Question

1. solve the equation $2x^3 - x^2 - 32x = -16$ in the complex number system.\

\bigcirc\\ $x = 2, x = 1, x = -1$\
\bigcirc\\ $x = \frac{2}{3}, x = -\frac{2}{3}, x = 1$\
\bigcirc\\ $x = \frac{1}{2}, x = 4, x = -4$\
\bigcirc\\ $x = 2, x = -2, x = 4$

Explanation:

Step 1: Rearrange the equation

First, we rewrite the given equation \( 2x^{3}-x^{2}-32x=- 16 \) in standard form by adding 16 to both sides:
\( 2x^{3}-x^{2}-32x + 16=0 \)

Step 2: Factor by grouping

Group the terms as follows:
\( (2x^{3}-x^{2})+(-32x + 16)=0 \)
Factor out the greatest common factor from each group. From the first group \( 2x^{3}-x^{2} \), we can factor out \( x^{2} \), and from the second group \( - 32x + 16 \), we can factor out - 16:
\( x^{2}(2x - 1)-16(2x - 1)=0 \)
Now, we can factor out the common binomial factor \( (2x - 1) \):
\( (2x - 1)(x^{2}-16)=0 \)

Step 3: Factor the difference of squares

We know that \( a^{2}-b^{2}=(a + b)(a - b) \). For \( x^{2}-16 \), where \( a=x \) and \( b = 4 \), we have:
\( x^{2}-16=(x + 4)(x - 4) \)
So the factored form of the equation is:
\( (2x - 1)(x + 4)(x - 4)=0 \)

Step 4: Solve for \( x \)

Set each factor equal to zero and solve for \( x \):

• For \( 2x-1=0 \), add 1 to both sides: \( 2x=1 \), then divide by 2: \( x=\frac{1}{2} \)
• For \( x + 4=0 \), subtract 4 from both sides: \( x=-4 \)
• For \( x - 4=0 \), add 4 to both sides: \( x = 4 \)

Answer:

\( x=\frac{1}{2},x = 4,x=-4 \)