A polynomial p has zeros when x=-2,x=(1)/(3), and x=3. What could be the equation of p ? Choose 1 answer: (A) p(x)=(x+2)(x+3)(3x+1) (B) p(x)=(x+2)(x+3)(3x-1) (C) p(x)=(x+2)(x-3)(3x-1) (D) p(x)=(x-2)(x+3)(3x+1)

Identify Zeros: Zeros are x=-2, x=1/3, and x=3. The factors of p(x) will be (x+2), (x-1/3), and (x-3).Eliminate Fraction: To get rid of the fraction in the second factor, multiply by 3 to get 3(x-1/3) which simplifies to (3x-1).Formulate Equation: The equation of p is p(x)=(x+2)(3x-1)(x-3).Match with Options: Match the equation with the given options. The correct option is (C) p(x)=(x+2)(x-3)(3x-1).

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A polynomial
p has zeros when
x=-2,x=(1)/(3), and
x=3.
What could be the equation of
p ?
Choose 1 answer:
(A)
p(x)=(x+2)(x+3)(3x+1)
(B)
p(x)=(x+2)(x+3)(3x-1)
(C)
p(x)=(x+2)(x-3)(3x-1)
(D)
p(x)=(x-2)(x+3)(3x+1)

A polynomial $p$ has zeros when $x=-2, x=\frac{1}{3}$ , and $x=3$ . $\newline$ What could be the equation of $p$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $p(x)=(x+2)(x+3)(3 x+1)$ $\newline$ (B) $p(x)=(x+2)(x+3)(3 x-1)$ $\newline$ (C) $p(x)=(x+2)(x-3)(3 x-1)$ $\newline$ (D) $p(x)=(x-2)(x+3)(3 x+1)$

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Math Problems

Algebra 2

Write a quadratic function from its x-intercepts and another point

Full solution

Q. A polynomial

p

has zeros when

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

, and

x=3

\newline

What could be the equation of

p

\newline

Choose

1

answer:

\newline

(A)

p(x)=(x+2)(x+3)(3 x+1)

\newline

(B)

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

\newline

(C)

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

\newline

(D)

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

Identify Zeros: Zeros are $x=-2$ , $x=\frac{1}{3}$ , and $x=3$ . The factors of $p(x)$ will be $(x+2)$ , $(x-\frac{1}{3})$ , and $(x-3)$ .
Eliminate Fraction: To get rid of the fraction in the second factor, multiply by $3$ to get $3(x-\frac{1}{3})$ which simplifies to $(3x-1)$ .
Formulate Equation: The equation of $p$ is $p(x)=(x+2)(3x-1)(x-3)$ .
Match with Options: Match the equation with the given options. The correct option is (C) $p(x)=(x+2)(x-3)(3x-1)$ .