Question

What is the remainder of the quotient x4+x3−64x2+76x+41÷x+9 ?(1 point) Responses −9 negative 9 - no response given 1 1 - no response given 5 5 - no response given 4

Answer 1

To find the remainder of the polynomial \( x^4 + x^3 - 64x^2 + 76x + 41 \) when divided by \( x + 9 \), we can apply the Remainder Theorem. The Remainder Theorem states that the remainder of the polynomial \( f(x) \) when divided by \( x - a \) is \( f(a) \).

In this case, we need to evaluate the polynomial at \( x = -9 \) (since we are dividing by \( x + 9 \)).

Let's denote the polynomial as:
\[
f(x) = x^4 + x^3 - 64x^2 + 76x + 41
\]

Now, we calculate \( f(-9) \):

\[
f(-9) = (-9)^4 + (-9)^3 - 64(-9)^2 + 76(-9) + 41
\]

Calculating each term:

1. \( (-9)^4 = 6561 \)
2. \( (-9)^3 = -729 \)
3. \( -64(-9)^2 = -64 \times 81 = -5184 \)
4. \( 76(-9) = -684 \)
5. Constant term = 41

Now, substitute these values into \( f(-9) \):

\[
f(-9) = 6561 - 729 - 5184 - 684 + 41
\]

Calculating this step-by-step:

1. \( 6561 - 729 = 5832 \)
2. \( 5832 - 5184 = 648 \)
3. \( 648 - 684 = -36 \)
4. \( -36 + 41 = 5 \)

Thus, the remainder when dividing \( x^4 + x^3 - 64x^2 + 76x + 41 \) by \( x + 9 \) is \( 5 \).

So the answer is \( \boxed{5} \).