(x%2B4)(x%2B5)-in-standard-form.jpeg "(x+3)(x+4)(x+5) In Standard Form")
Expanding (x+3)(x+4)(x+5) into Standard Form
This article explores how to expand the expression (x+3)(x+4)(x+5) into its standard form, which is a polynomial with terms arranged in descending order of their exponents.
Step 1: Expand the first two factors
First, we focus on multiplying the first two factors, (x+3) and (x+4):
(x+3)(x+4) = x(x+4) + 3(x+4)
Expanding further:
x(x+4) + 3(x+4) = x² + 4x + 3x + 12
Combining like terms:
x² + 4x + 3x + 12 = x² + 7x + 12
Step 2: Multiply the result by the remaining factor
Now, we multiply the result from Step 1 (x² + 7x + 12) by the remaining factor (x+5):
(x² + 7x + 12)(x+5) = x(x² + 7x + 12) + 5(x² + 7x + 12)
Expanding further:
x(x² + 7x + 12) + 5(x² + 7x + 12) = x³ + 7x² + 12x + 5x² + 35x + 60
Combining like terms:
x³ + 7x² + 12x + 5x² + 35x + 60 = x³ + 12x² + 47x + 60
Final Answer
Therefore, the standard form of (x+3)(x+4)(x+5) is x³ + 12x² + 47x + 60.