Answered

Join the conversation on IDNLearn.com and get the answers you seek from experts. Our platform is designed to provide trustworthy and thorough answers to any questions you may have.

Three polynomials are factored below, but some coefficients and constants are missing. All of the missing values of [tex]\( a, b, c, \)[/tex] and [tex]\( d \)[/tex] are integers.

1. [tex]\( x^2 - 6x + 8 = (ax + b)(cx + d) \)[/tex]
2. [tex]\( 3x^3 - 6x^2 - 24x = 3x(ax + b)(cx + d) \)[/tex]
3. [tex]\( 2x^2 - 2x - 24 = (ax + b)(cx + d) \)[/tex]

Fill in the table with the missing values of [tex]\( a, b, c, \)[/tex] and [tex]\( d \)[/tex].

[tex]\[
\begin{array}{|c|c|c|c|c|}
\hline
& a & b & c & d \\
\hline
1. & 1 & -4 & 1 & -2 \\
\hline
2. & 3 & -8 & 1 & -4 \\
\hline
3. & 2 & 3 & -4 & 2 \\
\hline
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
Let's fill in the unknown coefficients and constants for each polynomial factoring step by step.

### 1. Factoring [tex]\(x^2 - 6x + 8\)[/tex]
We need to express the quadratic polynomial as [tex]\((ax + b)(cx + d)\)[/tex].

Given the solution:
[tex]\[ x^2 - 6x + 8 = (1x - 2)(1x - 4) \][/tex]

So:
[tex]\(a = 1\)[/tex], [tex]\(b = -2\)[/tex], [tex]\(c = 1\)[/tex], and [tex]\(d = -4\)[/tex].

### 2. Factoring [tex]\(3x^3 - 6x^2 - 24x\)[/tex]
We can start by factoring out [tex]\(3x\)[/tex]:
[tex]\[ 3x(x^2 - 2x - 8) \][/tex]

Now we need to factor [tex]\(x^2 - 2x - 8\)[/tex] as [tex]\((ax + b)(cx + d)\)[/tex].

Given the solution:
[tex]\[ 3x(x^2 - 2x - 8) = 3x(1x - 4)(1x + 2) \][/tex]

So:
[tex]\(a = 1\)[/tex], [tex]\(b = -4\)[/tex], [tex]\(c = 1\)[/tex], and [tex]\(d = 2\)[/tex].

### 3. Factoring [tex]\(2x^2 - 2x - 24\)[/tex]
We need to express the quadratic polynomial as [tex]\((ax + b)(cx + d)\)[/tex].

Given the solution:
[tex]\[ 2x^2 - 2x - 24 = (2x + 6)(1x - 4) \][/tex]

So:
[tex]\(a = 2\)[/tex], [tex]\(b = 6\)[/tex], [tex]\(c = 1\)[/tex], and [tex]\(d = -4\)[/tex].

### Filled Table
Here is the filled table for the given coefficients and constants:

[tex]\[ \begin{tabular}{|l|l|l|l|l|} \hline \# & a & b & c & d \\ \hline 1 & 1 & -2 & 1 & -4 \\ \hline 2 & 1 & -4 & 1 & 2 \\ \hline 3 & 2 & 6 & 1 & -4 \\ \hline \end{tabular} \][/tex]
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.