Question

question
use synthetic division to find all the possible factors of this polynomial.
p(x) = x⁴ + 4x³ − 7x² − 22x + 24
select all the correct expressions in the table. for help, see this worked example

x - 1 \t x + 1 \t x - 2
x + 2 \t x - 3 \t x + 3
x - 4 \t x + 4 \t x - 6

Explanation:

Step1: Recall Rational Root Theorem

The possible rational roots of a polynomial \( p(x) = a_nx^n + \dots + a_1x + a_0 \) are of the form \( \frac{\text{factors of } a_0}{\text{factors of } a_n} \). For \( p(x)=x^4 + 4x^3 - 7x^2 - 22x + 24 \), \( a_n = 1 \) and \( a_0 = 24 \). So possible roots are \( \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24 \), and possible linear factors are \( x - k \) where \( k \) is a root. We use synthetic division to test these.

Step2: Test \( x - 1 \) (root \( k = 1 \))

Set up synthetic division:
Coefficients: \( 1, 4, -7, -22, 24 \)
Bring down 1. Multiply by 1: 1. Add to 4: 5. Multiply by 1: 5. Add to -7: -2. Multiply by 1: -2. Add to -22: -24. Multiply by 1: -24. Add to 24: 0. Wait, no, wait: Wait, \( 1\times1 = 1 \), \( 4 + 1 = 5 \); \( 5\times1 = 5 \), \( -7 + 5 = -2 \); \( -2\times1 = -2 \), \( -22 + (-2) = -24 \); \( -24\times1 = -24 \), \( 24 + (-24) = 0 \). Wait, that gives remainder 0? Wait, no, let's recalculate. Wait, \( p(1) = 1 + 4 - 7 - 22 + 24 = 0 \). So \( x - 1 \) is a factor. Wait, but let's check others.

Wait, maybe I made a mistake. Let's test \( x = 1 \): \( 1 + 4 -7 -22 +24 = (1+4) + (-7-22) +24 = 5 -29 +24 = 0 \). So \( x - 1 \) is a factor.

Test \( x = -1 \): \( p(-1) = 1 - 4 -7 +22 +24 = (1 -4) + (-7 +22) +24 = -3 +15 +24 = 36 eq 0 \), so \( x + 1 \) not a factor.

Test \( x = 2 \): \( p(2) = 16 + 32 - 28 - 44 +24 = (16+32) + (-28-44) +24 = 48 -72 +24 = 0 \). So \( x - 2 \) is a factor.

Test \( x = -2 \): \( p(-2) = 16 - 32 - 28 + 44 +24 = (16 -32) + (-28 +44) +24 = -16 +16 +24 = 24 eq 0 \)? Wait, no: \( (-2)^4 +4(-2)^3 -7(-2)^2 -22(-2) +24 = 16 -32 -28 +44 +24 = (16 -32) + (-28 +44) +24 = -16 +16 +24 = 24 eq 0 \). Wait, maybe miscalculation. Wait, \( (-2)^4 = 16 \), \( 4(-2)^3 = 4(-8) = -32 \), \( -7(-2)^2 = -7(4) = -28 \), \( -22(-2) = 44 \), \( +24 \). So 16 -32 = -16; -16 -28 = -44; -44 +44 = 0; 0 +24 = 24. So remainder 24, so \( x + 2 \) not a factor? Wait, but let's check \( x = -3 \): \( p(-3) = 81 + 4(-27) -7(9) -22(-3) +24 = 81 -108 -63 +66 +24 = (81 -108) + (-63 +66) +24 = -27 +3 +24 = 0 \). So \( x + 3 \) is a factor (since root is -3, factor is \( x - (-3) = x + 3 \)).

Test \( x = 3 \): \( p(3) = 81 + 4(27) -7(9) -22(3) +24 = 81 +108 -63 -66 +24 = (81+108) + (-63-66) +24 = 189 -129 +24 = 84 eq 0 \), so \( x - 3 \) not a factor.

Test \( x = 4 \): \( p(4) = 256 + 4(64) -7(16) -22(4) +24 = 256 +256 -112 -88 +24 = (256+256) + (-112-88) +24 = 512 -200 +24 = 336 eq 0 \), so \( x - 4 \) not a factor.

