Question
Write as a polynomial. \[ \begin{array}{l}(x+3)(x+7)= \\ (x-2)(x+3)= \\ (2 a-3)(3 a-4)=\end{array} \]
Ask by Johnson Blake. in the United States
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
\[
\begin{array}{l}
(x + 3)(x + 7) = x^2 + 10x + 21 \\
(x - 2)(x + 3) = x^2 + x - 6 \\
(2a - 3)(3a - 4) = 6a^2 - 17a + 12
\end{array}
\]
Solution
Sure! Let's expand each of the given expressions into polynomial form.
1. **\((x + 3)(x + 7)\)**
\[
\begin{align*}
(x + 3)(x + 7) &= x \cdot x + x \cdot 7 + 3 \cdot x + 3 \cdot 7 \\
&= x^2 + 7x + 3x + 21 \\
&= x^2 + 10x + 21
\end{align*}
\]
2. **\((x - 2)(x + 3)\)**
\[
\begin{align*}
(x - 2)(x + 3) &= x \cdot x + x \cdot 3 - 2 \cdot x - 2 \cdot 3 \\
&= x^2 + 3x - 2x - 6 \\
&= x^2 + x - 6
\end{align*}
\]
3. **\((2a - 3)(3a - 4)\)**
\[
\begin{align*}
(2a - 3)(3a - 4) &= 2a \cdot 3a + 2a \cdot (-4) - 3 \cdot 3a - 3 \cdot (-4) \\
&= 6a^2 - 8a - 9a + 12 \\
&= 6a^2 - 17a + 12
\end{align*}
\]
---
**Final Expanded Polynomials:**
\[
\begin{array}{l}
(x + 3)(x + 7) = x^2 + 10x + 21 \\
(x - 2)(x + 3) = x^2 + x - 6 \\
(2a - 3)(3a - 4) = 6a^2 - 17a + 12
\end{array}
\]
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Simplify this solution The Deep Dive
To write the given expressions as polynomials, we'll use the distributive property (also known as the FOIL method for binomials). 1. For \((x+3)(x+7)\): \[ (x+3)(x+7) = x^2 + 7x + 3x + 21 = x^2 + 10x + 21 \] 2. For \((x-2)(x+3)\): \[ (x-2)(x+3) = x^2 + 3x - 2x - 6 = x^2 + x - 6 \] 3. For \((2a-3)(3a-4)\): \[ (2a-3)(3a-4) = 2a \cdot 3a + 2a \cdot (-4) - 3 \cdot 3a + (-3) \cdot (-4) = 6a^2 - 8a - 9a + 12 = 6a^2 - 17a + 12 \] Putting it all together, we have: \[ \begin{array}{l} (x+3)(x+7) = x^2 + 10x + 21, \\ (x-2)(x+3) = x^2 + x - 6, \\ (2a-3)(3a-4) = 6a^2 - 17a + 12. \end{array} \]
