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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]



Answer :

GinnyAnswer
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]

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