Question

Determine the resulting polynomial:
f, of, x, equals, 5, x, minus, 2
f(x)=5x−2
g, of, x, equals, 10, x, squared, minus, 3, x, minus, 7
g(x)=10x
2
−3x−7
Find: , f, of, x, dot, g, of, x
Find: f(x)⋅g(x)

Answer 1

To find \( f(x) \cdot g(x) \), we need to first recall the expressions for \( f(x) \) and \( g(x) \):

\[
f(x) = 5x - 2
\]

\[
g(x) = 10x^2 - 3x - 7
\]

Now, we'll multiply \( f(x) \) and \( g(x) \):

\[
f(x) \cdot g(x) = (5x - 2)(10x^2 - 3x - 7)
\]

We'll distribute \( (5x - 2) \) across \( (10x^2 - 3x - 7) \).

1. **Multiply \( 5x \) by each term in \( g(x) \)**:

\[
5x \cdot 10x^2 = 50x^3
\]
\[
5x \cdot (-3x) = -15x^2
\]
\[
5x \cdot (-7) = -35x
\]

2. **Multiply \( -2 \) by each term in \( g(x) \)**:

\[
-2 \cdot 10x^2 = -20x^2
\]
\[
-2 \cdot (-3x) = 6x
\]
\[
-2 \cdot (-7) = 14
\]

Now, we can combine all the terms together:

\[
f(x) \cdot g(x) = 50x^3 + (-15x^2) + (-35x) + (-20x^2) + 6x + 14
\]

Next, we combine like terms:

- **Cubic term**: \( 50x^3 \)
- **Quadratic terms**: \( -15x^2 - 20x^2 = -35x^2 \)
- **Linear terms**: \( -35x + 6x = -29x \)
- **Constant term**: \( 14 \)

Putting it all together, we have:

\[
f(x) \cdot g(x) = 50x^3 - 35x^2 - 29x + 14
\]

Therefore, the resulting polynomial is:

\[
\boxed{50x^3 - 35x^2 - 29x + 14}
\]