Question

question 2 of 5
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when this polynomial is divided by ((x + 3)), the remainder is 0. what is the value of the polynomial’s constant term?
(3x^3 - 5x^2 - 47x + square)

Explanation:

Step1: Apply Remainder Theorem

Set $x + 3 = 0$, so $x = -3$. Substitute $x=-3$ into the polynomial, set equal to 0 (since remainder is 0).
Let the constant term be $k$. Then:
$$3(-3)^3 - 5(-3)^2 - 47(-3) + k = 0$$

Step2: Calculate each term

Compute each power and product:
$3(-3)^3 = 3(-27) = -81$
$-5(-3)^2 = -5(9) = -45$
$-47(-3) = 141$
Substitute back:
$$-81 - 45 + 141 + k = 0$$

Step3: Simplify left side

Combine the constants:
$-81 - 45 + 141 = 15$
So:
$$15 + k = 0$$

Step4: Solve for k

Isolate $k$ by subtracting 15:
$$k = -15$$

Answer:

$-15$