Math Problem Statement

βˆ’

(

2

π‘₯

3

π‘₯

2

)

3

(

π‘₯

3

βˆ’

4

π‘₯

2

)

βˆ’(2x

3

+x

2

)+3(x

3

βˆ’4x

2

)?

Solution

Let's simplify the given expression:

βˆ’(2x3+x2)+3(x3βˆ’4x2)-(2x^3 + x^2) + 3(x^3 - 4x^2)

Step 1: Distribute the negative sign across the first set of parentheses.

βˆ’2x3βˆ’x2-2x^3 - x^2

Step 2: Distribute the 3 across the second set of parentheses.

3x3βˆ’12x23x^3 - 12x^2

Step 3: Combine all the terms from Step 1 and Step 2.

βˆ’2x3βˆ’x2+3x3βˆ’12x2-2x^3 - x^2 + 3x^3 - 12x^2

Step 4: Combine like terms.

  • Combine the x3x^3 terms: βˆ’2x3+3x3=x3-2x^3 + 3x^3 = x^3
  • Combine the x2x^2 terms: βˆ’x2βˆ’12x2=βˆ’13x2-x^2 - 12x^2 = -13x^2

Final Expression:

x3βˆ’13x2x^3 - 13x^2

So, the simplified form of the expression is x3βˆ’13x2\boxed{x^3 - 13x^2}.

Would you like more details or have any questions about this solution?

Here are five related questions to expand your understanding:

  1. What would the expression look like if x3x^3 had a negative coefficient in both terms?
  2. How does the expression change if you factor it?
  3. Can you find the roots of the equation x3βˆ’13x2=0x^3 - 13x^2 = 0?
  4. What happens if you substitute a value for xx into the simplified expression?
  5. How does the distribution property apply when expanding expressions like this?

Tip: Always remember to distribute negative signs and coefficients carefully when simplifying expressions.

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Math Problem Analysis

Mathematical Concepts

Polynomial simplification
Distribution property

Formulas

-

Theorems

-

Suitable Grade Level

Grades 9-12