High School

Multiply: [tex](9x+7)(3x^2+5x-1)[/tex]

A. [tex]27x^3+66x^2+44x+7[/tex]
B. [tex]3x^2+14x+6[/tex]
C. [tex]27x^3+45x^2-7[/tex]
D. [tex]27x^3+66x^2+26x-7[/tex]

Answer :

To multiply the expressions

$$
(9x+7)(3x^2+5x-1),
$$

we can use the distributive property (also known as the FOIL method for binomials). Here is a step-by-step breakdown:

1. First, multiply the term $9x$ by each term in the second polynomial:

$$
9x \cdot 3x^2 = 27x^3,
$$
$$
9x \cdot 5x = 45x^2,
$$
$$
9x \cdot (-1) = -9x.
$$

So, the product of $9x$ and $(3x^2+5x-1)$ is:

$$
27x^3 + 45x^2 - 9x.
$$

2. Next, multiply the term $7$ by each term in the second polynomial:

$$
7 \cdot 3x^2 = 21x^2,
$$
$$
7 \cdot 5x = 35x,
$$
$$
7 \cdot (-1) = -7.
$$

So, the product of $7$ and $(3x^2+5x-1)$ is:

$$
21x^2 + 35x - 7.
$$

3. Now, add the two results together:

$$
(27x^3 + 45x^2 - 9x) + (21x^2 + 35x - 7).
$$

4. Combine like terms:

- The $x^3$ term: $27x^3$.
- The $x^2$ terms: $45x^2 + 21x^2 = 66x^2$.
- The $x$ terms: $-9x + 35x = 26x$.
- The constant term: $-7$.

This gives:

$$
27x^3 + 66x^2 + 26x - 7.
$$

Hence, the multiplied expression is:

$$
\boxed{27x^3 + 66x^2 + 26x - 7}.
$$