Test \( x = -4 \): \( p(-4) = 256 + 4(-64) -7(16) -22(-4) +24 = 256 -256 -112 +88 +24 = (256-256) + (-112+88) +24 = 0 -24 +24 = 0 \). Wait, \( p(-4) = 0 \)? Let's recalculate: \( (-4)^4 = 256 \), \( 4(-4)^3 = 4(-64) = -256 \), \( -7(-4)^2 = -7(16) = -112 \), \( -22(-4) = 88 \), \( +24 \). So 256 -256 = 0; 0 -112 = -112; -112 +88 = -24; -24 +24 = 0. So \( x + 4 \) is a factor (root -4, factor \( x - (-4) = x + 4 \))? Wait, but earlier when I tested \( x = -2 \), I got 24, but let's check again. Wait, maybe I missed some. Wait, let's list all possible roots and test:

Possible roots: \( \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24 \).

We found \( p(1) = 0 \), \( p(2) = 0 \), \( p(-3) = 0 \), \( p(-4) = 0 \). Let's confirm \( p(1) = 1 +4 -7 -22 +24 = 0 \), correct. \( p(2) = 16 + 32 -28 -44 +24 = 0 \), correct. \( p(-3) = 81 -108 -63 +66 +24 = 0 \), correct. \( p(-4) = 256 -256 -112 +88 +24 = 0 \), correct.

So the factors from these roots are \( x - 1 \) (root 1), \( x - 2 \) (root 2), \( x + 3 \) (root -3), \( x + 4 \) (root -4). Wait, but let's check the given options:

Options: \( x - 1 \), \( x + 1 \), \( x - 2 \), \( x + 2 \), \( x - 3 \), \( x + 3 \), \( x - 4 \), \( x + 4 \), \( x - 6 \).

So from our tests, \( x - 1 \) (root 1), \( x - 2 \) (root 2), \( x + 3 \) (root -3), \( x + 4 \) (root -4) are factors. Wait, but let's check \( x - 1 \): yes, \( p(1)=0 \). \( x - 2 \): \( p(2)=0 \). \( x + 3 \): \( p(-3)=0 \). \( x + 4 \): \( p(-4)=0 \). Let's check \( x - 4 \): \( p(4)=336 eq 0 \), so no. \( x + 2 \): \( p(-2)=24 eq 0 \), no. \( x - 3 \): \( p(3)=84 eq 0 \), no. \( x + 1 \): \( p(-1)=36 eq 0 \), no. \( x - 6 \): \( p(6)=1296 + 4(216) -7(36) -22(6) +24 = 1296 + 864 -252 -132 +24 = 1800 - 384 +24 = 1440 eq 0 \), no.

Wait, but let's do synthetic division for \( x - 1 \):

Coefficients: 1 (x⁴), 4 (x³), -7 (x²), -22 (x), 24 (constant)

Bring down 1. Multiply by 1: 1. Add to 4: 5. Multiply by 1: 5. Add to -7: -2. Multiply by 1: -2. Add to -22: -24. Multiply by 1: -24. Add to 24: 0. So remainder 0, so \( x - 1 \) is a factor.

Synthetic division for \( x - 2 \):

Bring down 1. Multiply by 2: 2. Add to 4: 6. Multiply by 2: 12. Add to -7: 5. Multiply by 2: 10. Add to -22: -12. Multiply by 2: -24. Add to 24: 0. Remainder 0, so \( x - 2 \) is a factor.

Synthetic division for \( x + 3 \) (root -3, so use -3 in synthetic division):

Bring down 1. Multiply by -3: -3. Add to 4: 1. Multiply by -3: -3. Add to -7: -10. Multiply by -3: 30. Add to -22: 8. Multiply by -3: -24. Add to 24: 0. Remainder 0, so \( x + 3 \) is a factor.

Synthetic division for \( x + 4 \) (root -4, use -4):

Bring down 1. Multiply by -4: -4. Add to 4: 0. Multiply by -4: 0. Add to -7: -7. Multiply by -4: 28. Add to -22: 6. Multiply by -4: -24. Add to 24: 0. Remainder 0, so \( x + 4 \) is a factor.

So the correct factors from the given options are \( x - 1 \), \( x - 2 \), \( x + 3 \), \( x + 4 \).

Answer:

A. \( x - 1 \)
C. \( x - 2 \)
F. \( x + 3 \)
H. \( x + 4 \)

(Note: Assuming the options are labeled as A: \( x - 1 \), B: \( x + 1 \), C: \( x - 2 \), D: \( x + 2 \), E: \( x - 3 \), F: \( x + 3 \), G: \( x - 4 \), H: \( x + 4 \), I: \( x - 6 \))