Factor $$ b^{2}-2 b^{2}+3 b-6 $$

Question

UpStudy AI · Accepted Answer

Let's factor the expression:

$$
b^{2} - 2b^{2} + 3b - 6
$$

First, simplify the expression by combining like terms:

$$
b^{2} - 2b^{2} + 3b - 6 = -b^{2} + 3b - 6
$$

To factor this quadratic expression, it's often helpful to factor out the leading coefficient (in this case, $$-1$$):

$$
-b^{2} + 3b - 6 = -\left(b^{2} - 3b + 6\right)
$$

Now, let's try to factor the quadratic inside the parentheses:

$$
b^{2} - 3b + 6
$$

To factor $$b^{2} - 3b + 6$$, we look for two numbers that multiply to $$6$$ and add to $$-3$$. However, there are no real numbers that satisfy these conditions because the discriminant ($$b^2 - 4ac$$) is negative:

$$
\text{Discriminant} = (-3)^2 - 4(1)(6) = 9 - 24 = -15
$$

Since the discriminant is negative, $$b^{2} - 3b + 6$$ does not factor over the real numbers.

**Conclusion:**

The expression cannot be factored further using real numbers. However, if you're working in the context of real numbers, you can express it as:

$$
-b^{2} + 3b - 6 = -\left(b^{2} - 3b + 6\right)
$$

If you need to factor it over the complex numbers, it would be:

$$
-b^{2} + 3b - 6 = -\left(b - \frac{3}{2} + \frac{\sqrt{15}}{2}i\right)\left(b - \frac{3}{2} - \frac{\sqrt{15}}{2}i\right)
$$

But typically, unless specified, factoring is done over the real numbers.

**Final Answer:**

The expression factors as –( b² – 3b + 6 ), which cannot be factored further using real numbers.
The expression $$ b^{2} - 2b^{2} + 3b - 6 $$ simplifies to $$ -b^{2} + 3b - 6 $$. This quadratic does not factor further using real numbers.

Factor \( b^{2}-2 b^{2}+3 b-6 \)

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