Let's solve the problem step-by-step by filling in the missing values for each polynomial factorization:
1. For the first polynomial [tex]\(x^2 + 2x - 15 = (ax + b)(cx + d)\)[/tex]:
[tex]\[
(1x - 3)(1x + 5) = x^2 + 5x - 3x - 15 = x^2 + 2x - 15
\][/tex]
From the calculation, we see that [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], [tex]\( c = 1 \)[/tex], and [tex]\( d = 5 \)[/tex].
2. For the second polynomial [tex]\(2x^3 + 6x^2 - 20x = 2x(ax + b)(cx + d)\)[/tex]:
[tex]\[
2x(x - 2)(x + 5) = 2x(x^2 + 3x - 10) = 2x^3 + 6x^2 - 20x
\][/tex]
From the calculation, we can deduce that [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], [tex]\( c = 1 \)[/tex], and [tex]\( d = 5 \)[/tex].
3. For the third polynomial [tex]\(9x^2 + 21x + 6 = (ax + b)(cx + d)\)[/tex]:
[tex]\[
(3x + 2)(3x + 1) = 9x^2 + 3x + 6x + 2 = 9x^2 + 21x + 6
\][/tex]
From the calculation, we find that [tex]\( a = 3 \)[/tex], [tex]\( b = 2 \)[/tex], [tex]\( c = 3 \)[/tex], and [tex]\( d = 1 \)[/tex].
Therefore, the complete table with the filled missing values is:
[tex]\[
\begin{tabular}{|c|c|c|c|}
\hline a & b & c & d \\
\hline 1. & 1 & -3 & 1 & 5 \\
\hline 2. & 1 & -2 & 1 & 5 \\
\hline 3. & 3 & 2 & 3 & 1 \\
\hline
\end{tabular}
\][/tex]
