Question

find the product of the integers below.
1 three more than four times a number and the same number minus 15.
2 the sum of two consecutive numbers and the smallest number plus its square minus two.

Explanation:

Sub - Question 1

Step 1: Define the variable

Let the number be \( x \). Then "three more than four times a number" can be written as \( 4x + 3 \), and "the same number minus 15" can be written as \( x-15 \).

Step 2: Find the product

We need to find the product of these two expressions, so we multiply them: \( (4x + 3)(x - 15) \)
Using the distributive property (FOIL method):
First terms: \( 4x\times x=4x^{2}\)
Outer terms: \( 4x\times(- 15)=-60x\)
Inner terms: \( 3\times x = 3x\)
Last terms: \( 3\times(-15)=-45\)
Now combine like terms: \( 4x^{2}-60x + 3x-45=4x^{2}-57x - 45\)

Step 1: Define the variable

Let the smallest of the two consecutive numbers be \( n \). Then the next consecutive number is \( n + 1 \).
"The sum of two consecutive numbers" is \( n+(n + 1)=2n + 1\)
"The smallest number plus its square minus two" is \( n + n^{2}-2=n^{2}+n - 2\)

Step 2: Find the product

We find the product of \( 2n + 1\) and \( n^{2}+n - 2\)
Using the distributive property:
\(2n(n^{2}+n - 2)+1\times(n^{2}+n - 2)\)
\(=2n^{3}+2n^{2}-4n+n^{2}+n - 2\)
Combine like terms: \(2n^{3}+(2n^{2}+n^{2})+(-4n + n)-2=2n^{3}+3n^{2}-3n - 2\)

Answer:

\(4x^{2}-57x - 45\)

Sub - Question 2