Question

Three polynomials are factored below, but some coefficients and constants are missing. All of the missing values of a, b, c, and d are integers.
x^2 - 6x + 8 = (ax + b)(cx + d)
3x^3 - 6x^2 - 24x = 3x(ax + b)(cx + d)
2x^2 - 2x - 24 = (ax + b)(cx + d)

Answer 1

To find the coefficients \(a\), \(b\), \(c\), and \(d\) in the given polynomial factorizations, let's analyze each polynomial one by one.

### 1. First Polynomial:
**Given:**
\[ x^2 - 6x + 8 = (ax + b)(cx + d) \]

**Expanding \( (ax + b)(cx + d) \):**
\[
(ac)x^2 + (ad + bc)x + (bd)
\]

**Comparing coefficients with \(x^2 - 6x + 8\):**
- Coefficient of \(x^2\): \(ac = 1\)
- Coefficient of \(x\): \(ad + bc = -6\)
- Constant term: \(bd = 8\)

From \(ac = 1\), we can have:
- \(a = 1, c = 1\) (or) \(a = -1, c = -1\)

Let's use \(a = 1\) and \(c = 1\) to find \(b\) and \(d\):
- This gives us \(1 \cdot d + 1 \cdot b = -6\) → \(d + b = -6\)
- \(bd = 8\)

From \(d + b = -6\),
let's express \(d\) in terms of \(b\):
\[
d = -6 - b
\]
Substituting into \(bd = 8\):
\[
b(-6 - b) = 8 \Rightarrow -6b - b^2 = 8 \Rightarrow b^2 + 6b + 8 = 0
\]
Using the quadratic formula:
\[
b = \frac{-6 \pm \sqrt{36 - 32}}{2} = \frac{-6 \pm 2}{2} = -2 \quad \text{or} \quad -4
\]

If \(b = -2\):
\[
d = -6 - (-2) = -4 \quad (b, d) = (-2, -4)
\]

If \(b = -4\):
\[
d = -6 - (-4) = -2 \quad (b, d) = (-4, -2)
\]

Thus, our values are either:
\((a, b, c, d) = (1, -2, 1, -4)\) or \((1, -4, 1, -2)\).

### 2. Second Polynomial:
**Given:**
\[ 3x^3 - 6x^2 - 24x = 3x(ax + b)(cx + d) \]

Factoring out the 3:
\[ x^3 - 2x^2 - 8x = (ax + b)(cx + d) \]

**Expanding:**
\[ acx^2 + (ad + bc)x + bd = x^3 - 2x^2 - 8x \]

**Comparing coefficients:**
- Coefficient of \(x^2\): \(ac = 1\)
- Coefficient of \(x\): \(ad + bc = -2\)
- Constant term: \(bd = 0\)

From \(bd = 0\), either \(b\) or \(d\) is zero. Let us set \(b = 0\) (we can check alternate cases later), yielding:
- Thus, \(bd = 0\) is satisfied.

Now substituting \(b = 0\) into equations:
1. \(ad = -2\)
2. \(ac = 1\)

From \(ac = 1\), with \(c = 1/a\), substituting in \(ad = -2\):
\[
ad = a\left(\frac{1}{a}\right)d = d = -2
\]
Here, we get \(d = -2\) when \(b = 0\). Therefore, from \(ac = 1\), we can choose \(a = 1\) and \(c = 1\), which gives:

The values become:
\((a, b, c, d) = (1, 0, 1, -2)\).

### 3. Third Polynomial:
**Given:**
\[ 2x^2 - 2x - 24 = (ax + b)(cx + d) \]

Dividing everything by 2, we have:
\[ x^2 - x - 12 = (ax + b)(cx + d) \]

**Expanding:**
\[ acx^2 + (ad + bc)x + bd \]

**Comparing coefficients:**
1. \(ac = 1\)
2. \(ad + bc = -1\)
3. \(bd = -12\)

Again, \(ac = 1\) implies \(a\) and \(c\) can be \(1\) or \(-1\). Using \(a = 1\), we yield:
\[
d + b = -1 \quad \text{(from \(ad+bc = -1\) results in this)}
\]

So, like before, let \(d = -1 - b\) and substituting:
\[
b(-1 - b) = -12 \Rightarrow -b - b^2 = -12 \Rightarrow b^2 + b - 12 = 0
\]

Factoring the quadratic gives:
\[
(b - 3)(b + 4) = 0
\]
Thus, \(b = 3\) or \(b = -4\).

If \(b = 3\):
\[
d = -1 - 3 = -4
\]

If \(b = -4\):
\[
d = -1 - (-4) = 3
\]

Now summarizing the values found:
1. \((1, -2, 1, -4)\) or \((1, -4, 1, -2)\) from the first.
2. \((1, 0, 1, -2)\) from the second.
3. \((1, 3, 1, -4)\) or \((1, -4, 1, 3)\) from the third.

**Final integers are:**
1. **From first equation**: \(a = 1\), \(b = -2\), \(c = 1\), \(d = -4\) (or similar variants).
2. **From second**: \(a = 1, b = 0, c = 1, d = -2\).
3. **From third**: Check \(a = 1, b = 3, d = -4\) as compatibly possible with previous results.

This concludes the evaluation of \(a, b, c, d\) for the polynomial equations presented.