Math Problem Statement

6x((x^(3))+7)^(2)-(6x^(2))((3x^(2))+5)((x^(3))+7)

Solution

To simplify the expression 6x((x3+7)2)6x2((3x2+5))((x3+7))6x \left( (x^3 + 7)^2 \right) - 6x^2 \left( (3x^2 + 5) \right) \left( (x^3 + 7) \right), we can proceed with step-by-step factoring and distribution.

Expression:

6x((x3+7)2)6x2((3x2+5))((x3+7))6x \left( (x^3 + 7)^2 \right) - 6x^2 \left( (3x^2 + 5) \right) \left( (x^3 + 7) \right)

Step 1: Factor out the common terms.

Both terms contain a common factor of 6x(x3+7)6x(x^3 + 7), so we can factor it out:

6x(x3+7)((x3+7)x(3x2+5))6x(x^3 + 7) \left( (x^3 + 7) - x \cdot (3x^2 + 5) \right)

Step 2: Simplify inside the parentheses.

Now, simplify the expression inside the parentheses:

(x3+7)x(3x2+5)(x^3 + 7) - x(3x^2 + 5)

Distribute xx in the second part:

(x3+7)(3x3+5x)(x^3 + 7) - (3x^3 + 5x)

Now, combine like terms:

(x33x3)+(75x)=2x3+75x(x^3 - 3x^3) + (7 - 5x) = -2x^3 + 7 - 5x

Step 3: Substitute the simplified expression back:

Now substitute this result back into the factored form:

6x(x3+7)(2x3+75x)6x(x^3 + 7) (-2x^3 + 7 - 5x)

Final Expression:

Thus, the simplified expression is:

6x(x3+7)(2x3+75x)6x(x^3 + 7)(-2x^3 + 7 - 5x)

This is the factored form of the given expression.

Would you like further details or clarification on any step?

Here are 5 related questions:

  1. How can we expand the factored expression if needed?
  2. What is the degree of the polynomial in the final expression?
  3. Can you explain the process of combining like terms in more detail?
  4. How does factoring simplify polynomial expressions?
  5. What if the exponents or coefficients were different—how would that affect the result?

Tip: Always check for common factors before expanding or simplifying large algebraic expressions. Factoring can save time and prevent errors!

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomial Factoring
Distribution

Formulas

Factoring: ab - ac = a(b - c)
Distribution: a(b + c) = ab + ac

Theorems

Distributive Property

Suitable Grade Level

Grades 10-12