Question

9 which expression is a factor of $x^3 + 2x^2 - 5x - 6$? a $x - 3$ b $x + 1$ c $x - 1$ d $x + 2$

Explanation:

Step1: Recall the Factor Theorem

The Factor Theorem states that if \( (x - a) \) is a factor of a polynomial \( f(x) \), then \( f(a)=0 \). For a factor of the form \( (x + b) \), this is equivalent to \( f(-b)=0 \) (since \( x + b=x-(-b) \)).

Step2: Define the polynomial

Let \( f(x)=x^{3}+2x^{2}-5x - 6 \). We will test each option by substituting the appropriate value into \( f(x) \) and checking if the result is 0.

Step3: Test Option A (\( x - 3 \))

For \( x - 3 \), we substitute \( x = 3 \) into \( f(x) \):
\( f(3)=(3)^{3}+2(3)^{2}-5(3)-6=27 + 18-15 - 6=24
eq0 \). So \( x - 3 \) is not a factor.

Step4: Test Option B (\( x + 1 \))

For \( x + 1 \), we substitute \( x=- 1 \) into \( f(x) \):
\( f(-1)=(-1)^{3}+2(-1)^{2}-5(-1)-6=-1 + 2 + 5-6=0 \). Since \( f(-1) = 0 \), by the Factor Theorem, \( x + 1 \) is a factor.

(We can also test the other options for completeness, but since we found a factor, we can stop here. However, for thoroughness:

Step5: Test Option C (\( x - 1 \))

Substitute \( x = 1 \) into \( f(x) \):
\( f(1)=(1)^{3}+2(1)^{2}-5(1)-6=1 + 2-5 - 6=-8
eq0 \). So \( x - 1 \) is not a factor.

Step6: Test Option D (\( x + 2 \))

Substitute \( x=-2 \) into \( f(x) \):
\( f(-2)=(-2)^{3}+2(-2)^{2}-5(-2)-6=-8 + 8 + 10-6=4
eq0 \). So \( x + 2 \) is not a factor.)

Answer:

B. \( x + 1 \)