Questions: 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. 1. x^2-2x-15=(ax+b)(cx+d) 2. 2x^3-6x^2-36x=2x(ax+b)(cx+d) 3. 4x^2+10x+4=(ax+b)(cx+d) Fill in the table with the missing values of a, b, c, and d. a b c d 1 -5 1 2. 1 -6 3. 1 2

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.
1. x^2-2x-15=(ax+b)(cx+d)
2. 2x^3-6x^2-36x=2x(ax+b)(cx+d)
3. 4x^2+10x+4=(ax+b)(cx+d)

Fill in the table with the missing values of a, b, c, and d.

a b c d
1 -5 1 
2. 1   -6
3.  1 2
Transcript text: 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. 1. $x^{2}-2 x-15=(a x+b)(c x+d)$ 2. $2 x^{3}-6 x^{2}-36 x=2 x(a x+b)(c x+d)$ 3. $4 x^{2}+10 x+4=(a x+b)(c x+d)$ Fill in the table with the missing values of $a, b, c$, and $d$. a b c d \begin{tabular}{|c|c|c|c|} \hline 1 & -5 & 1 & \\ \hline 2. 1 & & & -6 \\ \hline 3. & 1 & 2 & \\ \hline \end{tabular}
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Solution

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Solution Steps

To solve for the missing coefficients and constants in the given polynomial factorizations, we need to expand the factored forms and match the coefficients with the original polynomials. This will allow us to solve for the unknowns.

  1. For the first polynomial, expand \((a x + b)(c x + d)\) and match the coefficients with \(x^2 - 2x - 15\).
  2. For the second polynomial, factor out the common term and then expand \((a x + b)(c x + d)\) and match the coefficients with \(2 x^3 - 6 x^2 - 36 x\).
  3. For the third polynomial, expand \((a x + b)(c x + d)\) and match the coefficients with \(4 x^2 + 10 x + 4\).
Step 1: Expand and Match Coefficients for the First Polynomial

Given: \[ x^2 - 2x - 15 = (a x + b)(c x + d) \]

Expanding the right-hand side: \[ (a x + b)(c x + d) = a c x^2 + (a d + b c) x + b d \]

Matching coefficients with \( x^2 - 2x - 15 \): \[ a c = 1 \] \[ a d + b c = -2 \] \[ b d = -15 \]

From the given table, we have \( a = 1 \) and \( b = -5 \). Substituting these values: \[ 1 \cdot c = 1 \implies c = 1 \] \[ 1 \cdot d + (-5) \cdot 1 = -2 \implies d - 5 = -2 \implies d = 3 \]

Step 2: Expand and Match Coefficients for the Second Polynomial

Given: \[ 2 x^3 - 6 x^2 - 36 x = 2 x (a x + b)(c x + d) \]

Factoring out the common term: \[ 2 x (a x + b)(c x + d) = 2 x (a c x^2 + (a d + b c) x + b d) \]

Matching coefficients with \( 2 x^3 - 6 x^2 - 36 x \): \[ 2 a c = 2 \implies a c = 1 \] \[ 2 (a d + b c) = -6 \implies a d + b c = -3 \] \[ 2 b d = -36 \implies b d = -18 \]

From the given table, we have \( a = 1 \) and \( d = -6 \). Substituting these values: \[ 1 \cdot c = 1 \implies c = 1 \] \[ 1 \cdot (-6) + b \cdot 1 = -3 \implies -6 + b = -3 \implies b = 3 \]

Step 3: Expand and Match Coefficients for the Third Polynomial

Given: \[ 4 x^2 + 10 x + 4 = (a x + b)(c x + d) \]

Expanding the right-hand side: \[ (a x + b)(c x + d) = a c x^2 + (a d + b c) x + b d \]

Matching coefficients with \( 4 x^2 + 10 x + 4 \): \[ a c = 4 \] \[ a d + b c = 10 \] \[ b d = 4 \]

From the given table, we have \( b = 1 \) and \( c = 2 \). Substituting these values: \[ a \cdot 2 = 4 \implies a = 2 \] \[ 2 \cdot d + 1 \cdot 2 = 10 \implies 2 d + 2 = 10 \implies 2 d = 8 \implies d = 4 \]

Final Answer

\[ \begin{array}{|c|c|c|c|} \hline 1 & -5 & 1 & 3 \\ \hline 2 & 1 & 3 & -6 \\ \hline 3 & 2 & 1 & 4 \\ \hline \end{array} \]

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