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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]\(x^2 + 2x - 15 = (ax + b)(cx + d)\)[/tex]
2. [tex]\(2x^3 + 6x^2 - 20x = 2x(ax + b)(cx + d)\)[/tex]
3. [tex]\(9x^2 + 21x + 6 = (ax + b)(cx + d)\)[/tex]

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

[tex]\[
\begin{tabular}{|c|c|c|c|}
\hline
a & b & c & d \\
\hline
1. & 1 & -3 & 1 & 5 \\
\hline
2. & & & 5 \\
\hline
3. & 6 & 3 & \\
\hline
\end{tabular}
\][/tex]

Sagot :

GinnyAnsweridn
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]
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