Math Problem Statement

Factor the expression completely. (This type of expression arises in calculus when using the "Product Rule.") 5(x2 + 5)4(2x)(x − 3)4 + (x2 + 5)5(4)(x − 3)3

Solution

We are tasked with factoring the following expression completely:

5(x2+5)4(2x)(x3)4+(x2+5)5(4)(x3)35(x^2 + 5)^4(2x)(x - 3)^4 + (x^2 + 5)^5(4)(x - 3)^3

Step 1: Identify Common Factors

First, we look for common factors in both terms:

  • In the first term 5(x2+5)4(2x)(x3)45(x^2 + 5)^4(2x)(x - 3)^4:

    • The factors are 55, (x2+5)4(x^2 + 5)^4, 2x2x, and (x3)4(x - 3)^4.
  • In the second term (x2+5)5(4)(x3)3(x^2 + 5)^5(4)(x - 3)^3:

    • The factors are (x2+5)5(x^2 + 5)^5, 44, and (x3)3(x - 3)^3.

The common factors between the two terms are:

  • (x2+5)4(x^2 + 5)^4 (since the lower power is 44 and it appears in both terms)
  • (x3)3(x - 3)^3 (since the lower power is 33 and it appears in both terms)

We can factor out these common terms:

(x2+5)4(x3)3(x^2 + 5)^4(x - 3)^3

Step 2: Factor out the remaining terms

After factoring out (x2+5)4(x3)3(x^2 + 5)^4(x - 3)^3, we are left with:

(x2+5)4(x3)3[5(2x)(x3)+(x2+5)(4)](x^2 + 5)^4(x - 3)^3 \left[ 5(2x)(x - 3) + (x^2 + 5)(4) \right]

Step 3: Simplify the remaining expression

Now, simplify the expression inside the brackets:

5(2x)(x3)+(x2+5)(4)5(2x)(x - 3) + (x^2 + 5)(4)

Start by distributing:

  • 5(2x)(x3)=10x(x3)=10x230x5(2x)(x - 3) = 10x(x - 3) = 10x^2 - 30x
  • (x2+5)(4)=4x2+20(x^2 + 5)(4) = 4x^2 + 20

Now, add these two results:

10x230x+4x2+20=14x230x+2010x^2 - 30x + 4x^2 + 20 = 14x^2 - 30x + 20

Step 4: Final expression

The factored expression is:

(x2+5)4(x3)3(14x230x+20)(x^2 + 5)^4(x - 3)^3(14x^2 - 30x + 20)

Step 5: Optional Further Factoring

The quadratic term 14x230x+2014x^2 - 30x + 20 can be factored if possible. First, factor out a common factor of 2:

14x230x+20=2(7x215x+10)14x^2 - 30x + 20 = 2(7x^2 - 15x + 10)

We check whether 7x215x+107x^2 - 15x + 10 can be factored further. The discriminant is:

Δ=(15)24(7)(10)=225280=55\Delta = (-15)^2 - 4(7)(10) = 225 - 280 = -55

Since the discriminant is negative, the quadratic cannot be factored further using real numbers.

Final Factored Expression:

2(x2+5)4(x3)3(7x215x+10)2(x^2 + 5)^4(x - 3)^3(7x^2 - 15x + 10)

This is the completely factored form of the given expression.


Would you like more details or have any questions?

5 Related Questions:

  1. How do you apply the product rule in calculus?
  2. What are discriminants and how do they affect factoring?
  3. Can quadratic expressions with a negative discriminant be factored over the complex numbers?
  4. How do you recognize common factors in polynomial expressions?
  5. What are some examples of expressions that arise from applying the product rule?

Tip:

When factoring expressions with multiple terms, always check for common factors across all terms first before simplifying.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Factoring
Polynomial Expressions
Calculus

Formulas

(a^m)(a^n) = a^{m+n}
Distributive Property a(b + c) = ab + ac
Quadratic Formula: ax^2 + bx + c = 0

Theorems

Product Rule for Derivatives
Discriminant for Quadratic Equations

Suitable Grade Level

Grades 11-12, Calculus I