[48] Given that x = – 2 is a solution to 𝑥2+𝑏𝑥–6=0,x 2 +bx–6=0, which of the following polynomials is a factor of 𝑥2+𝑏𝑥–6?x 2 +bx–6?

1. Break Down the Problem

We are given that $x = -2$ is a solution to the polynomial equation $x^2 + bx - 6 = 0$. We need to identify a polynomial that is a factor of this equation.

2. Relevant Concepts

According to the Factor Theorem, if $x = r$ is a root of the polynomial $P(x)$, then $(x - r)$ is a factor of $P(x)$. Here, since $x = -2$ is a solution, we can express the factor associated with this solution as follows: $(x - (-2)) = (x + 2)$

3. Analysis and Detail

Since $x + 2$ is a factor, $P(x)$ can be expressed as: $P(x) = (x + 2)(Ax + B)$ where $A$ and $B$ are constants we need to determine.

Next, we can find $A$ and $B$ by expanding $(x + 2)(Ax + B)$ and equating it to $x^2 + bx - 6$:

1. Expanding: $P(x) = Ax^2 + Bx + 2Ax + 2B = Ax^2 + (B + 2A)x + 2B$

2. Since $P(x) = x^2 + bx - 6$, we have:

   • Coefficient of $x^2$: $A = 1$
   • Coefficient of $x$: $B + 2A = b$
   • Constant term: $2B = -6$

From $2B = -6$: $B = -3$ Substituting $A = 1$ into the second equation: $-3 + 2 \cdot 1 = b \implies b = -3 + 2 = -1$

Thus, the polynomial becomes: $x^2 - x - 6$

4. Verify and Summarize

To find if $x + 2$ is indeed a factor of $x^2 - x - 6$, we can factor $x^2 - x - 6$. The factors of $-6$ that add up to $-1$ are $-3$ and $2$: $x^2 - x - 6 = (x - 3)(x + 2)$

This shows that $x + 2$ is indeed a factor.

Final Answer

The polynomial $x + 2$ is a factor of $x^2 + bx - 6$.