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(3x-2)(x+3)(2x+1)=0
How many distinct roots does the given equation have?
Choose 1 answer:
(A) Zero
(B) One
(c) Two
(D) Three

$(3 x-2)(x+3)(2 x+1)=0$ $\newline$ How many distinct roots does the given equation have? $\newline$ Choose $1$ answer: $\newline$ (A) Zero $\newline$ (B) One $\newline$ (c) Two $\newline$ (D) Three

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Find the roots of factored polynomials

Full solution

Q.

(3 x-2)(x+3)(2 x+1)=0

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How many distinct roots does the given equation have?

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Choose

1

answer:

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(A) Zero

\newline

(B) One

\newline

(c) Two

\newline

(D) Three

Equation factors: The given equation is a product of three factors set equal to zero: $(3x-2)(x+3)(2x+1)=0$ . To find the roots, we need to set each factor equal to zero and solve for $x$ .
Solving first factor: First factor: $3x - 2 = 0$ $\newline$ To solve for $x$ , we add $2$ to both sides of the equation: $\newline$ $3x - 2 + 2 = 0 + 2$ $\newline$ $3x = 2$ $\newline$ Now, we divide both sides by $3$ to isolate $x$ : $\newline$ $\frac{3x}{3} = \frac{2}{3}$ $\newline$ $x = \frac{2}{3}$ $\newline$ We have found the first root, $x = \frac{2}{3}$ .
Solving second factor: Second factor: $x + 3 = 0$ $\newline$ To solve for $x$ , we subtract $3$ from both sides of the equation: $\newline$ $x + 3 - 3 = 0 - 3$ $\newline$ $x = -3$ $\newline$ We have found the second root, $x = -3$ .
Solving third factor: Third factor: $2x + 1 = 0$ $\newline$ To solve for $x$ , we subtract $1$ from both sides of the equation: $\newline$ $2x + 1 - 1 = 0 - 1$ $\newline$ $2x = -1$ $\newline$ Now, we divide both sides by $2$ to isolate $x$ : $\newline$ $\frac{2x}{2} = \frac{-1}{2}$ $\newline$ $x = -\frac{1}{2}$ $\newline$ We have found the third root, $x = -\frac{1}{2}$ .
Number of distinct roots: We have found three distinct roots for the equation: $x = \frac{2}{3}$ , $x = -3$ , and $x = -\frac{1}{2}$ . Therefore, the correct answer to the question of how many distinct roots the equation has is $3$ .

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