QuestionAugust 26, 2025

Write a possible equation for a polynomial whose graph has x-Axis intercepts at x=2,-(1)/(2),-3 P(x)=(x+2)(2x-1)(x-3) P(x)=(x+2)(x-2)(x-3) P(x)=(x-2)(2x+1)(x+3) P(x)=(x-3)(3x+1)(x+1)

Write a possible equation for a polynomial whose graph has x-Axis intercepts at x=2,-(1)/(2),-3 P(x)=(x+2)(2x-1)(x-3) P(x)=(x+2)(x-2)(x-3) P(x)=(x-2)(2x+1)(x+3) P(x)=(x-3)(3x+1)(x+1)
Write a possible equation for a polynomial whose graph
has x-Axis intercepts at x=2,-(1)/(2),-3
P(x)=(x+2)(2x-1)(x-3)
P(x)=(x+2)(x-2)(x-3)
P(x)=(x-2)(2x+1)(x+3)
P(x)=(x-3)(3x+1)(x+1)

Solution
4.5(290 votes)

Answer

P(x) = (x - 2)(2x + 1)(x + 3) Explanation 1. Identify the x-intercepts The given x-intercepts are x = 2, x = -\frac{1}{2}, and x = -3. 2. Formulate the polynomial equation Use the intercepts to form factors: (x - 2), (x + \frac{1}{2}), and (x + 3). 3. Construct the polynomial Multiply the factors: P(x) = (x - 2)(2x + 1)(x + 3).

Explanation

1. Identify the x-intercepts<br /> The given x-intercepts are $x = 2$, $x = -\frac{1}{2}$, and $x = -3$.<br /><br />2. Formulate the polynomial equation<br /> Use the intercepts to form factors: $(x - 2)$, $(x + \frac{1}{2})$, and $(x + 3)$.<br /><br />3. Construct the polynomial<br /> Multiply the factors: $P(x) = (x - 2)(2x + 1)(x + 3)$.
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