Question

use the remainder theorem to find which of the following is not a factor of the polynomial $x^3 + 3x^2 - 18x - 40$
a) $x + 5$
b) $x + 2$
c) $x - 4$
d) $x - 5$

Explanation:

Step1: Recall Remainder Theorem

For a polynomial \( f(x) \), if \( (x - a) \) is a factor, then \( f(a) = 0 \). For \( (x + b) \), it's equivalent to \( (x - (-b)) \), so we check \( f(-b) = 0 \). The polynomial is \( f(x)=x^{3}+3x^{2}-18x - 40 \).

Step2: Check Option A (\( x + 5 \), so \( a=-5 \))

Calculate \( f(-5) \):
\( f(-5)=(-5)^{3}+3(-5)^{2}-18(-5)-40 \)
\( = -125 + 75 + 90 - 40 \)
\( = (-125 + 75)+(90 - 40) \)
\( = -50 + 50 = 0 \). So \( x + 5 \) is a factor.

Step3: Check Option B (\( x + 2 \), so \( a=-2 \))

Calculate \( f(-2) \):
\( f(-2)=(-2)^{3}+3(-2)^{2}-18(-2)-40 \)
\( = -8 + 12 + 36 - 40 \)
\( = (-8 + 12)+(36 - 40) \)
\( = 4 - 4 = 0 \). So \( x + 2 \) is a factor.

Step4: Check Option C (\( x - 4 \), so \( a = 4 \))

Calculate \( f(4) \):
\( f(4)=(4)^{3}+3(4)^{2}-18(4)-40 \)
\( = 64 + 48 - 72 - 40 \)
\( = (64 + 48)-(72 + 40) \)
\( = 112 - 112 = 0 \). Wait, no, wait, recalculate:
Wait, \( 4^3=64 \), \( 3\times16 = 48 \), \( -18\times4=-72 \), \( -40 \). So \( 64 + 48 = 112 \), \( -72 - 40=-112 \), \( 112 - 112 = 0 \). Wait, but let's check Option D.

Step5: Check Option D (\( x - 5 \), so \( a = 5 \))

Calculate \( f(5) \):
\( f(5)=(5)^{3}+3(5)^{2}-18(5)-40 \)
\( = 125 + 75 - 90 - 40 \)
\( = (125 + 75)-(90 + 40) \)
\( = 200 - 130 = 70
eq 0 \). So \( x - 5 \) is not a factor. Wait, but earlier Option C: Wait, maybe I made a mistake. Wait the polynomial is \( x^{3}+3x^{2}-18x - 40 \). Let's factor it. Wait, maybe the original polynomial was misread. Wait the user's image: "x³+3x²-18x-40". Let's try factoring. Let's use rational root theorem. Possible roots: \( \pm1, \pm2, \pm4, \pm5, \pm8, \pm10, \pm20, \pm40 \). We saw \( f(-5)=0 \), \( f(-2)=0 \), \( f(4)=0 \)? Wait, no, \( f(4)=64 + 48 - 72 - 40 = 64+48=112; 72+40=112; 112-112=0 \). Wait, but then the factors would be \( (x + 5)(x + 2)(x - 4) \). Let's multiply: \( (x + 5)(x + 2)=x^{2}+7x + 10 \), then multiply by \( (x - 4) \): \( x^{3}-4x^{2}+7x^{2}-28x + 10x - 40 = x^{3}+3x^{2}-18x - 40 \). Yes, so \( x - 4 \) is a factor. Then Option D: \( x - 5 \), \( f(5)=125 + 75 - 90 - 40 = 70
eq 0 \). Wait, but the options are A)x+5, B)x+2, C)x-4, D)x-5. So the one that is not a factor is D)x - 5? Wait, but earlier when I checked Option C, I thought it was 0, but let's confirm. Wait, maybe the original problem had a typo, but according to the calculations:

Wait, let's recheck Option D: \( x - 5 \), so substitute \( x = 5 \) into \( f(x) \):

\( f(5) = 5^3 + 3(5)^2 - 185 - 40 = 125 + 75 - 90 - 40 = (125 + 75) - (90 + 40) = 200 - 130 = 70
eq 0 \).

Option A: \( x = -5 \), \( f(-5) = -125 + 75 + 90 - 40 = 0 \).

Option B: \( x = -2 \), \( f(-2) = -8 + 12 + 36 - 40 = 0 \).

Option C: \( x = 4 \), \( f(4) = 64 + 48 - 72 - 40 = 0 \).

So the one that is not a factor is D)x - 5. Wait, but the options are A)x+5, B)x+2, C)x-4, D)x-5. So the answer is D)x - 5.

Answer:

D. \( x - 5 \)