answer:

$x = \frac{1}{2}$, $x = -1$, and $x = -3$